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# Find the smallest missing number

• Difficulty Level : Medium
• Last Updated : 31 Mar, 2023

Given a sorted array of n distinct integers where each integer is in the range from 0 to m-1 and m > n. Find the smallest number that is missing from the array.

Examples:

```Input: {0, 1, 2, 6, 9}, n = 5, m = 10
Output: 3

Input: {4, 5, 10, 11}, n = 4, m = 12
Output: 0

Input: {0, 1, 2, 3}, n = 4, m = 5
Output: 4

Input: {0, 1, 2, 3, 4, 5, 6, 7, 10}, n = 9, m = 11
Output: 8```

Thanks to Ravichandra for suggesting following two methods.

Method 1 (Use Binary Search)
For i = 0 to m-1, do binary search for i in the array. If i is not present in the array then return i.
Time Complexity: O(m log n)

Method 2 (Linear Search
If arr[0] is not 0, return 0. Otherwise traverse the input array starting from index 0, and for each pair of elements a[i] and a[i+1], find the difference between them. if the difference is greater than 1 then a[i]+1 is the missing number.
Time Complexity: O(n)

Another approach using linear search involves no need of finding the difference between the elements a[i] and a[i+1]. Starting from arr[0] to arr[n-1] check until arr[i] != i. If the condition (arr[i] != i) is satisfied then ‘i’ is the smallest missing number. If the condition is not satisfied, then it means there is no missing number in the given arr[], then return the element arr[n-1]+1 which is same as ‘n’.

Time Complexity: O(n)

Method 3 (Use Modified Binary Search)
Thanks to yasein and Jams for suggesting this method.
In the standard Binary Search process, the element to be searched is compared with the middle element and on the basis of comparison result, we decide whether to search is over or to go to left half or right half.
In this method, we modify the standard Binary Search algorithm to compare the middle element with its index and make decision on the basis of this comparison.

• If the first element is not same as its index then return first index
• Else get the middle index say mid
• If arr[mid] greater than mid then the required element lies in left half.
• Else the required element lies in right half.

## C++

 `// C++ program to find the smallest elements` `// missing in a sorted array.` `#include` `using` `namespace` `std;`   `int` `findFirstMissing(``int` `array[], ` `                    ``int` `start, ``int` `end)` `{` `    ``if` `(start > end)` `        ``return` `end + 1;`   `    ``if` `(start != array[start])` `        ``return` `start;`   `    ``int` `mid = (start + end) / 2;`   `    ``// Left half has all elements ` `    ``// from 0 to mid` `    ``if` `(array[mid] == mid)` `        ``return` `findFirstMissing(array, ` `                            ``mid+1, end);`   `    ``return` `findFirstMissing(array, start, mid);` `}`   `// Driver code` `int` `main()` `{` `    ``int` `arr[] = {0, 1, 2, 3, 4, 5, 6, 7, 10};` `    ``int` `n = ``sizeof``(arr)/``sizeof``(arr[0]);` `    ``cout << ``"Smallest missing element is "` `<<` `        ``findFirstMissing(arr, 0, n-1) << endl;` `}`   `// This code is contributed by` `// Shivi_Aggarwal `

## C

 `// C program to find the smallest elements missing` `// in a sorted array.` `#include`   `int` `findFirstMissing(``int` `array[], ``int` `start, ``int` `end)` `{` `    ``if` `(start  > end)` `        ``return` `end + 1;`   `    ``if` `(start != array[start])` `        ``return` `start;`   `    ``int` `mid = (start + end) / 2;`   `    ``// Left half has all elements from 0 to mid` `    ``if` `(array[mid] == mid)` `        ``return` `findFirstMissing(array, mid+1, end);`   `    ``return` `findFirstMissing(array, start, mid);` `}`   `// driver program to test above function` `int` `main()` `{` `    ``int` `arr[] = {0, 1, 2, 3, 4, 5, 6, 7, 10};` `    ``int` `n = ``sizeof``(arr)/``sizeof``(arr[0]);` `    ``printf``(``"Smallest missing element is %d"``,` `           ``findFirstMissing(arr, 0, n-1));` `    ``return` `0;` `}`

