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# Merge two sorted arrays with O(1) extra space

We are given two sorted arrays. We need to merge these two arrays such that the initial numbers (after complete sorting) are in the first array and the remaining numbers are in the second array

Examples:

Input: ar1[] = {10}, ar2[] = {2, 3}
Output: ar1[] = {2}, ar2[] = {3, 10}

Input: ar1[] = {1, 5, 9, 10, 15, 20}, ar2[] = {2, 3, 8, 13}
Output: ar1[] = {1, 2, 3, 5, 8, 9}, ar2[] = {10, 13, 15, 20}

Recommended Practice

Note: This task is simple and O(m+n) if we are allowed to use extra space. But it becomes really complicated when extra space is not allowed and doesn’t look possible in less than O(m*n) worst-case time.  Though further optimizations are possible

Efficient Approach: To solve the problem follow the below idea:

The idea is to begin from the last element of ar2[] and search for it in ar1[]. If there is a greater element in ar1[], then we move the last element of ar1[] to ar2[]. To keep ar1[] and ar2[] sorted, we need to place the last element of ar2[] at the correct place in ar1[]. We can use the Insertion Sort for this

Follow the below steps to solve the problem:

• Iterate through every element of ar2[] starting from the last element
• Do the following for every element ar2[i]
• Store last element of ar1[]: last = ar1[m-1]
• Loop from the second last element of ar1[] while element ar1[j] is greater than ar2[i].
• ar1[j+1] = ar1[j] Move element one position ahead, then j–
• If last element of ar1[] is greater than ar2[i], then ar1[j+1] = ar2[i] and ar2[i] = last
• Print the arrays

Note: In the above loop, elements in ar1[] and ar2[] are always kept sorted.

Below is the implementation of the above approach:

## C++

 `// C++ program to merge two sorted arrays with O(1) extra ` `// space. ` `#include ` `using` `namespace` `std; ` ` `  `// Merge ar1[] and ar2[] with O(1) extra space ` `void` `merge(``int` `ar1[], ``int` `ar2[], ``int` `m, ``int` `n) ` `{ ` `    ``// Iterate through all elements ` `    ``// of ar2[] starting from the last element ` `    ``for` `(``int` `i = n - 1; i >= 0; i--) { ` `        ``// Find the smallest element greater than ar2[i]. ` `        ``// Move all elements one position ahead till the ` `        ``// smallest greater element is not found */ ` `        ``int` `j, last = ar1[m - 1]; ` `        ``for` `(j = m - 2; j >= 0 && ar1[j] > ar2[i]; j--) ` `            ``ar1[j + 1] = ar1[j]; ` ` `  `        ``// If there was a greater element ` `        ``if` `(last > ar2[i]) { ` `            ``ar1[j + 1] = ar2[i]; ` `            ``ar2[i] = last; ` `        ``} ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `ar1[] = { 1, 5, 9, 10, 15, 20 }; ` `    ``int` `ar2[] = { 2, 3, 8, 13 }; ` `    ``int` `m = ``sizeof``(ar1) / ``sizeof``(ar1); ` `    ``int` `n = ``sizeof``(ar2) / ``sizeof``(ar2); ` `    ``merge(ar1, ar2, m, n); ` ` `  `    ``cout << ``"After Merging \nFirst Array: "``; ` `    ``for` `(``int` `i = 0; i < m; i++) ` `        ``cout << ar1[i] << ``" "``; ` `    ``cout << ``"\nSecond Array: "``; ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``cout << ar2[i] << ``" "``; ` `    ``return` `0; ` `}`

## C

 `// C program to merge two sorted arrays with O(1) extra ` `// space. ` `#include ` ` `  `// Merge ar1[] and ar2[] with O(1) extra space ` `void` `merge(``int` `ar1[], ``int` `ar2[], ``int` `m, ``int` `n) ` `{ ` `    ``// Iterate through all elements ` `    ``// of ar2[] starting from the last element ` `    ``for` `(``int` `i = n - 1; i >= 0; i--) { ` `        ``// Find the smallest element greater than ar2[i]. ` `        ``// Move all elements one position ahead till the ` `        ``// smallest greater element is not found */ ` `        ``int` `j, last = ar1[m - 1]; ` `        ``for` `(j = m - 2; j >= 0 && ar1[j] > ar2[i]; j--) ` `            ``ar1[j + 1] = ar1[j]; ` ` `  `        ``// If there was a greater element ` `        ``if` `(last > ar2[i]) { ` `            ``ar1[j + 1] = ar2[i]; ` `            ``ar2[i] = last; ` `        ``} ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `ar1[] = { 1, 5, 9, 10, 15, 20 }; ` `    ``int` `ar2[] = { 2, 3, 8, 13 }; ` `    ``int` `m = ``sizeof``(ar1) / ``sizeof``(ar1); ` `    ``int` `n = ``sizeof``(ar2) / ``sizeof``(ar2); ` `    ``merge(ar1, ar2, m, n); ` ` `  `    ``printf``(``"After Merging \nFirst Array: "``); ` `    ``for` `(``int` `i = 0; i < m; i++) ` `        ``printf``(``"%d "``, ar1[i]); ` `    ``printf``(``"\nSecond Array: "``); ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``printf``(``"%d "``, ar2[i]); ` `    ``return` `0; ` `} ` ` `  `// This code is contributed by Aditya Kumar (adityakumar129)`

## Java

 `// Java program  to merge two ` `// sorted arrays with O(1) extra space. ` ` `  `import` `java.util.Arrays; ` ` `  `class` `Test { ` `    ``static` `int` `arr1[] = ``new` `int``[] { ``1``, ``5``, ``9``, ``10``, ``15``, ``20` `}; ` `    ``static` `int` `arr2[] = ``new` `int``[] { ``2``, ``3``, ``8``, ``13` `}; ` ` `  `    ``static` `void` `merge(``int` `m, ``int` `n) ` `    ``{ ` `        ``// Iterate through all elements of ar2[] starting ` `        ``// from the last element ` `        ``for` `(``int` `i = n - ``1``; i >= ``0``; i--) { ` `            ``/* Find the smallest element greater than ` `               ``ar2[i]. Move all elements one position ahead ` `               ``till the smallest greater element is not ` `               ``found */` `            ``int` `j, last = arr1[m - ``1``]; ` `            ``for` `(j = m - ``2``; j >= ``0` `&& arr1[j] > arr2[i]; ` `                 ``j--) ` `                ``arr1[j + ``1``] = arr1[j]; ` ` `  `            ``// If there was a greater element ` `            ``if` `(last > arr2[i]) { ` `                ``arr1[j + ``1``] = arr2[i]; ` `                ``arr2[i] = last; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``merge(arr1.length, arr2.length); ` `        ``System.out.print(``"After Merging \nFirst Array: "``); ` `        ``System.out.println(Arrays.toString(arr1)); ` `        ``System.out.print(``"Second Array:  "``); ` `        ``System.out.println(Arrays.toString(arr2)); ` `    ``} ` `}`

## Python3

 `# Python program to merge ` `# two sorted arrays ` `# with O(1) extra space. ` ` `  `# Merge ar1[] and ar2[] ` `# with O(1) extra space ` ` `  ` `  `def` `merge(ar1, ar2, m, n): ` ` `  `    ``# Iterate through all ` `    ``# elements of ar2[] starting from ` `    ``# the last element ` `    ``for` `i ``in` `range``(n``-``1``, ``-``1``, ``-``1``): ` ` `  `        ``# Find the smallest element ` `        ``# greater than ar2[i]. Move all ` `        ``# elements one position ahead ` `        ``# till the smallest greater ` `        ``# element is not found ` `        ``last ``=` `ar1[m``-``1``] ` `        ``j ``=` `m``-``2` `        ``while``(j >``=` `0` `and` `ar1[j] > ar2[i]): ` `            ``ar1[j``+``1``] ``=` `ar1[j] ` `            ``j ``-``=` `1` ` `  `        ``# If there was a greater element ` `        ``if` `(last > ar2[i]): ` ` `  `            ``ar1[j``+``1``] ``=` `ar2[i] ` `            ``ar2[i] ``=` `last ` ` `  `# Driver code ` `ar1 ``=` `[``1``, ``5``, ``9``, ``10``, ``15``, ``20``] ` `ar2 ``=` `[``2``, ``3``, ``8``, ``13``] ` `m ``=` `len``(ar1) ` `n ``=` `len``(ar2) ` ` `  `merge(ar1, ar2, m, n) ` ` `  `print``(``"After Merging \nFirst Array:"``, end``=``"") ` `for` `i ``in` `range``(m): ` `    ``print``(ar1[i], ``" "``, end``=``"") ` ` `  `print``(``"\nSecond Array: "``, end``=``"") ` `for` `i ``in` `range``(n): ` `    ``print``(ar2[i], ``" "``, end``=``"") ` ` `  `# This code is contributed ` `# by Anant Agarwal. `

## C#

 `// C# program  to merge two ` `// sorted arrays with O(1) extra space. ` `using` `System; ` ` `  `// Java program  to merge two ` `// sorted arrays with O(1) extra space. ` ` `  `public` `class` `Test { ` `    ``static` `int``[] arr1 = ``new` `int``[] { 1, 5, 9, 10, 15, 20 }; ` `    ``static` `int``[] arr2 = ``new` `int``[] { 2, 3, 8, 13 }; ` ` `  `    ``static` `void` `merge(``int` `m, ``int` `n) ` `    ``{ ` `        ``// Iterate through all elements of ar2[] starting ` `        ``// from the last element ` `        ``for` `(``int` `i = n - 1; i >= 0; i--) { ` `            ``/* Find the smallest element greater than ` `            ``ar2[i]. Move all elements one position ahead ` `            ``till the smallest greater element is not found ` `          ``*/` `            ``int` `j, last = arr1[m - 1]; ` `            ``for` `(j = m - 2; j >= 0 && arr1[j] > arr2[i]; ` `                 ``j--) ` `                ``arr1[j + 1] = arr1[j]; ` ` `  `            ``// If there was a greater element ` `            ``if` `(last > arr2[i]) { ` `                ``arr1[j + 1] = arr2[i]; ` `                ``arr2[i] = last; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// Driver method to test the above function ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``merge(arr1.Length, arr2.Length); ` `        ``Console.Write(``"After Merging \nFirst Array: "``); ` `        ``for` `(``int` `i = 0; i < arr1.Length; i++) { ` `            ``Console.Write(arr1[i] + ``" "``); ` `        ``} ` `        ``Console.Write(``"\nSecond Array: "``); ` `        ``for` `(``int` `i = 0; i < arr2.Length; i++) { ` `            ``Console.Write(arr2[i] + ``" "``); ` `        ``} ` `    ``} ` `} ` ` `  `/*This code is contributed by 29AjayKumar*/`

