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# Find the smallest positive number missing from an unsorted array | Set 1

• Difficulty Level : Hard
• Last Updated : 28 Dec, 2022

Given an unsorted array arr[] with both positive and negative elements, the task is to find the smallest positive number missing from the array.

Note: You can modify the original array.

Examples:

Input:  arr[] = {2, 3, 7, 6, 8, -1, -10, 15}
Output: 1

Input:  arr[] = { 2, 3, -7, 6, 8, 1, -10, 15 }
Output: 4

Input: arr[] = {1, 1, 0, -1, -2}
Output: 2

Naive Approach:

A naive method to solve this problem is to search all positive integers, starting from 1 in the given array.

Time Complexity: O(N2) because we may have to search at most n+1 numbers in the given array.
Auxiliary Space: O(1)

## Smallest positive number missing from an unsorted array by Marking Elements:

The idea is to mark the elements which are present in the array then traverse over the marked array and return the first element which is not marked.

Follow the steps below to solve the problem:

• Create a list full of 0’s with the size of the max value of the given array.
• Now, whenever we encounter any positive value in the original array, change the index value of the list to 1.
• After that simply iterate through the modified list, the first 0 encountered, (index value + 1) should be the answer.

Below is the implementation of the above approach.

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to return the first missing positive number from``// the given unsorted array``int` `firstMissingPos(``int` `A[], ``int` `n)``{` `    ``// To mark the occurrence of elements``    ``bool` `present[n + 1] = { ``false` `};` `    ``// Mark the occurrences``    ``for` `(``int` `i = 0; i < n; i++) {` `        ``// Only mark the required elements``        ``// All non-positive elements and the elements``        ``// greater n + 1 will never be the answer``        ``// For example, the array will be {1, 2, 3} in the``        ``// worst case and the result will be 4 which is n +``        ``// 1``        ``if` `(A[i] > 0 && A[i] <= n)``            ``present[A[i]] = ``true``;``    ``}` `    ``// Find the first element which didn't appear in the``    ``// original array``    ``for` `(``int` `i = 1; i <= n; i++)``        ``if` `(!present[i])``            ``return` `i;` `    ``// If the original array was of the type {1, 2, 3} in``    ``// its sorted form``    ``return` `n + 1;``}` `// Driver code``int` `main()``{` `    ``int` `arr[] = { 0, 10, 2, -10, -20 };``    ``int` `size = ``sizeof``(arr) / ``sizeof``(arr);``    ``cout << firstMissingPos(arr, size);``}` `// This code is contributed by Aditya Kumar (adityakumar129)`

## C

 `// C implementation of the approach``#include ``#include ` `// Function to return the first missing positive number from``// the given unsorted array``int` `firstMissingPos(``int` `A[], ``int` `n)``{` `    ``// To mark the occurrence of elements``    ``bool` `present[n + 1];``    ``for` `(``int` `i = 0; i < n; i++)``        ``present[i] = ``false``;` `    ``// Mark the occurrences``    ``for` `(``int` `i = 0; i < n; i++) {` `        ``// Only mark the required elements``        ``// All non-positive elements and the elements``        ``// greater n + 1 will never be the answer``        ``// For example, the array will be {1, 2, 3} in the``        ``// worst case and the result will be 4 which is n +``        ``// 1``        ``if` `(A[i] > 0 && A[i] <= n)``            ``present[A[i]] = ``true``;``    ``}` `    ``// Find the first element which didn't appear in the``    ``// original array``    ``for` `(``int` `i = 1; i <= n; i++)``        ``if` `(!present[i])``            ``return` `i;` `    ``// If the original array was of the type {1, 2, 3} in``    ``// its sorted form``    ``return` `n + 1;``}` `// Driver code``int` `main()``{` `    ``int` `arr[] = { 0, 10, 2, -10, -20 };``    ``int` `size = ``sizeof``(arr) / ``sizeof``(arr);``    ``printf``(``"%d"``, firstMissingPos(arr, size));``}` `// This code is contributed by Aditya Kumar (adityakumar129)`

