# Double the first element and move zero to end

• Difficulty Level : Easy
• Last Updated : 01 Dec, 2021

For a given array of n integers and assume that ‘0’ is an invalid number and all others as a valid number. Convert the array in such a way that if both current and next element is valid and both have same value then double current value and replace the next number with 0. After the modification, rearrange the array such that all 0’s shifted to the end.
Examples:

```Input : arr[] = {2, 2, 0, 4, 0, 8}
Output : 4 4 8 0 0 0

Input : arr[] = {0, 2, 2, 2, 0, 6, 6, 0, 0, 8}
Output :  4 2 12 8 0 0 0 0 0 0```

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Approach: First modify the array as mentioned, i.e., if the next valid number is the same as the current number, double its value and replace the next number with 0.
Algorithm for Modification:

```1. if n == 1
2.     return
3. for i = 0 to n-2
4.     if (arr[i] != 0) && (arr[i] == arr[i+1])
5.         arr[i] = 2 * arr[i]
6.       arr[i+1] = 0
7.       i++```

After modifying the array, Move all zeroes to the end of the array.

## C++

 `// C++ implementation to rearrange the array ` `// elements after modification` `#include `   `using` `namespace` `std;`   `// function which pushes all zeros to end of ` `// an array.` `void` `pushZerosToEnd(``int` `arr[], ``int` `n)` `{` `    ``// Count of non-zero elements` `    ``int` `count = 0;`   `    ``// Traverse the array. If element encountered` `    ``// is non-zero, then replace the element at ` `    ``// index 'count' with this element` `    ``for` `(``int` `i = 0; i < n; i++)` `        ``if` `(arr[i] != 0)`   `            ``// here count is incremented` `            ``arr[count++] = arr[i];`   `    ``// Now all non-zero elements have been shifted` `    ``// to front and 'count' is set as index of` `    ``// first 0. Make all elements 0 from count` `    ``// to end.` `    ``while` `(count < n)` `        ``arr[count++] = 0;` `}`   `// function to rearrange the array elements` `// after modification` `void` `modifyAndRearrangeArr(``int` `arr[], ``int` `n)` `{` `    ``// if 'arr[]' contains a single element` `    ``// only` `    ``if` `(n == 1)` `        ``return``;`   `    ``// traverse the array` `    ``for` `(``int` `i = 0; i < n - 1; i++) {`   `        ``// if true, perform the required modification` `        ``if` `((arr[i] != 0) && (arr[i] == arr[i + 1])) {`   `            ``// double current index value` `            ``arr[i] = 2 * arr[i];`   `            ``// put 0 in the next index` `            ``arr[i + 1] = 0;`   `            ``// increment by 1 so as to move two ` `            ``// indexes ahead during loop iteration` `            ``i++;` `        ``}` `    ``}`   `    ``// push all the zeros at the end of 'arr[]'` `    ``pushZerosToEnd(arr, n);` `}`   `// function to print the array elements` `void` `printArray(``int` `arr[], ``int` `n)` `{` `    ``for` `(``int` `i = 0; i < n; i++)` `        ``cout << arr[i] << ``" "``;` `}`   `// Driver program to test above` `int` `main()` `{` `    ``int` `arr[] = { 0, 2, 2, 2, 0, 6, 6, 0, 0, 8 };` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);`   `    ``cout << ``"Original array: "``;` `    ``printArray(arr, n);`   `    ``modifyAndRearrangeArr(arr, n);`   `    ``cout << ``"\nModified array: "``;` `    ``printArray(arr, n);`   `    ``return` `0;` `}`