## Java

 `import` `java.io.*;`   `class` `SmallestMissing ` `{` `    ``int` `findFirstMissing(``int` `array[], ``int` `start, ``int` `end) ` `    ``{` `        ``if` `(start > end)` `            ``return` `end + ``1``;`   `        ``if` `(start != array[start])` `            ``return` `start;`   `        ``int` `mid = (start + end) / ``2``;`   `        ``// Left half has all elements from 0 to mid` `        ``if` `(array[mid] == mid)` `            ``return` `findFirstMissing(array, mid+``1``, end);`   `        ``return` `findFirstMissing(array, start, mid);` `    ``}`   `    ``// Driver program to test the above function` `    ``public` `static` `void` `main(String[] args) ` `    ``{` `        ``SmallestMissing small = ``new` `SmallestMissing();` `        ``int` `arr[] = {``0``, ``1``, ``2``, ``3``, ``4``, ``5``, ``6``, ``7``, ``10``};` `        ``int` `n = arr.length;` `        ``System.out.println(``"First Missing element is : "` `                ``+ small.findFirstMissing(arr, ``0``, n - ``1``));` `    ``}` `}`

## Python3

 `# Python3 program to find the smallest` `# elements missing in a sorted array.`   `def` `findFirstMissing(array, start, end):`   `    ``if` `(start > end):` `        ``return` `end ``+` `1`   `    ``if` `(start !``=` `array[start]):` `        ``return` `start;`   `    ``mid ``=` `int``((start ``+` `end) ``/` `2``)`   `    ``# Left half has all elements` `    ``# from 0 to mid` `    ``if` `(array[mid] ``=``=` `mid):` `        ``return` `findFirstMissing(array,` `                          ``mid``+``1``, end)`   `    ``return` `findFirstMissing(array, ` `                          ``start, mid)`     `# driver program to test above function` `arr ``=` `[``0``, ``1``, ``2``, ``3``, ``4``, ``5``, ``6``, ``7``, ``10``]` `n ``=` `len``(arr)` `print``(``"Smallest missing element is"``,` `      ``findFirstMissing(arr, ``0``, n``-``1``))`   `# This code is contributed by Smitha Dinesh Semwal `

## C#

 `// C# program to find the smallest` `// elements missing in a sorted array.` `using` `System;`   `class` `GFG` `{` `    ``static` `int` `findFirstMissing(``int` `[]array,` `                            ``int` `start, ``int` `end) ` `    ``{` `        ``if` `(start > end)` `            ``return` `end + 1;`   `        ``if` `(start != array[start])` `            ``return` `start;`   `        ``int` `mid = (start + end) / 2;`   `        ``// Left half has all elements from 0 to mid` `        ``if` `(array[mid] == mid)` `            ``return` `findFirstMissing(array, mid+1, end);`   `        ``return` `findFirstMissing(array, start, mid);` `    ``}`   `    ``// Driver program to test the above function` `    ``public` `static` `void` `Main() ` `    ``{` `        ``int` `[]arr = {0, 1, 2, 3, 4, 5, 6, 7, 10};` `        ``int` `n = arr.Length;` `        `  `        ``Console.Write(``"smallest Missing element is : "` `                    ``+ findFirstMissing(arr, 0, n - 1));` `    ``}` `}`   `// This code is contributed by Sam007 `

## PHP

 ` ``\$end``)` `        ``return` `\$end` `+ 1;`   `    ``if` `(``\$start` `!= ``\$array``[``\$start``])` `        ``return` `\$start``;`   `    ``\$mid` `= (``\$start` `+ ``\$end``) / 2;`   `    ``// Left half has all ` `    ``// elements from 0 to mid` `    ``if` `(``\$array``[``\$mid``] == ``\$mid``)` `        ``return` `findFirstMissing(``\$array``, ` `                                ``\$mid` `+ 1, ` `                                ``\$end``);`   `    ``return` `findFirstMissing(``\$array``, ` `                            ``\$start``, ` `                            ``\$mid``);` `}`   `    ``// Driver Code` `    ``\$arr` `= ``array` `(0, 1, 2, 3, 4, 5, 6, 7, 10);` `    ``\$n` `= ``count``(``\$arr``);` `    ``echo` `"Smallest missing element is "` `,` `          ``findFirstMissing(``\$arr``, 2, ``\$n` `- 1);` `        `  `// This code Contributed by Ajit.` `?>`

## Javascript

 ``

Output

`Smallest missing element is 8`

Note: This method doesn’t work if there are duplicate elements in the array.