## Java

 `// Java Program to find the smallest positive missing number``import` `java.util.*;``public` `class` `GFG {` `    ``static` `int` `solution(``int``[] A)``    ``{``        ``int` `n = A.length;``        ``// Let this 1e6 be the maximum element provided in``        ``// the array;``        ``int` `N = ``1000010``;` `        ``// To mark the occurrence of elements``        ``boolean``[] present = ``new` `boolean``[N];` `        ``int` `maxele = Integer.MIN_VALUE;` `        ``// Mark the occurrences``        ``for` `(``int` `i = ``0``; i < n; i++) {` `            ``// Only mark the required elements``            ``// All non-positive elements and the elements``            ``// greater n + 1 will never be the answer``            ``// For example, the array will be {1, 2, 3} in``            ``// the worst case and the result will be 4 which``            ``// is n + 1``            ``if` `(A[i] > ``0` `&& A[i] <= n)``                ``present[A[i]] = ``true``;` `            ``// find the maximum element so that if all the``            ``// elements are in order can directly return the``            ``// next number``            ``maxele = Math.max(maxele, A[i]);``        ``}` `        ``// Find the first element which didn't``        ``// appear in the original array``        ``for` `(``int` `i = ``1``; i < N; i++)``            ``if` `(!present[i])``                ``return` `i;` `        ``// If the original array was of the``        ``// type {1, 2, 3} in its sorted form``        ``return` `maxele + ``1``;``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `arr[] = { ``0``, ``10``, ``2``, -``10``, -``20` `};``        ``System.out.println(solution(arr));``    ``}``}` `// This code is contributed by Aditya Kumar (adityakumar129)`

## Python3

 `# Python3 Program to find the smallest``# positive missing number`  `def` `solution(A):  ``# Our original array` `    ``m ``=` `max``(A)  ``# Storing maximum value``    ``if` `m < ``1``:` `        ``# In case all values in our array are negative``        ``return` `1``    ``if` `len``(A) ``=``=` `1``:` `        ``# If it contains only one element``        ``return` `2` `if` `A[``0``] ``=``=` `1` `else` `1``    ``l ``=` `[``0``] ``*` `m``    ``for` `i ``in` `range``(``len``(A)):``        ``if` `A[i] > ``0``:``            ``if` `l[A[i] ``-` `1``] !``=` `1``:` `                ``# Changing the value status at the index of our list``                ``l[A[i] ``-` `1``] ``=` `1``    ``for` `i ``in` `range``(``len``(l)):` `        ``# Encountering first 0, i.e, the element with least value``        ``if` `l[i] ``=``=` `0``:``            ``return` `i ``+` `1``            ``# In case all values are filled between 1 and m``    ``return` `i ``+` `2`  `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``arr ``=` `[``0``, ``10``, ``2``, ``-``10``, ``-``20``]``    ``print``(solution(arr))`

## C#

 `// C# Program to find the smallest``// positive missing number``using` `System;``using` `System.Linq;` `class` `GFG {``    ``static` `int` `solution(``int``[] A)``    ``{``        ``// Our original array` `        ``int` `m = A.Max(); ``// Storing maximum value` `        ``// In case all values in our array are negative``        ``if` `(m < 1) {``            ``return` `1;``        ``}``        ``if` `(A.Length == 1) {` `            ``// If it contains only one element``            ``if` `(A == 1) {``                ``return` `2;``            ``}``            ``else` `{``                ``return` `1;``            ``}``        ``}``        ``int` `i = 0;``        ``int``[] l = ``new` `int``[m];``        ``for` `(i = 0; i < A.Length; i++) {``            ``if` `(A[i] > 0) {``                ``// Changing the value status at the index of``                ``// our list``                ``if` `(l[A[i] - 1] != 1) {``                    ``l[A[i] - 1] = 1;``                ``}``            ``}``        ``}` `        ``// Encountering first 0, i.e, the element with least``        ``// value``        ``for` `(i = 0; i < l.Length; i++) {``            ``if` `(l[i] == 0) {``                ``return` `i + 1;``            ``}``        ``}` `        ``// In case all values are filled between 1 and m``        ``return` `i + 2;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``int``[] arr = { 0, 10, 2, -10, -20 };``        ``Console.WriteLine(solution(arr));``    ``}``}` `// This code is contributed by PrinciRaj1992`