## PHP

 `= 0; ``\$i``--) ` `    ``{ ` `        ``/* Find the smallest element greater than ar2[i]. Move all ` `           ``elements one position ahead till the smallest greater ` `           ``element is not found */` `        ``\$last` `= ``\$ar1``[``\$m``-1]; ` `        ``for` `(``\$j` `= ``\$m``-2; ``\$j` `>= 0 && ``\$ar1``[``\$j``] > ``\$ar2``[``\$i``]; ``\$j``--) ` `            ``\$ar1``[``\$j``+1] = ``\$ar1``[``\$j``]; ` `  `  `        ``// If there was a greater element ` `        ``if` `(``\$last` `> ``\$ar2``[``\$i``]) ` `        ``{ ` `            ``\$ar1``[``\$j``+1] = ``\$ar2``[``\$i``]; ` `            ``\$ar2``[``\$i``] = ``\$last``; ` `        ``} ` `    ``} ` `} ` `  `  `// Driver program ` ` `  `\$ar1` `= ``array``(1, 5, 9, 10, 15, 20); ` `\$ar2` `= ``array``(2, 3, 8, 13); ` `\$m` `= 6; ` `\$n` `= 4; ` `merge(``\$ar1``, ``\$ar2``, ``\$m``, ``\$n``); ` ` `  `echo` `"After Merging \nFirst Array: "``; ` `for` `(``\$i``=0; ``\$i``<``\$m``; ``\$i``++) ` `    ``echo` `\$ar1``[``\$i``] . ``" "``; ` `echo` `"\nSecond Array: "``; ` `for` `(``\$i``=0; ``\$i``<``\$n``; ``\$i``++) ` `    ``echo` `\$ar2``[``\$i``] .``" "``; ` `return` `0; ` `?>`

## Javascript

 ``

Output

```After Merging
First Array: 1 2 3 5 8 9
Second Array: 10 13 15 20 ```

Time Complexity: O(M * N)
Auxiliary Space: O(1)

Below is the illustration of the above approach: Initial Arrays:
ar1[] = {1, 5, 9, 10, 15, 20};
ar2[] = {2, 3, 8, 13};

=> After First Iteration:
ar1[] = {1, 5, 9, 10, 13, 15};
ar2[] = {2, 3, 8, 20};

20 is moved from ar1[] to ar2[]
13 from ar2[] is inserted in ar1[]

=> After Second Iteration:
ar1[] = {1, 5, 8, 9, 10, 13};
ar2[] = {2, 3, 15, 20};

15 is moved from ar1[] to ar2[]
8 from ar2[] is inserted in ar1[]

=> After Third Iteration:
ar1[] = {1, 3, 5, 8, 9, 10};
ar2[] = {2, 13, 15, 20};

13 is moved from ar1[] to ar2[]
3 from ar2[] is inserted in ar1[]

=> After Fourth Iteration:
ar1[] = {1, 2, 3, 5, 8, 9};
ar2[] = {10, 13, 15, 20};

10 is moved from ar1[] to ar2[]
2 from ar2[] is inserted in ar1[]

Efficient Approach: To solve the problem follow the below idea:

The solution can be further optimized by observing that while traversing the two sorted arrays parallelly, if we encounter the jth second array element being smaller than ith first array element, then the jth element is to be included and replace some kth element in the first array. This observation helps us with the following algorithm

Follow the below steps to solve the problem:

• Initialize i,j,k as 0,0,n-1 where n is the size of arr1
• Iterate through every element of arr1 and arr2 using two pointers i and j respectively
• if arr1[i] is less than arr2[j], then increment i
• else swap the arr2[j] and arr1[k]and increment j and decrement k
• Sort both arr1 and arr2

Below is the implementation of the above approach:

## C++

 `// CPP program for the above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to merge two arrays ` `void` `merge(``int` `arr1[], ``int` `arr2[], ``int` `n, ``int` `m) ` `{ ` `    ``int` `i = 0, j = 0, k = n - 1; ` ` `  `    ``// Until i less than equal to k ` `    ``// or j is less than m ` `    ``while` `(i <= k && j < m) { ` `        ``if` `(arr1[i] < arr2[j]) ` `            ``i++; ` `        ``else` `{ ` `            ``swap(arr2[j++], arr1[k--]); ` `        ``} ` `    ``} ` ` `  `    ``// Sort first array ` `    ``sort(arr1, arr1 + n); ` ` `  `    ``// Sort second array ` `    ``sort(arr2, arr2 + m); ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` ` `  `    ``int` `ar1[] = { 1, 5, 9, 10, 15, 20 }; ` `    ``int` `ar2[] = { 2, 3, 8, 13 }; ` `    ``int` `m = ``sizeof``(ar1) / ``sizeof``(ar1); ` `    ``int` `n = ``sizeof``(ar2) / ``sizeof``(ar2); ` `    ``merge(ar1, ar2, m, n); ` ` `  `    ``cout << ``"After Merging \nFirst Array: "``; ` `    ``for` `(``int` `i = 0; i < m; i++) ` `        ``cout << ar1[i] << ``" "``; ` `    ``cout << ``"\nSecond Array: "``; ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``cout << ar2[i] << ``" "``; ` `    ``return` `0; ` `}`

## Java

 `// Java program for the above approach ` `import` `java.util.Arrays; ` `import` `java.util.Collections; ` ` `  `class` `GFG { ` `    ``static` `int` `arr1[] = ``new` `int``[] { ``1``, ``5``, ``9``, ``10``, ``15``, ``20` `}; ` `    ``static` `int` `arr2[] = ``new` `int``[] { ``2``, ``3``, ``8``, ``13` `}; ` ` `  `    ``// Function to merge two arrays ` `    ``static` `void` `merge(``int` `m, ``int` `n) ` `    ``{ ` `        ``int` `i = ``0``, j = ``0``, k = m - ``1``; ` `        ``while` `(i <= k && j < n) { ` `            ``if` `(arr1[i] < arr2[j]) ` `                ``i++; ` `            ``else` `{ ` `                ``int` `temp = arr2[j]; ` `                ``arr2[j] = arr1[k]; ` `                ``arr1[k] = temp; ` `                ``j++; ` `                ``k--; ` `            ``} ` `        ``} ` `        ``Arrays.sort(arr1); ` `        ``Arrays.sort(arr2); ` `    ``} ` ` `  `      ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``merge(arr1.length, arr2.length); ` `        ``System.out.print(``"After Merging \nFirst Array: "``); ` `        ``System.out.println(Arrays.toString(arr1)); ` `        ``System.out.print(``"Second Array:  "``); ` `        ``System.out.println(Arrays.toString(arr2)); ` `    ``} ` `}`

## Python3

 `# Python program for the above approach ` `arr1 ``=` `[``1``, ``5``, ``9``, ``10``, ``15``, ``20``] ` `arr2 ``=` `[``2``, ``3``, ``8``, ``13``] ` ` `  `# Function to merge two arrays ` ` `  ` `  `def` `merge(n, m): ` `    ``i ``=` `0` `    ``j ``=` `0` `    ``k ``=` `n ``-` `1` `    ``while` `(i <``=` `k ``and` `j < m): ` `        ``if` `(arr1[i] < arr2[j]): ` `            ``i ``+``=` `1` `        ``else``: ` `            ``temp ``=` `arr2[j] ` `            ``arr2[j] ``=` `arr1[k] ` `            ``arr1[k] ``=` `temp ` `            ``j ``+``=` `1` `            ``k ``-``=` `1` ` `  `    ``arr1.sort() ` `    ``arr2.sort() ` ` `  ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``merge(``len``(arr1), ``len``(arr2)) ` `    ``print``(``"After Merging \nFirst Array: "``, ``','``.join(``str``(x) ``for` `x ``in` `arr1)) ` `    ``print``(``"Second Array: "``, ``','``.join(``str``(x) ``for` `x ``in` `arr2)) ` ` ``# This code is contributed by gauravrajput1 `

## C#

 `// C# program for the above approach ` `using` `System; ` ` `  `public` `class` `GFG { ` `    ``static` `int``[] arr1 = { 1, 5, 9, 10, 15, 20 }; ` ` `  `    ``static` `int``[] arr2 = { 2, 3, 8, 13 }; ` ` `  `    ``// Function to merge two arrays ` `    ``static` `void` `merge(``int` `m, ``int` `n) ` `    ``{ ` `        ``int` `i = 0, j = 0, k = n - 1; ` `        ``while` `(i <= k && j < m) { ` `            ``if` `(arr1[i] < arr2[j]) ` `                ``i++; ` `            ``else` `{ ` `                ``int` `temp = arr2[j]; ` `                ``arr2[j] = arr1[k]; ` `                ``arr1[k] = temp; ` `                ``j++; ` `                ``k--; ` `            ``} ` `        ``} ` `        ``Array.Sort(arr1); ` `        ``Array.Sort(arr2); ` `    ``} ` ` `  `    ``public` `static` `void` `Main(String[] args) ` `    ``{ ` `        ``merge(arr1.Length, arr2.Length); ` `        ``Console.Write(``"After Merging \nFirst Array: "``); ` `        ``foreach``(``int` `i ``in` `arr1) { Console.Write(i + ``" "``); } ` `        ``Console.Write(``"\nSecond Array:  "``); ` `        ``foreach``(``int` `i ``in` `arr2) { Console.Write(i + ``" "``); } ` `    ``} ` `} ` ` `  `// This code is contributed by gauravrajput1`

## Javascript

 ``

Output

```After Merging
First Array: 1 2 3 5 8 9
Second Array: 10 13 15 20 ```

Time Complexity: O((N+M) * log(N+M))
Auxiliary Space: O(1)

Efficient Approach: To solve the problem follow the below idea:

We can compare the last element of array one with first element of array two and if the last element is greater than first element the swap the elements and sort the second array as the elements of first array should be less than or equal to elements in second array. Repeating this process while this condition holds true will give us two sorted arrays

Follow the below steps to solve the problem:

• Initialize i with 0
• Iterate while loop until the last element of array 1 is greater than the first element of array 2
• if arr1[i] greater than first element of arr2
• swap arr1[i] with arr2
• sort arr2
• Incrementing i by 1
• Print the arrays