Time Complexity: O(Log n)
Auxiliary Space : O(Log n)

Another Method: The idea is to use Recursive Binary Search to find the smallest missing number. Below is the illustration with the help of steps:

• If the first element of the array is not 0, then the smallest missing number is 0.
• If the last elements of the array is N-1, then the smallest missing number is N.
• Otherwise, find the middle element from the first and last index and check if the middle element is equal to the desired element. i.e. first + middle_index.
• If the middle element is the desired element, then the smallest missing element is in the right search space of the middle.
• Otherwise, the smallest missing number is in the left search space of the middle.

Below is the implementation of the above approach:

## C++

 `//C++ program for the above approach` `#include `   `using` `namespace` `std;`   `// Program to find missing element` `int` `findFirstMissing(vector<``int``> arr , ``int` `start ,` `                        ``int`  `end,``int` `first)` `{`   `  ``if` `(start < end)` `  ``{` `    ``int` `mid = (start + end) / 2;`   `    ``/** Index matches with value` `      ``at that index, means missing` `      ``element cannot be upto that po*/` `    ``if` `(arr[mid] != mid+first)` `      ``return` `findFirstMissing(arr, start,` `                                 ``mid , first);` `    ``else` `      ``return` `findFirstMissing(arr, mid + 1,` `                                 ``end , first);` `  ``}` `  ``return` `start + first;`   `}`   `// Program to find Smallest` `// Missing in Sorted Array` `int` `findSmallestMissinginSortedArray(vector<``int``> arr)` `{` `  `  `  ``// Check if 0 is missing` `  ``// in the array` `  ``if``(arr[0] != 0)` `    ``return` `0;`   `  ``// Check is all numbers 0 to n - 1` `  ``// are present in array` `  ``if``(arr[arr.size() - 1] == arr.size() - 1)` `    ``return` `arr.size();`   `  ``int` `first = arr[0];`   `  ``return` `findFirstMissing(arr, 0, arr.size() - 1, first);` `}`     `// Driver program to test the above function` `int` `main()` `{` `    ``vector<``int``> arr = {0, 1, 2, 3, 4, 5, 7};` `    ``int` `n = arr.size();`   `    ``// Function Call` `    ``cout<<``"First Missing element is : "``<

## Java

 `// Java Program for above approach` `import` `java.io.*;`   `class` `GFG ` `{` `  `  `    ``// Program to find Smallest ` `    ``// Missing in Sorted Array` `    ``int` `findSmallestMissinginSortedArray(` `                              ``int``[] arr) ` `    ``{ ` `      ``// Check if 0 is missing ` `      ``// in the array` `      ``if``(arr[``0``] != ``0``)` `        ``return` `0``;` `      `  `      ``// Check is all numbers 0 to n - 1 ` `      ``// are present in array` `      ``if``(arr[arr.length-``1``] == arr.length - ``1``)` `        ``return` `arr.length;` `      `  `      ``int` `first = arr[``0``];`   `      ``return` `findFirstMissing(arr,``0``,` `                       ``arr.length-``1``,first);` `    ``}` `    `  `    ``// Program to find missing element ` `    ``int` `findFirstMissing(``int``[] arr , ``int` `start , ` `                              ``int` `end, ``int` `first) ` `    ``{` `      `  `      ``if` `(start < end) ` `      ``{` `        ``int` `mid = (start+end)/``2``;`   `        ``/** Index matches with value ` `          ``at that index, means missing` `          ``element cannot be upto that point */` `        ``if` `(arr[mid] != mid+first)` `          ``return` `findFirstMissing(arr, start, ` `                                     ``mid , first);` `        ``else` `          ``return` `findFirstMissing(arr, mid+``1``, ` `                                     ``end , first);` `      ``}` `      ``return` `start+first;`   `    ``}` `  `  `    ``// Driver program to test the above function` `    ``public` `static` `void` `main(String[] args) ` `    ``{` `        ``GFG small = ``new` `GFG();` `        ``int` `arr[] = {``0``, ``1``, ``2``, ``3``, ``4``, ``5``, ``7``};` `        ``int` `n = arr.length;` `        `  `        ``// Function Call` `        ``System.out.println(``"First Missing element is : "` `            ``+ small.findSmallestMissinginSortedArray(arr));` `    ``}` `}`