## PHP

 ` 0)``        ``{``            ``if` `(``\$l``[``\$A``[``\$i``] - 1] != 1)``            ``{``                ` `                ``//Changing the value status at the index of our list``                ``\$l``[``\$A``[``\$i``] - 1] = 1;``            ``}``        ``}``    ``}``    ``for` `(``\$i` `= 0;``\$i` `< sizeof(``\$l``); ``\$i``++)``    ``{``         ` `        ``//Encountering first 0, i.e, the element with least value``        ``if` `(``\$l``[``\$i``] == 0)``            ``return` `\$i``+1;``    ``}``            ``//In case all values are filled between 1 and m``    ``return` `\$i``+2;   ``}` `// Driver Code``\$arr` `= ``array``(0, 10, 2, -10, -20);``echo` `solution(``\$arr``);``return` `0;``?>`

## Javascript

 ``

Output

`1`

Time Complexity: O(N), Only two traversals are needed.
Auxiliary Space: O(N), using the list will require extra space

## Smallest positive number missing from an unsorted Array by using array elements as Index:

The idea is to use array elements as an index. To mark the presence of an element x, change the value at the index x to negative. But this approach doesn’t work if there are non-positive (-ve and 0) numbers.

So segregate positive from negative numbers as the first step and then apply the approach.

Follow the steps below to solve the problem:

• Segregate positive numbers from others i.e., move all non-positive numbers to the left side.
• Now ignore non-positive elements and consider only the part of the array which contains all positive elements.
• Traverse the array containing all positive numbers and to mark the presence of an element x, change the sign of value at index x to negative.
• Traverse the array again and print the first index which has a positive value.

Below is the implementation of the above approach.

v

Output

`1`

Time Complexity: O(N), Traversing the array of size N.
Auxiliary Space: O(1)

## Smallest positive number missing from an unsorted array by changing the input Array

The idea is to mark the elements in the array which are greater than N and less than 1 with 1.

Follow the steps below to solve the problem:

• The smallest positive integer is 1. First, we will check if 1 is present in the array or not. If it is not present then 1 is the answer.
• If present then, again traverse the array. The largest possible answer is N+1 where N is the size of the array
• When traversing the array, if we find any number less than 1 or greater than N, change it to 1.
• This will not change anything as the answer will always be between 1 to N+1. Now our array has elements from 1 to N.
• Now, for every ith number, increase arr[ (arr[i]-1) ] by N. But this will increase the value more than N. So, we will access the array by arr[(arr[i]-1)%N].
• We will find now which index has a value less than N+1. Then i+1 will be our answer.

Below is the implementation of the above approach.

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function for finding the first missing positive number` `int` `firstMissingPositive(``int` `arr[], ``int` `n)``{``    ``int` `ptr = 0;` `    ``// Check if 1 is present in array or not``    ``for` `(``int` `i = 0; i < n; i++) {``        ``if` `(arr[i] == 1) {``            ``ptr = 1;``            ``break``;``        ``}``    ``}` `    ``// If 1 is not present``    ``if` `(ptr == 0)``        ``return` `1;` `    ``// Changing values to 1``    ``for` `(``int` `i = 0; i < n; i++)``        ``if` `(arr[i] <= 0 || arr[i] > n)``            ``arr[i] = 1;` `    ``// Updating indices according to values``    ``for` `(``int` `i = 0; i < n; i++)``        ``arr[(arr[i] - 1) % n] += n;` `    ``// Finding which index has value less than n``    ``for` `(``int` `i = 0; i < n; i++)``        ``if` `(arr[i] <= n)``            ``return` `i + 1;` `    ``// If array has values from 1 to n``    ``return` `n + 1;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 0, 10, 2, -10, -20 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);` `    ``int` `ans = firstMissingPositive(arr, n);` `    ``cout << ans;` `    ``return` `0;``}`

## C

 `// C program for the above approach``#include ``#include ` `// Function for finding the first``// missing positive number``int` `firstMissingPositive(``int` `arr[], ``int` `n)``{``    ``int` `ptr = 0;` `    ``// Check if 1 is present in array or not``    ``for` `(``int` `i = 0; i < n; i++) {``        ``if` `(arr[i] == 1) {``            ``ptr = 1;``            ``break``;``        ``}``    ``}` `    ``// If 1 is not present``    ``if` `(ptr == 0)``        ``return` `1;` `    ``// Changing values to 1``    ``for` `(``int` `i = 0; i < n; i++)``        ``if` `(arr[i] <= 0 || arr[i] > n)``            ``arr[i] = 1;` `    ``// Updating indices according to values``    ``for` `(``int` `i = 0; i < n; i++)``        ``arr[(arr[i] - 1) % n] += n;` `    ``// Finding which index has value less than n``    ``for` `(``int` `i = 0; i < n; i++)``        ``if` `(arr[i] <= n)``            ``return` `i + 1;` `    ``// If array has values from 1 to n``    ``return` `n + 1;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 0, 10, 2, -10, -20 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);``    ``int` `ans = firstMissingPositive(arr, n);` `    ``printf``(``"%d"``, ans);` `    ``return` `0;``}` `// This code is contributed by shailjapriya`