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach ` ` `  `#include ` `using` `namespace` `std; ` ` `  `void` `merge(``int` `arr1[], ``int` `arr2[], ``int` `n, ``int` `m) ` `{ ` `    ``int` `i = 0; ` `    ``// while loop till last element of array 1(sorted) is ` `    ``// greater than first element of array 2(sorted) ` `    ``while` `(arr1[n - 1] > arr2) { ` `        ``if` `(arr1[i] > arr2) { ` `            ``// swap arr1[i] with first element ` `            ``// of arr2 and sorting the updated ` `            ``// arr2(arr1 is already sorted) ` `            ``swap(arr1[i], arr2); ` `            ``sort(arr2, arr2 + m); ` `        ``} ` `        ``i++; ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` ` `  `    ``int` `ar1[] = { 1, 5, 9, 10, 15, 20 }; ` `    ``int` `ar2[] = { 2, 3, 8, 13 }; ` `    ``int` `m = ``sizeof``(ar1) / ``sizeof``(ar1); ` `    ``int` `n = ``sizeof``(ar2) / ``sizeof``(ar2); ` `    ``merge(ar1, ar2, m, n); ` ` `  `    ``cout << ``"After Merging \nFirst Array: "``; ` `    ``for` `(``int` `i = 0; i < m; i++) ` `        ``cout << ar1[i] << ``" "``; ` `    ``cout << ``"\nSecond Array: "``; ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``cout << ar2[i] << ``" "``; ` `    ``return` `0; ` `}`

## Java

 `// Java program for the above approach ` ` `  `import` `java.io.*; ` `import` `java.util.Arrays; ` `import` `java.util.Collections; ` ` `  `class` `GFG { ` ` `  `    ``static` `int` `arr1[] = ``new` `int``[] { ``1``, ``5``, ``9``, ``10``, ``15``, ``20` `}; ` `    ``static` `int` `arr2[] = ``new` `int``[] { ``2``, ``3``, ``8``, ``13` `}; ` ` `  `    ``static` `void` `merge(``int` `n, ``int` `m) ` `    ``{ ` `        ``int` `i = ``0``; ` `        ``int` `temp = ``0``; ` ` `  `        ``// While loop till last element ` `        ``// of array 1(sorted) ` `        ``// is greater than first element ` `        ``// of array 2(sorted) ` `        ``while` `(arr1[n - ``1``] > arr2[``0``]) { ` `            ``if` `(arr1[i] > arr2[``0``]) { ` ` `  `                ``// Swap arr1[i] with first element ` `                ``// of arr2 and sorting the updated ` `                ``// arr2(arr1 is already sorted) ` `                ``// swap(arr1[i],arr2); ` `                ``temp = arr1[i]; ` `                ``arr1[i] = arr2[``0``]; ` `                ``arr2[``0``] = temp; ` `                ``Arrays.sort(arr2); ` `            ``} ` `            ``i++; ` `        ``} ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``merge(arr1.length, arr2.length); ` ` `  `        ``System.out.print(``"After Merging \nFirst Array: "``); ` `        ``System.out.println(Arrays.toString(arr1)); ` ` `  `        ``System.out.print(``"Second Array:  "``); ` `        ``System.out.println(Arrays.toString(arr2)); ` `    ``} ` `} ` ` `  `// This code is contributed by Aakash Tiwari(nighteagle)`

## Python3

 `# Python3 program for the above approach ` ` `  `arr1 ``=` `[``1``, ``5``, ``9``, ``10``, ``15``, ``20``] ` `arr2 ``=` `[``2``, ``3``, ``8``, ``13``] ` ` `  ` `  `def` `merge(n, m): ` `    ``i ``=` `0` `    ``temp ``=` `0` ` `  `    ``# While loop till last element ` `    ``# of array 1(sorted) ` `    ``# is greater than first element ` `    ``# of array 2(sorted) ` `    ``while` `(arr1[n ``-` `1``] > arr2[``0``]): ` `        ``if` `(arr1[i] > arr2[``0``]): ` ` `  `            ``# Swap arr1[i] with first element ` `            ``# of arr2 and sorting the updated ` `            ``# arr2(arr1 is already sorted) ` `            ``# swap(arr1[i],arr2); ` `            ``temp ``=` `arr1[i] ` `            ``arr1[i] ``=` `arr2[``0``] ` `            ``arr2[``0``] ``=` `temp ` `            ``arr2.sort() ` ` `  `        ``i ``+``=` `1` ` `  ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``merge(``len``(arr1), ``len``(arr2)) ` ` `  `    ``print``(``"After Merging \nFirst Array: "``, arr1) ` ` `  `    ``print``(``"Second Array: "``, arr2) ` ` `  ` `  `# This code contributed by gauravrajput1 `

## C#

 `// C# program for the above approach ` ` `  `using` `System; ` ` `  `public` `class` `GFG { ` ` `  `    ``static` `int``[] arr1 = ``new` `int``[] { 1, 5, 9, 10, 15, 20 }; ` `    ``static` `int``[] arr2 = ``new` `int``[] { 2, 3, 8, 13 }; ` ` `  `    ``static` `void` `merge(``int` `n, ``int` `m) ` `    ``{ ` `        ``int` `i = 0; ` `        ``int` `temp = 0; ` ` `  `        ``// While loop till last element ` `        ``// of array 1(sorted) ` `        ``// is greater than first element ` `        ``// of array 2(sorted) ` `        ``while` `(arr1[n - 1] > arr2) { ` `            ``if` `(arr1[i] > arr2) { ` ` `  `                ``// Swap arr1[i] with first element ` `                ``// of arr2 and sorting the updated ` `                ``// arr2(arr1 is already sorted) ` `                ``// swap(arr1[i],arr2); ` `                ``temp = arr1[i]; ` `                ``arr1[i] = arr2; ` `                ``arr2 = temp; ` `                ``Array.Sort(arr2); ` `            ``} ` `            ``i++; ` `        ``} ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main(String[] args) ` `    ``{ ` `        ``merge(arr1.Length, arr2.Length); ` ` `  `        ``Console.Write(``"After Merging \nFirst Array: "``); ` `        ``foreach``(``int` `i ``in` `arr1) Console.Write(i + ``" "``); ` `        ``Console.Write(``"\nSecond Array:  "``); ` `        ``foreach``(``int` `i ``in` `arr2) Console.Write(i + ``" "``); ` `    ``} ` `} ` ` `  `// This code is contributed by gauravrajput1`