## Java

 `// Java implementation to rearrange the ` `// array elements after modification` `class` `GFG {`   `    ``// function which pushes all ` `    ``// zeros to end of an array.` `    ``static` `void` `pushZerosToEnd(``int` `arr[], ``int` `n)` `    ``{` `        ``// Count of non-zero elements` `        ``int` `count = ``0``;`   `        ``// Traverse the array. If element ` `        ``// encountered is non-zero, then` `        ``// replace the element at index` `        ``// 'count' with this element` `        ``for` `(``int` `i = ``0``; i < n; i++)` `            ``if` `(arr[i] != ``0``)`   `                ``// here count is incremented` `                ``arr[count++] = arr[i];`   `        ``// Now all non-zero elements ` `        ``// have been shifted to front and ` `        ``// 'count' is set as index of first 0. ` `        ``// Make all elements 0 from count to end.` `        ``while` `(count < n)` `            ``arr[count++] = ``0``;` `    ``}`   `    ``// function to rearrange the array` `    ``//  elements after modification` `    ``static` `void` `modifyAndRearrangeArr(``int` `arr[], ``int` `n)` `    ``{` `        ``// if 'arr[]' contains a single element` `        ``// only` `        ``if` `(n == ``1``)` `            ``return``;`   `        ``// traverse the array` `        ``for` `(``int` `i = ``0``; i < n - ``1``; i++) {`   `            ``// if true, perform the required modification` `            ``if` `((arr[i] != ``0``) && (arr[i] == arr[i + ``1``]))` `            ``{`   `                ``// double current index value` `                ``arr[i] = ``2` `* arr[i];`   `                ``// put 0 in the next index` `                ``arr[i + ``1``] = ``0``;`   `                ``// increment by 1 so as to move two` `                ``// indexes ahead during loop iteration` `                ``i++;` `            ``}` `        ``}`   `        ``// push all the zeros at ` `        ``// the end of 'arr[]'` `        ``pushZerosToEnd(arr, n);` `    ``}`   `    ``// function to print the array elements` `    ``static` `void` `printArray(``int` `arr[], ``int` `n)` `    ``{` `        ``for` `(``int` `i = ``0``; i < n; i++)` `            ``System.out.print(arr[i] + ``" "``);` `        ``System.out.println();` `    ``}`   `    ``// Driver program to test above` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``int` `arr[] = { ``0``, ``2``, ``2``, ``2``, ``0``, ``6``, ``6``, ``0``, ``0``, ``8` `};` `        ``int` `n = arr.length;`   `        ``System.out.print(``"Original array: "``);` `        ``printArray(arr, n);`   `        ``modifyAndRearrangeArr(arr, n);`   `        ``System.out.print(``"Modified array: "``);` `        ``printArray(arr, n);` `    ``}` `}`   `// This code is contributed  ` `// by prerna saini`

## Python3

 `# Python3 implementation to rearrange ` `# the array elements after modification`   `# function which pushes all zeros ` `# to end of an array.` `def` `pushZerosToEnd(arr, n):`   `    ``# Count of non-zero elements` `    ``count ``=` `0`   `    ``# Traverse the array. If element ` `    ``# encountered is non-zero, then ` `    ``# replace the element at index ` `    ``# 'count' with this element` `    ``for` `i ``in` `range``(``0``, n):` `        ``if` `arr[i] !``=` `0``:`   `            ``# here count is incremented` `            ``arr[count] ``=` `arr[i]` `            ``count``+``=``1`   `    ``# Now all non-zero elements have been ` `    ``# shifted to front and 'count' is set` `    ``# as index of first 0. Make all ` `    ``# elements 0 from count to end.` `    ``while` `(count < n):` `        ``arr[count] ``=` `0` `        ``count``+``=``1`     `# function to rearrange the array` `# elements after modification` `def` `modifyAndRearrangeArr(ar, n):`   `    ``# if 'arr[]' contains a single` `    ``# element only` `    ``if` `n ``=``=` `1``:` `        ``return`   `    ``# traverse the array` `    ``for` `i ``in` `range``(``0``, n ``-` `1``):`   `        ``# if true, perform the required modification` `        ``if` `(arr[i] !``=` `0``) ``and` `(arr[i] ``=``=` `arr[i ``+` `1``]):`   `            ``# double current index value` `            ``arr[i] ``=` `2` `*` `arr[i]`   `            ``# put 0 in the next index` `            ``arr[i ``+` `1``] ``=` `0`   `            ``# increment by 1 so as to move two ` `            ``# indexes ahead during loop iteration` `            ``i``+``=``1`   `    `    `    ``# push all the zeros at the end of 'arr[]'` `    ``pushZerosToEnd(arr, n)`     `# function to print the array elements` `def` `printArray(arr, n):`   `    ``for` `i ``in` `range``(``0``, n):` `        ``print``(arr[i],end``=``" "``)`     `# Driver program to test above` `arr ``=` `[ ``0``, ``2``, ``2``, ``2``, ``0``, ``6``, ``6``, ``0``, ``0``, ``8` `]` `n ``=` `len``(arr) `   `print``(``"Original array:"``,end``=``" "``)` `printArray(arr, n)`   `modifyAndRearrangeArr(arr, n)`   `print``(``"\nModified array:"``,end``=``" "``)` `printArray(arr, n)`   `# This code is contributed by Smitha Dinesh Semwal`