## Java

 `// Java program for the above approach``import` `java.util.Arrays;` `class` `GFG {` `    ``// Function for finding the first``    ``// missing positive number``    ``static` `int` `firstMissingPositive(``int` `arr[], ``int` `n)``    ``{``        ``int` `ptr = ``0``;` `        ``// Check if 1 is present in array or not``        ``for` `(``int` `i = ``0``; i < n; i++) {``            ``if` `(arr[i] == ``1``) {``                ``ptr = ``1``;``                ``break``;``            ``}``        ``}` `        ``// If 1 is not present``        ``if` `(ptr == ``0``)``            ``return` `(``1``);` `        ``// Changing values to 1``        ``for` `(``int` `i = ``0``; i < n; i++)``            ``if` `(arr[i] <= ``0` `|| arr[i] > n)``                ``arr[i] = ``1``;` `        ``// Updating indices according to values``        ``for` `(``int` `i = ``0``; i < n; i++)``            ``arr[(arr[i] - ``1``) % n] += n;` `        ``// Finding which index has value less than n``        ``for` `(``int` `i = ``0``; i < n; i++)``            ``if` `(arr[i] <= n)``                ``return` `(i + ``1``);` `        ``// If array has values from 1 to n``        ``return` `(n + ``1``);``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `arr[] = { ``0``, ``10``, ``2``, -``10``, -``20` `};``        ``int` `n = arr.length;``        ``int` `ans = firstMissingPositive(arr, n);` `        ``System.out.println(ans);``    ``}``}` `// This code is contributed by shailjapriya`

## Python3

 `# Python3 program for the above approach` `# Function for finding the first missing``# positive number``def` `firstMissingPositive(arr, n):` `    ``ptr ``=` `0` `    ``# Check if 1 is present in array or not``    ``for` `i ``in` `range``(n):``        ``if` `arr[i] ``=``=` `1``:``            ``ptr ``=` `1``            ``break` `    ``# If 1 is not present``    ``if` `ptr ``=``=` `0``:``        ``return``(``1``)` `    ``# Changing values to 1``    ``for` `i ``in` `range``(n):``        ``if` `arr[i] <``=` `0` `or` `arr[i] > n:``            ``arr[i] ``=` `1` `    ``# Updating indices according to values``    ``for` `i ``in` `range``(n):``        ``arr[(arr[i] ``-` `1``) ``%` `n] ``+``=` `n` `    ``# Finding which index has value less than n``    ``for` `i ``in` `range``(n):``        ``if` `arr[i] <``=` `n:``            ``return``(i ``+` `1``)` `    ``# If array has values from 1 to n``    ``return``(n ``+` `1``)` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``# Given array``    ``A ``=` `[``0``, ``10``, ``2``, ``-``10``, ``-``20``]``    ` `    ``# Size of the array``    ``N ``=` `len``(A)``    ` `    ``# Function call``    ``print``(firstMissingPositive(A, N))` `# This code is contributed by shailjapriya`

## C#

 `// C# program for the above approach``using` `System;``using` `System.Linq;` `class` `GFG {` `    ``// Function for finding the first missing``    ``// positive number``    ``static` `int` `firstMissingPositive(``int``[] arr, ``int` `n)``    ``{``        ``int` `ptr = 0;` `        ``// Check if 1 is present in array or not``        ``for` `(``int` `i = 0; i < n; i++) {``            ``if` `(arr[i] == 1) {``                ``ptr = 1;``                ``break``;``            ``}``        ``}` `        ``// If 1 is not present``        ``if` `(ptr == 0)``            ``return` `1;` `        ``// Changing values to 1``        ``for` `(``int` `i = 0; i < n; i++)``            ``if` `(arr[i] <= 0 || arr[i] > n)``                ``arr[i] = 1;` `        ``// Updating indices according to values``        ``for` `(``int` `i = 0; i < n; i++)``            ``arr[(arr[i] - 1) % n] += n;` `        ``// Finding which index has value less than n``        ``for` `(``int` `i = 0; i < n; i++)``            ``if` `(arr[i] <= n)``                ``return` `i + 1;` `        ``// If array has values from 1 to n``        ``return` `n + 1;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``int``[] A = { 0, 10, 2, -10, -20 };``        ``int` `n = A.Length;``        ``int` `ans = firstMissingPositive(A, n);` `        ``Console.WriteLine(ans);``    ``}``}` `// This code is contributed by shailjapriya`