## Javascript

 ``

Output

```After Merging
First Array: 1 2 3 5 8 9
Second Array: 10 13 15 20 ```

Time Complexity: O(M * (N * logN))
Auxiliary Space: O(1)

Approach: Follow the below steps to solve the problem

Note: Let the length of the shorter array be ‘M’ and the larger array be ‘N’

• Select the shorter array and find the index at which the partition should be done.
• Partition the shorter array at its median (l1)
• Select the first N-l1 elements from the second array • Compare the border elements i.e.
• if l1 < r2 and l2 < r2 we have found the index
• else if l1 > r2 we have to search in the left subarray
• else we have to search in the right subarray

Note: This step will store all the smallest elements in the shorter array

• Swap all the elements right to the index(i) of the shorter array with the first N-i elements of the larger array
• Sort both arrays

Note: if length(arr1) > length(arr2) all the smallest elements are stored in arr2 so we have to move all the elements in arr1 since we have to print arr1 first.

• Rotate the larger array (arr1) M times counter-clockwise
• Swap the first M elements of both arrays

Below is the implementation of the above approach:

## C++

 `#include ` `using` `namespace` `std; ` ` `  `void` `swap(``int``& a, ``int``& b) ` `{ ` `    ``int` `temp = a; ` `    ``a = b; ` `    ``b = temp; ` `} ` ` `  `void` `rotate(``int` `a[], ``int` `n, ``int` `idx) ` `{ ` `    ``int` `i; ` `    ``for` `(i = 0; i < idx / 2; i++) ` `        ``swap(a[i], a[idx - 1 - i]); ` `    ``for` `(i = idx; i < (n + idx) / 2; i++) ` `        ``swap(a[i], a[n - 1 - (i - idx)]); ` `    ``for` `(i = 0; i < n / 2; i++) ` `        ``swap(a[i], a[n - 1 - i]); ` `} ` ` `  `void` `sol(``int` `a1[], ``int` `a2[], ``int` `n, ``int` `m) ` `{ ` `    ``int` `l = 0, h = n - 1, idx = 0; ` `    ``//--------------------------------------------------------- ` `    ``while` `(l <= h) { ` `        ``// select the median of the remaining subarray ` `        ``int` `c1 = (l + h) / 2; ` `        ``// select the first elements from the larger array ` `        ``// equal to the size of remaining portion to the ` `        ``// right of the smaller array ` `        ``int` `c2 = n - c1 - 1; ` `        ``int` `l1 = a1[c1]; ` `        ``int` `l2 = a2[c2 - 1]; ` `        ``int` `r1 = c1 == n - 1 ? INT_MAX : a1[c1 + 1]; ` `        ``int` `r2 = c2 == m ? INT_MAX : a2[c2]; ` `        ``// compare the border elements and check for the ` `        ``// target index ` `        ``if` `(l1 > r2) { ` `            ``h = c1 - 1; ` `            ``if` `(h == -1) ` `                ``idx = 0; ` `        ``} ` `        ``else` `if` `(l2 > r1) { ` `            ``l = c1 + 1; ` `            ``if` `(l == n - 1) ` `                ``idx = n; ` `        ``} ` `        ``else` `{ ` `            ``idx = c1 + 1; ` `            ``break``; ` `        ``} ` `    ``} ` ` `  `    ``for` `(``int` `i = idx; i < n; i++) ` `        ``swap(a1[i], a2[i - idx]); ` ` `  `    ``sort(a1, a1 + n); ` ` `  `    ``sort(a2, a2 + m); ` `} ` ` `  `void` `merge(``int` `arr1[], ``int` `arr2[], ``int` `n, ``int` `m) ` `{ ` `    ``// code here ` `    ``if` `(n > m) { ` `        ``sol(arr2, arr1, m, n); ` `        ``rotate(arr1, n, n - m); ` ` `  `        ``for` `(``int` `i = 0; i < m; i++) ` `            ``swap(arr2[i], arr1[i]); ` `    ``} ` `    ``else` `{ ` `        ``sol(arr1, arr2, n, m); ` `    ``} ` `} ` ` `  `int` `main() ` `{ ` ` `  `    ``int` `ar1[] = { 1, 5, 9, 10, 15, 20 }; ` `    ``int` `ar2[] = { 2, 3, 8, 13 }; ` `    ``int` `m = ``sizeof``(ar1) / ``sizeof``(ar1); ` `    ``int` `n = ``sizeof``(ar2) / ``sizeof``(ar2); ` `    ``merge(ar1, ar2, m, n); ` ` `  `    ``cout << ``"After Merging \nFirst Array: "``; ` `    ``for` `(``int` `i = 0; i < m; i++) ` `        ``cout << ar1[i] << ``" "``; ` `    ``cout << ``"\nSecond Array: "``; ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``cout << ar2[i] << ``" "``; ` `    ``return` `0; ` `}`

## Java

 `// Java program for the above approach ` `import` `java.util.*; ` ` `  `class` `GFG { ` ` `  `    ``static` `void` `swap(``int` `a, ``int` `b) ` `    ``{ ` `        ``int` `temp = a; ` `        ``a = b; ` `        ``b = temp; ` `    ``} ` ` `  `    ``static` `void` `rotate(``int` `a[], ``int` `n, ``int` `idx) ` `    ``{ ` `        ``int` `i; ` `        ``for` `(i = ``0``; i < idx / ``2``; i++) ` `            ``swap(a[i], a[idx - ``1` `- i]); ` `        ``for` `(i = idx; i < (n + idx) / ``2``; i++) ` `            ``swap(a[i], a[n - ``1` `- (i - idx)]); ` `        ``for` `(i = ``0``; i < n / ``2``; i++) ` `            ``swap(a[i], a[n - ``1` `- i]); ` `    ``} ` ` `  `    ``static` `void` `sol(``int` `a1[], ``int` `a2[], ``int` `n, ``int` `m) ` `    ``{ ` `        ``int` `l = ``0``, h = n - ``1``, idx = ``0``; ` `        ``//--------------------------------------------------------- ` `        ``while` `(l <= h) { ` `            ``// select the median of the remaining subarray ` `            ``int` `c1 = (``int``)(l + h) / ``2``; ` `            ``// select the first elements from the larger ` `            ``// array equal to the size of remaining portion ` `            ``// to the right of the smaller array ` `            ``int` `c2 = n - c1 - ``1``; ` `            ``int` `l1 = a1[c1]; ` `            ``int` `l2 = a2[c2 - ``1``]; ` `            ``int` `r1 = (c1 == n - ``1``) ? Integer.MAX_VALUE ` `                                   ``: a1[c1 + ``1``]; ` `            ``int` `r2 = (c2 == m) ? Integer.MAX_VALUE : a2[c2]; ` ` `  `            ``// compare the border elements and check for the ` `            ``// target index ` `            ``if` `(l1 > r2) { ` `                ``h = c1 - ``1``; ` `                ``if` `(h == -``1``) ` `                    ``idx = ``0``; ` `            ``} ` `            ``else` `if` `(l2 > r1) { ` `                ``l = c1 + ``1``; ` `                ``if` `(l == n - ``1``) ` `                    ``idx = n; ` `            ``} ` `            ``else` `{ ` `                ``idx = c1 + ``1``; ` `                ``break``; ` `            ``} ` `        ``} ` ` `  `        ``for` `(``int` `i = idx; i < n; i++) ` `            ``swap(a1[i], a2[i - idx]); ` ` `  `        ``Arrays.sort(a1); ` ` `  `        ``Arrays.sort(a2); ` `    ``} ` ` `  `    ``static` `void` `merge(``int` `arr1[], ``int` `arr2[], ``int` `n, ``int` `m) ` `    ``{ ` `        ``// code here ` `        ``if` `(n > m) { ` `            ``sol(arr2, arr1, m, n); ` `            ``rotate(arr1, n, n - m); ` ` `  `            ``for` `(``int` `i = ``0``; i < m; i++) ` `                ``swap(arr2[i], arr1[i]); ` `        ``} ` `        ``else` `{ ` `            ``sol(arr1, arr2, n, m); ` `        ``} ` `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `ar1[] =  {``1``, ``2``, ``3``, ``5``, ``8``, ``9``}; ` `        ``int` `ar2[] = {``10``, ``13``, ``15``, ``20``}; ` `        ``int` `m = ar1.length; ` `        ``int` `n = ar2.length; ` `        ``merge(ar1, ar2, m, n); ` ` `  `        ``System.out.print(``"After Merging \nFirst Array: "``); ` `        ``for` `(``int` `i = ``0``; i < m; i++) ` `            ``System.out.print(ar1[i] + ``" "``); ` `        ``System.out.print(``"\nSecond Array: "``); ` `        ``for` `(``int` `i = ``0``; i < n; i++) ` `            ``System.out.print(ar2[i] + ``" "``); ` `    ``} ` `} ` ` `  `// This code is contributed by sanjoy_62.`

## Python3

 `# Python program to merge ` `# two sorted arrays ` `# with O(1) extra space. ` ` `  `# Merge ar1[] and ar2[] ` `# with O(1) extra space ` ` `  ` `  `def` `rotate(a, n, idx): ` `    ``for` `i ``in` `range``((``int``)(idx``/``2``)): ` `        ``a[i], a[idx``-``1``-``i] ``=` `a[idx``-``1``-``i], a[i] ` `    ``for` `i ``in` `range``(idx, (``int``)((n``+``idx)``/``2``)): ` `        ``a[i], a[n``-``1``-``(i``-``idx)] ``=` `a[n``-``1``-``(i``-``idx)], a[i] ` `    ``for` `i ``in` `range``((``int``)(n``/``2``)): ` `        ``a[i], a[n``-``1``-``i] ``=` `a[n``-``1``-``i], a[i] ` ` `  ` `  `def` `sol(a1, a2, n, m): ` `    ``l ``=` `0` `    ``h ``=` `n``-``1` `    ``idx ``=` `0` `    ``while` `(l <``=` `h): ` `        ``c1 ``=` `(``int``)((l``+``h)``/``2``) ` `        ``c2 ``=` `n``-``c1``-``1` `        ``l1 ``=` `a1[c1] ` `        ``l2 ``=` `a2[c2``-``1``] ` `        ``r1 ``=` `sys.maxint ``if` `c1 ``=``=` `n``-``1` `else` `a1[c1``+``1``] ` `        ``r2 ``=` `sys.maxint ``if` `c2 ``=``=` `m ``else` `a2[c2] ` `        ``if` `l1 > r2: ` `            ``h ``=` `c1``-``1` `            ``if` `h ``=``=` `-``1``: ` `                ``idx ``=` `0` `        ``elif` `l2 > r1: ` `            ``l ``=` `c1``+``1` `            ``if` `l ``=``=` `n``-``1``: ` `                ``idx ``=` `n ` `        ``else``: ` `            ``idx ``=` `c1``+``1` `            ``break` `    ``for` `i ``in` `range``(idx, n): ` `        ``a1[i], a2[i``-``idx] ``=` `a2[i``-``idx], a1[i] ` ` `  `    ``a1.sort() ` `    ``a2.sort() ` ` `  ` `  `def` `merge(a1, a2, n, m): ` `    ``if` `n > m: ` `        ``sol(a2, a1, m, n) ` `        ``rotate(a1, n, n``-``m) ` `        ``for` `i ``in` `range``(m): ` `            ``a1[i], a2[i] ``=` `a2[i], a1[i] ` `    ``else``: ` `        ``sol(a1, a2, n, m) ` `# Driver program ` ` `  ` `  `ar1 ``=` `[``1``, ``5``, ``9``, ``10``, ``15``, ``20``] ` `ar2 ``=` `[``2``, ``3``, ``8``, ``13``] ` `m ``=` `len``(ar1) ` `n ``=` `len``(ar2) ` ` `  `merge(ar1, ar2, m, n) ` ` `  `print``(``"After Merging \nFirst Array:"``, end``=``"") ` `for` `i ``in` `range``(m): ` `    ``print``(ar1[i], ``" "``, end``=``"") ` `print``(``"\nSecond Array: "``, end``=``"") ` `for` `i ``in` `range``(n): ` `    ``print``(ar2[i], ``" "``, end``=``"") ` `# This code is contributed ` `# by Aditya Anand. `

## C#

 `// C# program for the above approach ` `using` `System; ` `using` `System.Collections.Generic; ` ` `  `public` `class` `GFG { ` ` `  `    ``static` `void` `swap(``int` `a, ``int` `b) ` `    ``{ ` `        ``int` `temp = a; ` `        ``a = b; ` `        ``b = temp; ` `    ``} ` ` `  `    ``static` `void` `rotate(``int``[] a, ``int` `n, ``int` `idx) ` `    ``{ ` `        ``int` `i; ` `        ``for` `(i = 0; i < idx / 2; i++) ` `            ``swap(a[i], a[idx - 1 - i]); ` `        ``for` `(i = idx; i < (n + idx) / 2; i++) ` `            ``swap(a[i], a[n - 1 - (i - idx)]); ` `        ``for` `(i = 0; i < n / 2; i++) ` `            ``swap(a[i], a[n - 1 - i]); ` `    ``} ` ` `  `    ``static` `void` `sol(``int``[] a1, ``int``[] a2, ``int` `n, ``int` `m) ` `    ``{ ` `        ``int` `l = 0, h = n - 1, idx = 0; ` `        ``// --------------------------------------------------------- ` `        ``while` `(l <= h) { ` `            ``// select the median of the remaining subarray ` `            ``int` `c1 = (``int``)(l + h) / 2; ` `            ``// select the first elements from the larger ` `            ``// array equal to the size of remaining portion ` `            ``// to the right of the smaller array ` `            ``int` `c2 = n - c1 - 1; ` `            ``int` `l1 = a1[c1]; ` `            ``int` `l2 = a2[c2 - 1]; ` `            ``int` `r1 ` `                ``= (c1 == n - 1) ? ``int``.MaxValue : a1[c1 + 1]; ` `            ``int` `r2 = (c2 == m) ? ``int``.MaxValue : a2[c2]; ` ` `  `            ``// compare the border elements and check for the ` `            ``// target index ` `            ``if` `(l1 > r2) { ` `                ``h = c1 - 1; ` `                ``if` `(h == -1) ` `                    ``idx = 0; ` `            ``} ` `            ``else` `if` `(l2 > r1) { ` `                ``l = c1 + 1; ` `                ``if` `(l == n - 1) ` `                    ``idx = n; ` `            ``} ` `            ``else` `{ ` `                ``idx = c1 + 1; ` `                ``break``; ` `            ``} ` `        ``} ` ` `  `        ``for` `(``int` `i = idx; i < n; i++) ` `            ``swap(a1[i], a2[i - idx]); ` ` `  `        ``Array.Sort(a1); ` ` `  `        ``Array.Sort(a2); ` `    ``} ` ` `  `    ``static` `void` `merge(``int``[] arr1, ``int``[] arr2, ``int` `n, ``int` `m) ` `    ``{ ` `        ``// code here ` `        ``if` `(n > m) { ` `            ``sol(arr2, arr1, m, n); ` `            ``rotate(arr1, n, n - m); ` ` `  `            ``for` `(``int` `i = 0; i < m; i++) ` `                ``swap(arr2[i], arr1[i]); ` `        ``} ` `        ``else` `{ ` `            ``sol(arr1, arr2, n, m); ` `        ``} ` `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `Main(String[] args) ` `    ``{ ` `        ``int``[] ar1 = {1, 2, 3, 5, 8, 9}; ` `        ``int``[] ar2 = {10, 13, 15, 20}; ` `        ``int` `m = ar1.Length; ` `        ``int` `n = ar2.Length; ` `        ``merge(ar1, ar2, m, n); ` ` `  `        ``Console.Write(``"After Merging \nFirst Array: "``); ` `        ``for` `(``int` `i = 0; i < m; i++) ` `            ``Console.Write(ar1[i] + ``" "``); ` `        ``Console.Write(``"\nSecond Array: "``); ` `        ``for` `(``int` `i = 0; i < n; i++) ` `            ``Console.Write(ar2[i] + ``" "``); ` `    ``} ` `} ` ` `  `// This code contributed by gauravrajput1`

## Javascript

 ``

Output