## C#

 `// C# implementation to rearrange the` `// array elements after modification` `using` `System;`   `class` `GFG {`   `    ``// function which pushes all` `    ``// zeros to end of an array.` `    ``static` `void` `pushZerosToEnd(``int``[] arr, ``int` `n)` `    ``{` `        ``// Count of non-zero elements` `        ``int` `count = 0;`   `        ``// Traverse the array. If element` `        ``// encountered is non-zero, then` `        ``// replace the element at index` `        ``// 'count' with this element` `        ``for` `(``int` `i = 0; i < n; i++)` `            ``if` `(arr[i] != 0)`   `                ``// here count is incremented` `                ``arr[count++] = arr[i];`   `        ``// Now all non-zero elements` `        ``// have been shifted to front and` `        ``// 'count' is set as index of first 0.` `        ``// Make all elements 0 from count to end.` `        ``while` `(count < n)` `            ``arr[count++] = 0;` `    ``}`   `    ``// function to rearrange the array` `    ``// elements after modification` `    ``static` `void` `modifyAndRearrangeArr(``int``[] arr, ``int` `n)` `    ``{` `        ``// if 'arr[]' contains a single element` `        ``// only` `        ``if` `(n == 1)` `            ``return``;`   `        ``// traverse the array` `        ``for` `(``int` `i = 0; i < n - 1; i++) {`   `            ``// if true, perform the required modification` `            ``if` `((arr[i] != 0) && (arr[i] == arr[i + 1])) {`   `                ``// double current index value` `                ``arr[i] = 2 * arr[i];`   `                ``// put 0 in the next index` `                ``arr[i + 1] = 0;`   `                ``// increment by 1 so as to move two` `                ``// indexes ahead during loop iteration` `                ``i++;` `            ``}` `        ``}`   `        ``// push all the zeros at` `        ``// the end of 'arr[]'` `        ``pushZerosToEnd(arr, n);` `    ``}`   `    ``// function to print the array elements` `    ``static` `void` `printArray(``int``[] arr, ``int` `n)` `    ``{` `        ``for` `(``int` `i = 0; i < n; i++)` `            ``Console.Write(arr[i] + ``" "``);` `        ``Console.WriteLine();` `    ``}`   `    ``// Driver program to test above` `    ``public` `static` `void` `Main()` `    ``{` `        ``int``[] arr = { 0, 2, 2, 2, 0, 6, 6, 0, 0, 8 };` `        ``int` `n = arr.Length;`   `        ``Console.Write(``"Original array: "``);` `        ``printArray(arr, n);`   `        ``modifyAndRearrangeArr(arr, n);`   `        ``Console.Write(``"Modified array: "``);` `        ``printArray(arr, n);` `    ``}` `}`   `// This code is contributed by Sam007`

## Javascript

 ``

Output:

```Original array: 0 2 2 2 0 6 6 0 0 8
Modified array: 4 2 12 8 0 0 0 0 0 0```

Time Complexity: O(n).

Approach with efficient zero shiftings:

Although the above solution is efficient, we can further optimise it in shifting zero algorithms by reducing the number of operations.

In the above shifting algorithms, we scan some elements twice when we set the count index to last index element to zero.

Efficient Zero Shifting Algorithms:

```int lastSeenPositiveIndex = 0;
for( index = 0; index < n; index++)
{
if(array[index] != 0)
{
swap(array[index], array[lastSeenPositiveIndex]);
lastSeenPositiveIndex++;
}
}```

## C++

 `// Utility Function For Swaping Two Element Of An Array` `void` `swap(``int``& a, ``int``& b) { a = b + a - (b = a); }`   `// shift all zero to left side of an array` `void` `shiftAllZeroToLeft(``int` `array[], ``int` `n)` `{` `    ``// Maintain last index with positive value` `    ``int` `lastSeenNonZero = 0;`   `    ``for` `(index = 0; index < n; index++) ` `    ``{` `        ``// If Element is non-zero` `        ``if` `(array[index] != 0) ` `        ``{` `            ``// swap current index, with lastSeen non-zero` `            ``swap(array[index], array[lastSeenNonZero]);`   `            ``// next element will be last seen non-zero` `            ``lastSeenNonZero++;` `        ``}` `    ``}` `}`   `// This snippet is contributed By: Faizanur Rahman`

## Java

 `class` `GFG {` `    ``// Function For Swaping Two Element Of An Array` `    ``public` `static` `void` `swap(``int``[] A, ``int` `i, ``int` `j)` `    ``{` `        ``int` `temp = A[i];` `        ``A[i] = A[j];` `        ``A[j] = temp;` `    ``}`   `    ``// shift all zero to left side of an array` `    ``static` `void` `shiftAllZeroToLeft(``int` `array[], ``int` `n)` `    ``{` `        ``// Maintain last index with positive value` `        ``int` `lastSeenNonZero = ``0``;`   `        ``for` `(``int` `index = ``0``; index < n; index++) {` `            ``// If Element is non-zero` `            ``if` `(array[index] != ``0``) {` `                ``// swap current index, with lastSeen` `                ``// non-zero` `                ``swap(array, array[index],` `                     ``array[lastSeenNonZero]);`   `                ``// next element will be last seen non-zero` `                ``lastSeenNonZero++;` `            ``}` `        ``}` `    ``}` `}`   `// This code is contributed By sam_2200`

## Python3

 `# Maintain last index with positive value` `def` `shiftAllZeroToLeft(arr, n):` `    ``lastSeenNonZero ``=` `0` `    ``for` `index ``in` `range``(``0``, n):` `      `  `        ``# If Element is non-zero` `        ``if` `(array[index] !``=` `0``):` `          `  `            ``# swap current index, with lastSeen` `            ``# non-zero` `            ``array[index], array[lastSeenNonZero] ``=` `array[lastSeenNonZero], array[index]` `            `  `            ``# next element will be last seen non-zero` `            ``lastSeenNonZero``+``+`   `            ``# This code is contributed By sam_2200`

## C#

 `using` `System;` `class` `GFG` `{` `  `  `    ``// Function For Swaping Two Element Of An Array` `    ``public` `static` `void` `swap(``int``[] A, ``int` `i, ``int` `j)` `    ``{` `        ``int` `temp = A[i];` `        ``A[i] = A[j];` `        ``A[j] = temp;` `    ``}`   `    ``// shift all zero to left side of an array` `    ``static` `void` `shiftAllZeroToLeft(``int``[] array, ``int` `n)` `    ``{` `      `  `        ``// Maintain last index with positive value` `        ``int` `lastSeenNonZero = 0;`   `        ``for` `(``int` `index = 0; index < n; index++)` `        ``{` `          `  `            ``// If Element is non-zero` `            ``if` `(array[index] != 0)` `            ``{` `              `  `                ``// swap current index, with lastSeen` `                ``// non-zero` `                ``swap(array, array[index],` `                     ``array[lastSeenNonZero]);`   `                ``// next element will be last seen non-zero` `                ``lastSeenNonZero++;` `            ``}` `        ``}` `    ``}` `}`   `// This code is contributed By Saurabh Jaiswal`

## Javascript

 ``

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