## Javascript

 ``

Output

`1`

Time Complexity: O(N), Traversing over the array
Auxiliary Space:  O(1)

## Smallest positive number missing from an unsorted array by Swapping:

The idea is to swap the elements which are in the range 1 to N should be placed at their respective indexes.

Follow the steps below to solve the problem:

1. Traverse the array, Ignore the elements which are greater than N and less than 1.
2. While traversing, check if a[i] ≠ a[a[i]-1] holds true or not .
1. If the above condition is true then swap a[i] and a[a[i] – 1]  and swap until (a[i] ≠ a[a[i] – 1]) condition fails.
3. Traverse the array and check whether a[i] ≠ i + 1 then return i + 1.
4. If all are equal to its index then return N+1.

Below is the implementation of the above approach.

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function for finding the first``// missing positive number``int` `firstMissingPositive(``int` `arr[], ``int` `n)``{` `    ``// Loop to traverse the whole array``    ``for` `(``int` `i = 0; i < n; i++) {` `        ``// Loop to check boundary``        ``// condition and for swapping``        ``while` `(arr[i] >= 1 && arr[i] <= n``               ``&& arr[i] != arr[arr[i] - 1]) {``            ``swap(arr[i], arr[arr[i] - 1]);``        ``}``    ``}` `    ``// Checking any element which``    ``// is not equal to i+1``    ``for` `(``int` `i = 0; i < n; i++) {``        ``if` `(arr[i] != i + 1) {``            ``return` `i + 1;``        ``}``    ``}` `    ``// Nothing is present return last index``    ``return` `n + 1;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 0, 10, 2, -10, -20 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);` `    ``int` `ans = firstMissingPositive(arr, n);` `    ``cout << ans;` `    ``return` `0;``}``// This code is contributed by Harsh kedia`

## Java

 `// Java program for the above approach``import` `java.util.Arrays;` `class` `GFG {` `    ``// Function for finding the first``    ``// missing positive number``    ``static` `int` `firstMissingPositive(``int` `arr[], ``int` `n)``    ``{` `        ``// Check if 1 is present in array or not``        ``for` `(``int` `i = ``0``; i < n; i++) {` `            ``// Loop to check boundary``            ``// condition and for swapping``            ``while` `(arr[i] >= ``1` `&& arr[i] <= n``                   ``&& arr[i] != arr[arr[i] - ``1``]) {` `                ``int` `temp = arr[arr[i] - ``1``];``                ``arr[arr[i] - ``1``] = arr[i];``                ``arr[i] = temp;``            ``}``        ``}` `        ``// Finding which index has value less than n``        ``for` `(``int` `i = ``0``; i < n; i++)``            ``if` `(arr[i] != i + ``1``)``                ``return` `(i + ``1``);` `        ``// If array has values from 1 to n``        ``return` `(n + ``1``);``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `arr[] = { ``0``, ``10``, ``2``, -``10``, -``20` `};``        ``int` `n = arr.length;``        ``int` `ans = firstMissingPositive(arr, n);` `        ``System.out.println(ans);``    ``}``}` `// This code is contributed by mohit kumar 29.`