## Python3

 `# Python3 program for above approach `   `# Function to find Smallest  ` `# Missing in Sorted Array` `def` `findSmallestMissinginSortedArray(arr):` `    `  `    ``# Check if 0 is missing  ` `    ``# in the array ` `    ``if` `(arr[``0``] !``=` `0``):` `        ``return` `0` `    `  `    ``# Check is all numbers 0 to n - 1  ` `    ``# are present in array ` `    ``if` `(arr[``-``1``] ``=``=` `len``(arr) ``-` `1``):` `        ``return` `len``(arr)` `    `  `    ``first ``=` `arr[``0``]` `    `  `    ``return` `findFirstMissing(arr, ``0``, ` `            ``len``(arr) ``-` `1``, first)`   `# Function to find missing element  ` `def` `findFirstMissing(arr, start, end, first):` `    `  `    ``if` `(start < end):` `        ``mid ``=` `int``((start ``+` `end) ``/` `2``)` `        `  `        ``# Index matches with value  ` `        ``# at that index, means missing ` `        ``# element cannot be upto that point ` `        ``if` `(arr[mid] !``=` `mid ``+` `first):` `            ``return` `findFirstMissing(arr, start,` `                                    ``mid, first)` `        ``else``:` `            ``return` `findFirstMissing(arr, mid ``+` `1``,  ` `                                    ``end, first)` `    `  `    ``return` `start ``+` `first`   `# Driver code` `arr ``=` `[ ``0``, ``1``, ``2``, ``3``, ``4``, ``5``, ``7` `]` `n ``=` `len``(arr)`   `# Function Call ` `print``(``"First Missing element is :"``,` `      ``findSmallestMissinginSortedArray(arr))`   `# This code is contributed by rag2127`

## C#

 `// C# program for above approach ` `using` `System;`   `class` `GFG{` `    `  `// Program to find Smallest  ` `// Missing in Sorted Array ` `int` `findSmallestMissinginSortedArray(``int``[] arr)  ` `{  ` `    `  `    ``// Check if 0 is missing  ` `    ``// in the array ` `    ``if` `(arr[0] != 0) ` `        ``return` `0; ` `    `  `    ``// Check is all numbers 0 to n - 1  ` `    ``// are present in array ` `    ``if` `(arr[arr.Length - 1] == arr.Length - 1) ` `        ``return` `arr.Length; ` `    `  `    ``int` `first = arr[0]; ` `    `  `    ``return` `findFirstMissing(arr, 0, ` `           ``arr.Length - 1,first); ` `} ` `    `  `// Program to find missing element  ` `int` `findFirstMissing(``int``[] arr , ``int` `start ,  ` `                     ``int` `end, ``int` `first)  ` `{ ` `    `  `    ``if` `(start < end)  ` `    ``{ ` `        ``int` `mid = (start + end) / 2; ` `        `  `        ``/*Index matches with value  ` `        ``at that index, means missing ` `        ``element cannot be upto that point */` `        ``if` `(arr[mid] != mid+first) ` `            ``return` `findFirstMissing(arr, start,  ` `                                    ``mid, first); ` `        ``else` `            ``return` `findFirstMissing(arr, mid + 1,  ` `                                    ``end, first); ` `    ``} ` `    ``return` `start + first; ` `}`   `// Driver code` `static` `public` `void` `Main ()` `{` `    ``GFG small = ``new` `GFG(); ` `    ``int``[] arr = {0, 1, 2, 3, 4, 5, 7}; ` `    ``int` `n = arr.Length; ` `    `  `    ``// Function Call ` `    ``Console.WriteLine(``"First Missing element is : "` `+ ` `    ``small.findSmallestMissinginSortedArray(arr)); ` `}` `}`   `// This code is contributed by avanitrachhadiya2155`

## Javascript

 ``

Output

`First Missing element is : 6`

Time Complexity: O(Log n)
Auxiliary Space : O(Log n)

Method-4(Using Hash Vector)

Make a vector of size m and initialize that with 0, so that every element of the array can be treated as an index of the vector.