```After Merging
First Array: 1 2 3 5 8 9
Second Array: 10 13 15 20 ```

Time Complexity: O(max(N * logN, M * logM))
Auxiliary Space: O(1)

## Merge two sorted arrays with O(1) extra space using Insertion Sort with Simultaneous Merge:

To solve the problem follow the below idea:

We can use the insertion sort in the third approach above to reduce the time complexity as the array is already sorted, so we can place swapped element in its correct position by just performing a single traversal on the second array

Follow the below steps to solve the problem

• Sort list 1 by always comparing with the head/first of list 2 and swap if required
• After each head/first swap, perform insertion of the swapped element into the correct position in list 2 which will eventually sort list 2 at the end
• For every swapped item from list 1, perform insertion sort in list 2 to find its correct position so that when list 1 is sorted, list 2 is also sorted

Below is the implementation of the above approach:

## C++

 `#include ` `using` `namespace` `std; ` ` `  `void` `merge(``int` `arr1[], ``int` `arr2[], ``int` `n, ``int` `m) ` `{ ` `    ``// two pointer to iterate ` `    ``int` `i = 0; ` `    ``int` `j = 0; ` `    ``while` `(i < n && j < m) { ` `        ``// if arr1[i] <= arr2[j] then both array is already ` `        ``// sorted ` `        ``if` `(arr1[i] <= arr2[j]) { ` `            ``i++; ` `        ``} ` `        ``else` `if` `(arr1[i] > arr2[j]) { ` `            ``// if arr1[i]>arr2[j] then first we swap both ` `            ``// element so that arr1[i] become smaller means ` `            ``// arr1[] become sorted then we check that ` `            ``// arr2[j] is smaller than all other element in ` `            ``// right side of arr2[j] if arr2[] is not sorted ` `            ``// then we linearly do sorting ` `            ``// means while adjacent element are less than ` `            ``// new arr2[j] we do sorting like by changing ` `            ``// position of element by shifting one position ` `            ``// toward left ` `            ``swap(arr1[i], arr2[j]); ` `            ``i++; ` `            ``if` `(j < m - 1 && arr2[j + 1] < arr2[j]) { ` `                ``int` `temp = arr2[j]; ` `                ``int` `tempj = j + 1; ` `                ``while` `(arr2[tempj] < temp && tempj < m) { ` `                    ``arr2[tempj - 1] = arr2[tempj]; ` `                    ``tempj++; ` `                ``} ` `                ``arr2[tempj - 1] = temp; ` `            ``} ` `        ``} ` `    ``} ` `} ` ` `  `int` `main() ` `{ ` ` `  `    ``int` `ar1[] = { 1, 5, 9, 10, 15, 20 }; ` `    ``int` `ar2[] = { 2, 3, 8, 13 }; ` `    ``int` `m = ``sizeof``(ar1) / ``sizeof``(ar1); ` `    ``int` `n = ``sizeof``(ar2) / ``sizeof``(ar2); ` `    ``merge(ar1, ar2, m, n); ` ` `  `    ``cout << ``"After Merging \nFirst Array: "``; ` `    ``for` `(``int` `i = 0; i < m; i++) ` `        ``cout << ar1[i] << ``" "``; ` `    ``cout << ``"\nSecond Array: "``; ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``cout << ar2[i] << ``" "``; ` `    ``return` `0; ` `}`

## Java

 `import` `java.util.*; ` `class` `GFG { ` ` `  `    ``static` `void` `merge(``int` `arr1[], ``int` `arr2[], ``int` `n, ``int` `m) ` `    ``{ ` `        ``// two pointer to iterate ` `        ``int` `i = ``0``; ` `        ``int` `j = ``0``; ` `        ``while` `(i < n && j < m) { ` ` `  `            ``// if arr1[i] <= arr2[j] then both array is ` `            ``// already sorted ` `            ``if` `(arr1[i] <= arr2[j]) { ` `                ``i++; ` `            ``} ` `            ``else` `if` `(arr1[i] > arr2[j]) { ` ` `  `                ``// if arr1[i]>arr2[j] then first we swap ` `                ``// both element so that arr1[i] become ` `                ``// smaller means arr1[] become sorted then ` `                ``// we check that arr2[j] is smaller than all ` `                ``// other element in right side of arr2[j] if ` `                ``// arr2[] is not sorted then we linearly do ` `                ``// sorting means while adjacent element are ` `                ``// less than new arr2[j] we do sorting like ` `                ``// by changing position of element by ` `                ``// shifting one position toward left ` `                ``int` `t = arr1[i]; ` `                ``arr1[i] = arr2[j]; ` `                ``arr2[j] = t; ` `                ``i++; ` `                ``if` `(j < m - ``1` `&& arr2[j + ``1``] < arr2[j]) { ` `                    ``int` `temp = arr2[j]; ` `                    ``int` `tempj = j + ``1``; ` `                    ``while` `(tempj < m && arr2[tempj] < temp ` `                           ``&& tempj < m) { ` `                        ``arr2[tempj - ``1``] = arr2[tempj]; ` `                        ``tempj++; ` `                    ``} ` `                    ``arr2[tempj - ``1``] = temp; ` `                ``} ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` ` `  `        ``int` `ar1[] = { ``1``, ``5``, ``9``, ``10``, ``15``, ``20` `}; ` `        ``int` `ar2[] = { ``2``, ``3``, ``8``, ``13` `}; ` `        ``int` `m = ar1.length; ` `        ``int` `n = ar2.length; ` `        ``merge(ar1, ar2, m, n); ` ` `  `        ``System.out.print(``"After Merging \nFirst Array: "``); ` `        ``for` `(``int` `i = ``0``; i < m; i++) ` `            ``System.out.print(ar1[i] + ``" "``); ` `        ``System.out.print(``"\nSecond Array: "``); ` `        ``for` `(``int` `i = ``0``; i < n; i++) ` `            ``System.out.print(ar2[i] + ``" "``); ` `    ``} ` `} ` ` `  `// This code is contributed by gauravrajput1`

## Python3

 `# code contributed by mahee96 ` ` `  `# "Insertion sort of list 2 with swaps from list 1" ` `# ` `# swap elements to get list 1 correctly, meanwhile ` `# place the swapped item in correct position of list 2 ` `# eventually list 2 is also sorted ` `# Time = O(m*n) or O(n*m) ` `# AUX = O(1) ` ` `  ` `  `def` `merge(arr1, arr2): ` `    ``x ``=` `arr1 ` `    ``y ``=` `arr2 ` `    ``end ``=` `len``(arr1) ` `    ``i ``=` `0` `    ``while``(i < end):                 ``# O(m) or O(n) ` `        ``if``(x[i] > y[``0``]): ` `            ``swap(x, y, i, ``0``) ` `            ``insert(y, ``0``)             ``# O(n) or O(m) number of shifts ` `        ``i ``+``=` `1` ` `  `# O(n): ` ` `  ` `  `def` `insert(y, i): ` `    ``orig ``=` `y[i] ` `    ``i ``+``=` `1` `    ``while` `(i < ``len``(y) ``and` `y[i] < orig): ` `        ``y[i``-``1``] ``=` `y[i] ` `        ``i ``+``=` `1` `    ``y[i``-``1``] ``=` `orig ` ` `  ` `  `def` `swap(x, y, i, j): ` `    ``temp ``=` `x[i] ` `    ``x[i] ``=` `y[j] ` `    ``y[j] ``=` `temp ` ` `  ` `  `def` `test(): ` `    ``c1 ``=` `[``2``, ``3``, ``8``, ``13``] ` `    ``c2 ``=` `[``1``, ``5``, ``9``, ``10``, ``15``, ``20``] ` `    ``c1, c2 ``=` `c2, c1 ` `    ``merge(c1, c2) ` `    ``print``(c1, c2) ` `         `  ` `  ` `  `test() `

## C#

 `using` `System; ` ` `  `public` `class` `GFG { ` ` `  `    ``static` `void` `merge(``int``[] arr1, ``int``[] arr2, ``int` `n, ``int` `m) ` `    ``{ ` `        ``// two pointer to iterate ` `        ``int` `i = 0; ` `        ``int` `j = 0; ` `        ``while` `(i < n && j < m) { ` ` `  `            ``// if arr1[i] <= arr2[j] then both array is ` `            ``// already sorted ` `            ``if` `(arr1[i] <= arr2[j]) { ` `                ``i++; ` `            ``} ` `            ``else` `if` `(arr1[i] > arr2[j]) { ` ` `  `                ``// if arr1[i]>arr2[j] then first we swap ` `                ``// both element so that arr1[i] become ` `                ``// smaller means arr1[] become sorted then ` `                ``// we check that arr2[j] is smaller than all ` `                ``// other element in right side of arr2[j] if ` `                ``// arr2[] is not sorted then we linearly do ` `                ``// sorting means while adjacent element are ` `                ``// less than new arr2[j] we do sorting like ` `                ``// by changing position of element by ` `                ``// shifting one position toward left ` `                ``int` `t = arr1[i]; ` `                ``arr1[i] = arr2[j]; ` `                ``arr2[j] = t; ` `                ``i++; ` `                ``if` `(j < m - 1 && arr2[j + 1] < arr2[j]) { ` `                    ``int` `temp = arr2[j]; ` `                    ``int` `tempj = j + 1; ` `                    ``while` `(tempj < m && arr2[tempj] < temp ` `                           ``&& tempj < m) { ` `                        ``arr2[tempj - 1] = arr2[tempj]; ` `                        ``tempj++; ` `                    ``} ` `                    ``arr2[tempj - 1] = temp; ` `                ``} ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``public` `static` `void` `Main(String[] args) ` `    ``{ ` ` `  `        ``int``[] ar1 = { 1, 5, 9, 10, 15, 20 }; ` `        ``int``[] ar2 = { 2, 3, 8, 13 }; ` `        ``int` `m = ar1.Length; ` `        ``int` `n = ar2.Length; ` `        ``merge(ar1, ar2, m, n); ` ` `  `        ``Console.Write(``"After Merging \nFirst Array: "``); ` `        ``for` `(``int` `i = 0; i < m; i++) ` `            ``Console.Write(ar1[i] + ``" "``); ` `        ``Console.Write(``"\nSecond Array: "``); ` `        ``for` `(``int` `i = 0; i < n; i++) ` `            ``Console.Write(ar2[i] + ``" "``); ` `    ``} ` `} ` ` `  `// This code is contributed by gauravrajput1`