Now traverse the array and mark the value as 1 in vector at a position equal to the element of the array. Now traverse the vector and find the first index where we get a value of 0, then that index is the smallest missing number.

Code-

## C++

 `// C++ program to find the smallest element` `// missing in a sorted array.` `#include` `using` `namespace` `std;`   `int` `findFirstMissing(``int` `arr[],``int` `n ,``int` `m)` `{` ` `  `  ``vector<``int``> vec(m,0);` `  `  `  ``for``(``int` `i=0;i

## Java

 `// Java program to find the smallest element` `// missing in a sorted array.`   `import` `java.util.*;`   `class` `Main {` `    `  `    ``static` `int` `findFirstMissing(``int` `arr[], ``int` `n, ``int` `m) {` `        ``int` `vec[] = ``new` `int``[m];` `        `  `        ``for` `(``int` `i = ``0``; i < n; i++) {` `            ``vec[arr[i]] = ``1``;` `        ``}` `        `  `        ``for` `(``int` `i = ``0``; i < m; i++) {` `            ``if` `(vec[i] == ``0``) {` `                ``return` `i;` `            ``}` `        ``}` `        ``return` `m;` `    ``}` `// Driver code  ` `    `  `    ``public` `static` `void` `main(String[] args) {` `        ``int` `arr[] = {``0``, ``1``, ``2``, ``3``, ``4``, ``5``, ``6``, ``7``, ``10``};` `        ``int` `n = arr.length;` `        ``int` `m = ``11``;` `        ``System.out.println(``"Smallest missing element is "` `+ findFirstMissing(arr, n, m));` `    ``}` `}`

## Python3

 `# Python program to find the smallest element missing in a sorted array.`   `def` `findFirstMissing(arr, n, m):` `    ``vec ``=` `[``0``] ``*` `m`   `    ``for` `i ``in` `range``(n):` `        ``vec[arr[i]] ``=` `1` `        `  `    ``for` `i ``in` `range``(m):` `        ``if` `vec[i] ``=``=` `0``:` `            ``return` `i` `            `  `    ``return` `m`   `# Driver code` `arr ``=` `[``0``, ``1``, ``2``, ``3``, ``4``, ``5``, ``6``, ``7``, ``10``]` `n ``=` `len``(arr)` `m ``=` `11` `print``(``"Smallest missing element is"``, findFirstMissing(arr, n, m))`

## C#

 `using` `System;`   `public` `class` `Program` `{` `    ``// function to find the smallest missing element in a sorted array` `    ``public` `static` `int` `FindFirstMissing(``int``[] arr, ``int` `n, ``int` `m)` `    ``{` `        ``// create an auxiliary array of size m with all elements set to 0` `        ``int``[] vec = ``new` `int``[m];`   `        ``// iterate over the input array and set the corresponding element in vec to 1 if it is present in arr` `        ``for` `(``int` `i = 0; i < n; i++)` `        ``{` `            ``vec[arr[i]] = 1;` `        ``}`   `        ``// iterate over the range 0 to m and return the first index where the corresponding element in vec is 0` `        ``for` `(``int` `i = 0; i < m; i++)` `        ``{` `            ``if` `(vec[i] == 0)` `            ``{` `                ``return` `i;` `            ``}` `        ``}`   `        ``// if all elements in the range are present, return m (the upper bound of the range)` `        ``return` `m;` `    ``}`   `    ``public` `static` `void` `Main()` `    ``{` `        ``// initialize an example array, compute its length, and set the upper bound of the range of elements` `        ``int``[] arr = { 0, 1, 2, 3, 4, 5, 6, 7, 10 };` `        ``int` `n = arr.Length;` `        ``int` `m = 11;`   `        ``// call the FindFirstMissing function and output the result to the console` `        ``Console.WriteLine(``"Smallest missing element is "` `+ FindFirstMissing(arr, n, m));` `    ``}` `}`

Output-

`Smallest missing element is 8`

Time Complexity: O(m+n), where n is the size of the array and m is the range of elements in the array
Auxiliary Space : O(m), where n is the size of the array

Please write comments if you find the above codes/algorithms incorrect, or find other ways to solve the same problem.

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