## Javascript

 ``

Output

```After Merging
First Array: 1 2 3 5 8 9
Second Array: 10 13 15 20 ```

Time Complexity: O(M * N)
Auxiliary Space: O(1)

## Merge two sorted arrays with O(1) extra space using Euclidean Division Lemma:

To solve the problem follow the below idea:

We can merge the two arrays as in merge sort and simultaneously use Euclidean Division Lemma i.e. (((Operation on array) % x) * x).

Follow the below steps to solve the problem:

• Merge the two arrays as we do in merge sort, while simultaneously using Euclidean Division Lemma, i.e. (((Operation on the array) % x) * x)
• After merging divide both arrays with x
• Where x needs to be a number greater than all elements of the array
• Here in this case x, (according to the constraints) can be 10e7 + 1

Below is the implementation of the above approach:

## C++

 `// C++ program to merge two sorted arrays without using ` `// extra space ` `#include ` `using` `namespace` `std; ` ` `  `void` `merge(``int` `arr1[], ``int` `arr2[], ``int` `n, ``int` `m) ` `{ ` `    ``// three pointers to iterate ` `    ``int` `i = 0, j = 0, k = 0; ` `    ``// for euclid's division lemma ` `    ``int` `x = 10e7 + 1; ` `    ``// in this loop we are rearranging the elements of arr1 ` `    ``while` `(i < n && (j < n && k < m)) { ` `        ``// if both arr1 and arr2 elements are modified ` `        ``if` `(arr1[j] >= x && arr2[k] >= x) { ` `            ``if` `(arr1[j] % x <= arr2[k] % x) { ` `                ``arr1[i] += (arr1[j++] % x) * x; ` `            ``} ` `            ``else` `{ ` `                ``arr1[i] += (arr2[k++] % x) * x; ` `            ``} ` `        ``} ` `        ``// if only arr1 elements are modified ` `        ``else` `if` `(arr1[j] >= x) { ` `            ``if` `(arr1[j] % x <= arr2[k]) { ` `                ``arr1[i] += (arr1[j++] % x) * x; ` `            ``} ` `            ``else` `{ ` `                ``arr1[i] += (arr2[k++] % x) * x; ` `            ``} ` `        ``} ` `        ``// if only arr2 elements are modified ` `        ``else` `if` `(arr2[k] >= x) { ` `            ``if` `(arr1[j] <= arr2[k] % x) { ` `                ``arr1[i] += (arr1[j++] % x) * x; ` `            ``} ` `            ``else` `{ ` `                ``arr1[i] += (arr2[k++] % x) * x; ` `            ``} ` `        ``} ` `        ``// if none elements are modified ` `        ``else` `{ ` `            ``if` `(arr1[j] <= arr2[k]) { ` `                ``arr1[i] += (arr1[j++] % x) * x; ` `            ``} ` `            ``else` `{ ` `                ``arr1[i] += (arr2[k++] % x) * x; ` `            ``} ` `        ``} ` `        ``i++; ` `    ``} ` `    ``// we can copy the elements directly as the other array ` `    ``// is exhausted ` `    ``while` `(j < n && i < n) { ` `        ``arr1[i++] += (arr1[j++] % x) * x; ` `    ``} ` `    ``while` `(k < m && i < n) { ` `        ``arr1[i++] += (arr2[k++] % x) * x; ` `    ``} ` `    ``// we need to reset i ` `    ``i = 0; ` `    ``// in this loop we are rearranging the elements of arr2 ` `    ``while` `(i < m && (j < n && k < m)) { ` `        ``// if both arr1 and arr2 elements are modified ` `        ``if` `(arr1[j] >= x && arr2[k] >= x) { ` `            ``if` `(arr1[j] % x <= arr2[k] % x) { ` `                ``arr2[i] += (arr1[j++] % x) * x; ` `            ``} ` `            ``else` `{ ` `                ``arr2[i] += (arr2[k++] % x) * x; ` `            ``} ` `        ``} ` `        ``// if only arr1 elements are modified ` `        ``else` `if` `(arr1[j] >= x) { ` `            ``if` `(arr1[j] % x <= arr2[k]) { ` `                ``arr2[i] += (arr1[j++] % x) * x; ` `            ``} ` `            ``else` `{ ` `                ``arr2[i] += (arr2[k++] % x) * x; ` `            ``} ` `        ``} ` `        ``// if only arr2 elements are modified ` `        ``else` `if` `(arr2[k] >= x) { ` `            ``if` `(arr1[j] <= arr2[k] % x) { ` `                ``arr2[i] += (arr1[j++] % x) * x; ` `            ``} ` `            ``else` `{ ` `                ``arr2[i] += (arr2[k++] % x) * x; ` `            ``} ` `        ``} ` `        ``else` `{ ` `            ``// if none elements are modified ` `            ``if` `(arr1[j] <= arr2[k]) { ` `                ``arr2[i] += (arr1[j++] % x) * x; ` `            ``} ` `            ``else` `{ ` `                ``arr2[i] += (arr2[k++] % x) * x; ` `            ``} ` `        ``} ` `        ``i++; ` `    ``} ` `    ``// we can copy the elements directly as the other array ` `    ``// is exhausted ` `    ``while` `(j < n && i < m) { ` `        ``arr2[i++] += (arr1[j++] % x) * x; ` `    ``} ` `    ``while` `(k < m && i < m) { ` `        ``arr2[i++] += (arr2[k++] % x) * x; ` `    ``} ` `    ``// we need to reset i ` `    ``i = 0; ` `    ``// we need to divide the whole arr1 by x ` `    ``while` `(i < n) { ` `        ``arr1[i++] /= x; ` `    ``} ` `    ``// we need to reset i ` `    ``i = 0; ` `    ``// we need to divide the whole arr2 by x ` `    ``while` `(i < m) { ` `        ``arr2[i++] /= x; ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` ` `  `    ``int` `ar1[] = { 1, 5, 9, 10, 15, 20 }; ` `    ``int` `ar2[] = { 2, 3, 8, 13 }; ` `    ``int` `m = ``sizeof``(ar1) / ``sizeof``(ar1); ` `    ``int` `n = ``sizeof``(ar2) / ``sizeof``(ar2); ` `    ``merge(ar1, ar2, m, n); ` ` `  `    ``cout << ``"After Merging \nFirst Array: "``; ` `    ``for` `(``int` `i = 0; i < m; i++) ` `        ``cout << ar1[i] << ``" "``; ` `    ``cout << ``"\nSecond Array: "``; ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``cout << ar2[i] << ``" "``; ` `    ``return` `0; ` `} ` `// This code is contributed by @ancientmoon8 (Mayank ` `// Kashyap)`

## Java

 `// Java program to merge two sorted arrays without using ` `// extra space ` ` `  `import` `java.util.*; ` `class` `GFG { ` ` `  `    ``static` `void` `merge(``int` `arr1[], ``int` `arr2[], ``int` `n, ``int` `m) ` `    ``{ ` `        ``// three pointers to iterate ` `        ``int` `i = ``0``, j = ``0``, k = ``0``; ` ` `  `        ``// for euclid's division lemma ` `        ``int` `x = ``10000000` `+ ``7``; ` ` `  `        ``// in this loop we are rearranging the elements of ` `        ``// arr1 ` `        ``while` `(i < n && (j < n && k < m)) { ` ` `  `            ``// if both arr1 and arr2 elements are modified ` `            ``if` `(arr1[j] >= x && arr2[k] >= x) { ` `                ``if` `(arr1[j] % x <= arr2[k] % x) { ` `                    ``arr1[i] += (arr1[j++] % x) * x; ` `                ``} ` `                ``else` `{ ` `                    ``arr1[i] += (arr2[k++] % x) * x; ` `                ``} ` `            ``} ` ` `  `            ``// if only arr1 elements are modified ` `            ``else` `if` `(arr1[j] >= x) { ` `                ``if` `(arr1[j] % x <= arr2[k]) { ` `                    ``arr1[i] += (arr1[j++] % x) * x; ` `                ``} ` `                ``else` `{ ` `                    ``arr1[i] += (arr2[k++] % x) * x; ` `                ``} ` `            ``} ` ` `  `            ``// if only arr2 elements are modified ` `            ``else` `if` `(arr2[k] >= x) { ` `                ``if` `(arr1[j] <= arr2[k] % x) { ` `                    ``arr1[i] += (arr1[j++] % x) * x; ` `                ``} ` `                ``else` `{ ` `                    ``arr1[i] += (arr2[k++] % x) * x; ` `                ``} ` `            ``} ` ` `  `            ``// if none elements are modified ` `            ``else` `{ ` `                ``if` `(arr1[j] <= arr2[k]) { ` `                    ``arr1[i] += (arr1[j++] % x) * x; ` `                ``} ` `                ``else` `{ ` `                    ``arr1[i] += (arr2[k++] % x) * x; ` `                ``} ` `            ``} ` `            ``i++; ` `        ``} ` ` `  `        ``// we can copy the elements directly as the other ` `        ``// array is exchausted ` `        ``while` `(j < n && i < n) { ` `            ``arr1[i++] += (arr1[j++] % x) * x; ` `        ``} ` `        ``while` `(k < m && i < n) { ` `            ``arr1[i++] += (arr2[k++] % x) * x; ` `        ``} ` ` `  `        ``// we need to reset i ` `        ``i = ``0``; ` ` `  `        ``// in this loop we are rearranging the elements of ` `        ``// arr2 ` `        ``while` `(i < m && (j < n && k < m)) { ` ` `  `            ``// if both arr1 and arr2 elements are modified ` `            ``if` `(arr1[j] >= x && arr2[k] >= x) { ` `                ``if` `(arr1[j] % x <= arr2[k] % x) { ` `                    ``arr2[i] += (arr1[j++] % x) * x; ` `                ``} ` `                ``else` `{ ` `                    ``arr2[i] += (arr2[k++] % x) * x; ` `                ``} ` `            ``} ` ` `  `            ``// if only arr1 elements are modified ` `            ``else` `if` `(arr1[j] >= x) { ` `                ``if` `(arr1[j] % x <= arr2[k]) { ` `                    ``arr2[i] += (arr1[j++] % x) * x; ` `                ``} ` `                ``else` `{ ` `                    ``arr2[i] += (arr2[k++] % x) * x; ` `                ``} ` `            ``} ` ` `  `            ``// if only arr2 elements are modified ` `            ``else` `if` `(arr2[k] >= x) { ` `                ``if` `(arr1[j] <= arr2[k] % x) { ` `                    ``arr2[i] += (arr1[j++] % x) * x; ` `                ``} ` `                ``else` `{ ` `                    ``arr2[i] += (arr2[k++] % x) * x; ` `                ``} ` `            ``} ` `            ``else` `{ ` ` `  `                ``// if none elements are modified ` `                ``if` `(arr1[j] <= arr2[k]) { ` `                    ``arr2[i] += (arr1[j++] % x) * x; ` `                ``} ` `                ``else` `{ ` `                    ``arr2[i] += (arr2[k++] % x) * x; ` `                ``} ` `            ``} ` `            ``i++; ` `        ``} ` ` `  `        ``// we can copy the elements directly as the other ` `        ``// array is exhausted ` `        ``while` `(j < n && i < m) { ` `            ``arr2[i++] += (arr1[j++] % x) * x; ` `        ``} ` `        ``while` `(k < m && i < m) { ` `            ``arr2[i++] += (arr2[k++] % x) * x; ` `        ``} ` ` `  `        ``// we need to reset i ` `        ``i = ``0``; ` ` `  `        ``// we need to divide the whole arr1 by x ` `        ``while` `(i < n) { ` `            ``arr1[i++] /= x; ` `        ``} ` ` `  `        ``// we need to reset i ` `        ``i = ``0``; ` ` `  `        ``// we need to divide the whole arr2 by x ` `        ``while` `(i < m) { ` `            ``arr2[i++] /= x; ` `        ``} ` `    ``} ` ` `  `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` ` `  `        ``int` `ar1[] = { ``1``, ``5``, ``9``, ``10``, ``15``, ``20` `}; ` `        ``int` `ar2[] = { ``2``, ``3``, ``8``, ``13` `}; ` `        ``int` `m = ar1.length; ` `        ``int` `n = ar2.length; ` `        ``merge(ar1, ar2, m, n); ` ` `  `        ``System.out.print(``"After Merging \nFirst Array: "``); ` `        ``for` `(``int` `i = ``0``; i < m; i++) ` `            ``System.out.print(ar1[i] + ``" "``); ` `        ``System.out.print(``"\nSecond Array: "``); ` `        ``for` `(``int` `i = ``0``; i < n; i++) ` `            ``System.out.print(ar2[i] + ``" "``); ` `    ``} ` `} ` ` `  `// This code is contributed by Aarti_Rathi`

## Python3

 `# Python3 program to merge two sorted arrays without using extra space ` ` `  ` `  `def` `merge(arr1, arr2, n, m): ` `    ``# three pointers to iterate ` `    ``i ``=` `0` `    ``j ``=` `0` `    ``k ``=` `0` `    ``# for euclid's division lemma ` `    ``x ``=` `10e7` `+` `1` `    ``# in this loop we are rearranging the elements of arr1 ` `    ``while` `i < n ``and` `(j < n ``and` `k < m): ` `        ``# if both arr1 and arr2 elements are modified ` `        ``if` `arr1[j] >``=` `x ``and` `arr2[k] >``=` `x: ` `            ``if` `arr1[j] ``%` `x <``=` `arr2[k] ``%` `x: ` `                ``arr1[i] ``+``=` `(arr1[j] ``%` `x) ``*` `x ` `                ``j ``+``=` `1` `            ``else``: ` `                ``arr1[i] ``+``=` `(arr2[k] ``%` `x) ``*` `x ` `                ``k ``+``=` `1` `        ``# if only arr1 elements are modified ` `        ``elif` `arr1[j] >``=` `x: ` `            ``if` `arr1[j] ``%` `x <``=` `arr2[k]: ` `                ``arr1[i] ``+``=` `(arr1[j] ``%` `x) ``*` `x ` `                ``j ``+``=` `1` `            ``else``: ` `                ``arr1[i] ``+``=` `(arr2[k] ``%` `x) ``*` `x ` `                ``k ``+``=` `1` `        ``# if only arr2 elements are modified ` `        ``elif` `arr2[k] >``=` `x: ` `            ``if` `arr1[j] <``=` `arr2[k] ``%` `x: ` `                ``arr1[i] ``+``=` `(arr1[j] ``%` `x) ``*` `x ` `                ``j ``+``=` `1` `            ``else``: ` `                ``arr1[i] ``+``=` `(arr2[k] ``%` `x) ``*` `x ` `                ``k ``+``=` `1` `        ``# if none elements are modified ` `        ``else``: ` `            ``if` `arr1[j] <``=` `arr2[k]: ` `                ``arr1[i] ``+``=` `(arr1[j] ``%` `x) ``*` `x ` `                ``j ``+``=` `1` `            ``else``: ` `                ``arr1[i] ``+``=` `(arr2[k] ``%` `x) ``*` `x ` `                ``k ``+``=` `1` `        ``i ``+``=` `1` ` `  `    ``#  we can copy the elements directly as the other array ` `    ``#  is exchausted ` `    ``while` `j < n ``and` `i < n: ` `        ``arr1[i] ``+``=` `(arr1[j] ``%` `x) ``*` `x ` `        ``i ``+``=` `1` `        ``j ``+``=` `1` `    ``while` `k < m ``and` `i < n: ` `        ``arr1[i] ``+``=` `(arr2[k] ``%` `x) ``*` `x ` `        ``i ``+``=` `1` `        ``k ``+``=` `1` `    ``#  we need to reset i ` `    ``i ``=` `0` ` `  `    ``# in this loop we are rearranging the elements of arr2 ` `    ``while` `i < m ``and` `(j < n ``and` `k < m): ` `        ``# if both arr1 and arr2 elements are modified ` `        ``if` `arr1[j] >``=` `x ``and` `arr2[k] >``=` `x: ` `            ``if` `arr1[j] ``%` `x <``=` `arr2[k] ``%` `x: ` `                ``arr2[i] ``+``=` `(arr1[j] ``%` `x) ``*` `x ` `                ``j ``+``=` `1` ` `  `            ``else``: ` `                ``arr2[i] ``+``=` `(arr2[k] ``%` `x) ``*` `x ` `                ``k ``+``=` `1` ` `  `        ``# if only arr1 elements are modified ` `        ``elif` `arr1[j] >``=` `x: ` `            ``if` `arr1[j] ``%` `x <``=` `arr2[k]: ` `                ``arr2[i] ``+``=` `(arr1[j] ``%` `x) ``*` `x ` `                ``j ``+``=` `1` ` `  `            ``else``: ` `                ``arr2[i] ``+``=` `(arr2[k] ``%` `x) ``*` `x ` `                ``k ``+``=` `1` ` `  `        ``# if only arr2 elements are modified ` `        ``elif` `arr2[k] >``=` `x: ` `            ``if` `arr1[j] <``=` `arr2[k] ``%` `x: ` `                ``arr2[i] ``+``=` `(arr1[j] ``%` `x) ``*` `x ` `                ``j ``+``=` `1` ` `  `            ``else``: ` `                ``arr2[i] ``+``=` `(arr2[k] ``%` `x) ``*` `x ` `                ``k ``+``=` `1` ` `  `        ``else``: ` `            ``# if none elements are modified ` `            ``if` `arr1[j] <``=` `arr2[k]: ` `                ``arr2[i] ``+``=` `(arr1[j] ``%` `x) ``*` `x ` `                ``j ``+``=` `1` ` `  `            ``else``: ` `                ``arr2[i] ``+``=` `(arr2[k] ``%` `x) ``*` `x ` `                ``k ``+``=` `1` ` `  `        ``i ``+``=` `1` `    ``# we can copy the elements directly as the other array ` `    ``# is exhausted ` `    ``while` `j < n ``and` `i < m: ` `        ``arr2[i] ``+``=` `(arr1[j] ``%` `x) ``*` `x ` `        ``i ``+``=` `1` `        ``j ``+``=` `1` ` `  `    ``while` `k < m ``and` `i < m: ` `        ``arr2[i] ``+``=` `(arr2[k] ``%` `x) ``*` `x ` `        ``i ``+``=` `1` `        ``k ``+``=` `1` ` `  `    ``# we need to reset i ` `    ``i ``=` `0` `    ``# we need to divide the whole arr1 by x ` `    ``while` `i < n: ` `        ``arr1[i] ``/``=` `x ` `        ``i ``+``=` `1` ` `  `    ``# we need to reset i ` `    ``i ``=` `0` `    ``# we need to divide the whole arr2 by x ` `    ``while` `i < m: ` `        ``arr2[i] ``/``=` `x ` `        ``i ``+``=` `1` ` `  `# Driver program ` ` `  ` `  `ar1 ``=` `[``1``, ``5``, ``9``, ``10``, ``15``, ``20``] ` `ar2 ``=` `[``2``, ``3``, ``8``, ``13``] ` `m ``=` `len``(ar1) ` `n ``=` `len``(ar2) ` ` `  `merge(ar1, ar2, m, n) ` ` `  `print``(``"After Merging \nFirst Array:"``, end``=``" "``) ` `for` `i ``in` `range``(m): ` `    ``print``(``int``(ar1[i]), end``=``" "``) ` `print``(``"\nSecond Array:"``, end``=``" "``) ` `for` `i ``in` `range``(n): ` `    ``print``(``int``(ar2[i]), end``=``" "``) ` ` `  `# This code is contributed by Tapesh(tapeshdua420) `

## C#

 `// C# program to merge two sorted arrays without using ` `// extra space ` ` `  `using` `System; ` ` `  `public` `class` `GFG { ` ` `  `    ``static` `void` `merge(``int``[] arr1, ``int``[] arr2, ``int` `n, ``int` `m) ` `    ``{ ` ` `  `        ``// three pointers to iterate ` `        ``int` `i = 0, j = 0, k = 0; ` ` `  `        ``// for euclid's division lemma ` `        ``int` `x = 10000000 + 1; ` ` `  `        ``// in this loop we are rearranging the elements of ` `        ``// arr1 ` `        ``while` `(i < n && (j < n && k < m)) { ` ` `  `            ``// if both arr1 and arr2 elements are modified ` `            ``if` `(arr1[j] >= x && arr2[k] >= x) { ` `                ``if` `(arr1[j] % x <= arr2[k] % x) { ` `                    ``arr1[i] += (arr1[j++] % x) * x; ` `                ``} ` `                ``else` `{ ` `                    ``arr1[i] += (arr2[k++] % x) * x; ` `                ``} ` `            ``} ` ` `  `            ``// if only arr1 elements are modified ` `            ``else` `if` `(arr1[j] >= x) { ` `                ``if` `(arr1[j] % x <= arr2[k]) { ` `                    ``arr1[i] += (arr1[j++] % x) * x; ` `                ``} ` `                ``else` `{ ` `                    ``arr1[i] += (arr2[k++] % x) * x; ` `                ``} ` `            ``} ` ` `  `            ``// if only arr2 elements are modified ` `            ``else` `if` `(arr2[k] >= x) { ` `                ``if` `(arr1[j] <= arr2[k] % x) { ` `                    ``arr1[i] += (arr1[j++] % x) * x; ` `                ``} ` `                ``else` `{ ` `                    ``arr1[i] += (arr2[k++] % x) * x; ` `                ``} ` `            ``} ` ` `  `            ``// if none elements are modified ` `            ``else` `{ ` `                ``if` `(arr1[j] <= arr2[k]) { ` `                    ``arr1[i] += (arr1[j++] % x) * x; ` `                ``} ` `                ``else` `{ ` `                    ``arr1[i] += (arr2[k++] % x) * x; ` `                ``} ` `            ``} ` `            ``i++; ` `        ``} ` ` `  `        ``// we can copy the elements directly as the other ` `        ``// array is exchausted ` `        ``while` `(j < n && i < n) { ` `            ``arr1[i++] += (arr1[j++] % x) * x; ` `        ``} ` `        ``while` `(k < m && i < n) { ` `            ``arr1[i++] += (arr2[k++] % x) * x; ` `        ``} ` ` `  `        ``// we need to reset i ` `        ``i = 0; ` ` `  `        ``// in this loop we are rearranging the elements of ` `        ``// arr2 ` `        ``while` `(i < m && (j < n && k < m)) { ` ` `  `            ``// if both arr1 and arr2 elements are modified ` `            ``if` `(arr1[j] >= x && arr2[k] >= x) { ` `                ``if` `(arr1[j] % x <= arr2[k] % x) { ` `                    ``arr2[i] += (arr1[j++] % x) * x; ` `                ``} ` `                ``else` `{ ` `                    ``arr2[i] += (arr2[k++] % x) * x; ` `                ``} ` `            ``} ` ` `  `            ``// if only arr1 elements are modified ` `            ``else` `if` `(arr1[j] >= x) { ` `                ``if` `(arr1[j] % x <= arr2[k]) { ` `                    ``arr2[i] += (arr1[j++] % x) * x; ` `                ``} ` `                ``else` `{ ` `                    ``arr2[i] += (arr2[k++] % x) * x; ` `                ``} ` `            ``} ` ` `  `            ``// if only arr2 elements are modified ` `            ``else` `if` `(arr2[k] >= x) { ` `                ``if` `(arr1[j] <= arr2[k] % x) { ` `                    ``arr2[i] += (arr1[j++] % x) * x; ` `                ``} ` `                ``else` `{ ` `                    ``arr2[i] += (arr2[k++] % x) * x; ` `                ``} ` `            ``} ` `            ``else` `{ ` ` `  `                ``// if none elements are modified ` `                ``if` `(arr1[j] <= arr2[k]) { ` `                    ``arr2[i] += (arr1[j++] % x) * x; ` `                ``} ` `                ``else` `{ ` `                    ``arr2[i] += (arr2[k++] % x) * x; ` `                ``} ` `            ``} ` `            ``i++; ` `        ``} ` ` `  `        ``// we can copy the elements directly as the other ` `        ``// array is exhausted ` `        ``while` `(j < n && i < m) { ` `            ``arr2[i++] += (arr1[j++] % x) * x; ` `        ``} ` `        ``while` `(k < m && i < m) { ` `            ``arr2[i++] += (arr2[k++] % x) * x; ` `        ``} ` ` `  `        ``// we need to reset i ` `        ``i = 0; ` ` `  `        ``// we need to divide the whole arr1 by x ` `        ``while` `(i < n) { ` `            ``arr1[i++] /= x; ` `        ``} ` ` `  `        ``// we need to reset i ` `        ``i = 0; ` ` `  `        ``// we need to divide the whole arr2 by x ` `        ``while` `(i < m) { ` `            ``arr2[i++] /= x; ` `        ``} ` `    ``} ` ` `  `    ``public` `static` `void` `Main(String[] args) ` `    ``{ ` ` `  `        ``int``[] ar1 = { 1, 5, 9, 10, 15, 20 }; ` `        ``int``[] ar2 = { 2, 3, 8, 13 }; ` `        ``int` `m = ar1.Length; ` `        ``int` `n = ar2.Length; ` `        ``merge(ar1, ar2, m, n); ` ` `  `        ``Console.Write(``"After Merging \nFirst Array: "``); ` `        ``for` `(``int` `i = 0; i < m; i++) ` `            ``Console.Write(ar1[i] + ``" "``); ` `        ``Console.Write(``"\nSecond Array: "``); ` `        ``for` `(``int` `i = 0; i < n; i++) ` `            ``Console.Write(ar2[i] + ``" "``); ` `    ``} ` `} ` ` `  `// This code is contributed by Aarti_Rathi`

## Javascript

 `// JavaScript program to merge two sorted arrays without using extra space ` `static merge(arr1, arr2, n, m) ` `{ ` `    ``// three pointers to iterate ` `    ``var` `i = 0; ` `    ``var` `j = 0; ` `    ``var` `k = 0; ` `     `  `    ``// for euclid's division lemma ` `    ``var` `x = 10000000 + 7; ` `     `  `    ``// in this loop we are rearranging the elements of arr1 ` `    ``while` `(i < n && (j < n && k < m)) ` `    ``{ ` `     `  `        ``// if both arr1 and arr2 elements are modified ` `        ``if` `(arr1[j] >= x && arr2[k] >= x) ` `        ``{ ` `            ``if` `(arr1[j] % x <= arr2[k] % x) ` `            ``{ ` `                ``arr1[i] += (arr1[j++] % x) * x; ` `            ``} ` `            ``else`  `            ``{ ` `                ``arr1[i] += (arr2[k++] % x) * x; ` `            ``} ` `        ``} ` `        ``else` `if` `(arr1[j] >= x) ` `        ``{ ` `            ``if` `(arr1[j] % x <= arr2[k]) ` `            ``{ ` `                ``arr1[i] += (arr1[j++] % x) * x; ` `            ``} ` `            ``else`  `            ``{ ` `                ``arr1[i] += (arr2[k++] % x) * x; ` `            ``} ` `        ``} ` `        ``else` `if` `(arr2[k] >= x) ` `        ``{ ` `            ``if` `(arr1[j] <= arr2[k] % x) ` `            ``{ ` `                ``arr1[i] += (arr1[j++] % x) * x; ` `            ``} ` `            ``else`  `            ``{ ` `                ``arr1[i] += (arr2[k++] % x) * x; ` `            ``} ` `        ``} ` `        ``else`  `        ``{ ` `            ``if` `(arr1[j] <= arr2[k]) ` `            ``{ ` `                ``arr1[i] += (arr1[j++] % x) * x; ` `            ``} ` `            ``else`  `            ``{ ` `                ``arr1[i] += (arr2[k++] % x) * x; ` `            ``} ` `        ``} ` `        ``i++; ` `    ``} ` `     `  `    ``// we can copy the elements directly as the other array ` `    ``// is exchausted ` `    ``while` `(j < n && i < n) ` `    ``{ ` `        ``arr1[i++] += (arr1[j++] % x) * x; ` `    ``} ` `    ``while` `(k < m && i < n) ` `    ``{ ` `        ``arr1[i++] += (arr2[k++] % x) * x; ` `    ``} ` `     `  `    ``// we need to reset i ` `    ``i = 0; ` `     `  `    ``// in this loop we are rearranging the elements of arr2 ` `    ``while` `(i < m && (j < n && k < m)) ` `    ``{ ` `     `  `        ``// if both arr1 and arr2 elements are modified ` `        ``if` `(arr1[j] >= x && arr2[k] >= x) ` `        ``{ ` `            ``if` `(arr1[j] % x <= arr2[k] % x) ` `            ``{ ` `                ``arr2[i] += (arr1[j++] % x) * x; ` `            ``} ` `            ``else`  `            ``{ ` `                ``arr2[i] += (arr2[k++] % x) * x; ` `            ``} ` `        ``} ` `        ``else` `if` `(arr1[j] >= x) ` `        ``{ ` `            ``if` `(arr1[j] % x <= arr2[k]) ` `            ``{ ` `                ``arr2[i] += (arr1[j++] % x) * x; ` `            ``} ` `            ``else`  `            ``{ ` `                ``arr2[i] += (arr2[k++] % x) * x; ` `            ``} ` `        ``} ` `        ``else` `if` `(arr2[k] >= x) ` `        ``{ ` `            ``if` `(arr1[j] <= arr2[k] % x) ` `            ``{ ` `                ``arr2[i] += (arr1[j++] % x) * x; ` `            ``} ` `            ``else`  `            ``{ ` `                ``arr2[i] += (arr2[k++] % x) * x; ` `            ``} ` `        ``} ` `        ``else`  `        ``{ ` `         `  `            ``// if none elements are modified ` `            ``if` `(arr1[j] <= arr2[k]) ` `            ``{ ` `                ``arr2[i] += (arr1[j++] % x) * x; ` `            ``} ` `            ``else`  `            ``{ ` `                ``arr2[i] += (arr2[k++] % x) * x; ` `            ``} ` `        ``} ` `        ``i++; ` `    ``} ` `     `  `    ``// we can copy the elements directly as the other array ` `    ``// is exhausted ` `    ``while` `(j < n && i < m) ` `    ``{ ` `        ``arr2[i++] += (arr1[j++] % x) * x; ` `    ``} ` `    ``while` `(k < m && i < m) ` `    ``{ ` `        ``arr2[i++] += (arr2[k++] % x) * x; ` `    ``} ` `     `  `    ``// we need to reset i ` `    ``i = 0; ` `     `  `    ``// we need to divide the whole arr1 by x ` `    ``while` `(i < n) ` `    ``{ ` `        ``arr1[i++] /= x; ` `    ``} ` `     `  `    ``// we need to reset i ` `    ``i = 0; ` `     `  `    ``// we need to divide the whole arr2 by x ` `    ``while` `(i < m) ` `    ``{ ` `        ``arr2[i++] /= x; ` `    ``} ` `} ` ` `  `var` `ar1 = [1, 5, 9, 10, 15, 20]; ` `var` `ar2 = [2, 3, 8, 13]; ` `var` `m = ar1.length; ` `var` `n = ar2.length; ` `GFG.merge(ar1, ar2, m, n); ` `console.log(``"After Merging \nFirst Array: "``); ` `for` `(``var` `i=0; i < m; i++) ` `{ ` `    ``console.log(parseInt(ar1[i]) + ``" "``); ` `} ` `console.log(``"\nSecond Array: "``); ` `for` `(``var` `i=0; i < n; i++) ` `{ ` `    ``console.log(parseInt(ar2[i]) + ``" "``); ` `} ` ` `  `// This code is contributed by Aarti_Rathi`

Output

```After Merging
First Array: 1 2 3 5 8 9
Second Array: 10 13 15 20 ```

Time Complexity: O(M + N)
Auxiliary Space: O(1), since no extra space has been taken

Thanks to Shubham Chauhan for suggesting 1st solution and Himanshu Kaushik for the 2nd solution. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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