## Python3

 `# Python program for the above approach`  `# Function for finding the first``# missing positive number``def` `firstMissingPositive(arr, n):` `    ``# Loop to traverse the whole array``    ``for` `i ``in` `range``(n):` `        ``# Loop to check boundary``        ``# condition and for swapping``        ``while` `(arr[i] >``=` `1` `and` `arr[i] <``=` `n``               ``and` `arr[i] !``=` `arr[arr[i] ``-` `1``]):``            ``temp ``=` `arr[i]``            ``arr[i] ``=` `arr[arr[i] ``-` `1``]``            ``arr[temp ``-` `1``] ``=` `temp` `    ``# Checking any element which``    ``# is not equal to i + 1``    ``for` `i ``in` `range``(n):``        ``if` `(arr[i] !``=` `i ``+` `1``):``            ``return` `i ``+` `1` `    ``# Nothing is present return last index``    ``return` `n ``+` `1`  `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ``arr ``=` `[``0``, ``10``, ``2``, ``-``10``, ``-``20``]``    ``n ``=` `len``(arr)``    ``ans ``=` `firstMissingPositive(arr, n)``    ``print``(ans)` `# This code is contributed by shivanisinghss2110`

## C#

 `// C# program for the above approach``using` `System;``public` `class` `GFG {` `    ``// Function for finding the first``    ``// missing positive number``    ``static` `int` `firstMissingPositive(``int``[] arr, ``int` `n)``    ``{` `        ``// Check if 1 is present in array or not``        ``for` `(``int` `i = 0; i < n; i++) {` `            ``// Loop to check boundary``            ``// condition and for swapping``            ``while` `(arr[i] >= 1 && arr[i] <= n``                   ``&& arr[i] != arr[arr[i] - 1]) {` `                ``int` `temp = arr[arr[i] - 1];``                ``arr[arr[i] - 1] = arr[i];``                ``arr[i] = temp;``            ``}``        ``}` `        ``// Finding which index has value less than n``        ``for` `(``int` `i = 0; i < n; i++)``            ``if` `(arr[i] != i + 1)``                ``return` `(i + 1);` `        ``// If array has values from 1 to n``        ``return` `(n + 1);``    ``}` `    ``// Driver Code` `    ``static` `public` `void` `Main()``    ``{` `        ``int``[] arr = { 0, 10, 2, -10, -20 };``        ``int` `n = arr.Length;``        ``int` `ans = firstMissingPositive(arr, n);` `        ``Console.WriteLine(ans);``    ``}``}` `// This code is contributed by ab2127`

## Javascript

 ``

Output

`1`

Time Complexity: O(N), Only two traversals are needed.
Auxiliary Space: O(1), No extra space is needed

## Smallest positive number missing from an unsorted array using Sorting:

The idea is to sort the array and then check for the smallest missing number (start from 1) if it is present then increment it.

Follow the steps below to solve the problem:

• First sort the array and the smallest positive integer is 1.
• So, take ans=1 and iterate over the array once and check whether arr[i] = ans (Checking for value from 1 up to the missing number).
• By iterating if that condition meets where arr[i] = ans then increment ans by 1 and again check for the same condition until the size of the array.
• After one scan of the array, the missing number is stored in ans variable.
• Now return that ans to the function.

Below is the implementation of the above approach:

## C++

 `#include ``using` `namespace` `std;` `// Function to find first positive missing number``int` `firstMissingPositive(vector<``int``>& nums)``{``    ``sort(nums.begin(), nums.end());``    ``int` `ans = 1;``    ``for` `(``int` `i = 0; i < nums.size(); i++) {``        ``if` `(nums[i] == ans) {``            ``ans++;``        ``}``    ``}``    ``return` `ans;``}` `// Driver code``int` `main()``{``    ``vector<``int``> arr = { 0, 10, 2, -10, -20 };``    ``// Function call``    ``cout << firstMissingPositive(arr);``    ``return` `0;``}`

## Java

 `/*package whatever // do not write package name here */``import` `java.io.*;``import` `java.util.Arrays;``class` `GFG {``    ``public` `static` `int` `firstMissingPositive(``int``[] nums,``                                           ``int` `n)``    ``{``        ``Arrays.sort(nums);``        ``int` `ans = ``1``;``        ``for` `(``int` `i = ``0``; i < n; i++) {``            ``if` `(nums[i] == ans)``                ``ans++;``        ``}``        ``return` `ans;``    ``}``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `arr[] = { ``0``, ``10``, ``2``, -``10``, -``20` `};``        ``int` `n = arr.length;``        ``int` `ans = firstMissingPositive(arr, n);``        ``System.out.println(ans);``    ``}``}`

## Python3

 `# Python code for the same approach``from` `functools ``import` `cmp_to_key`  `def` `cmp``(a, b):``    ``return` `(a ``-` `b)`  `def` `firstMissingPositive(nums):` `    ``nums.sort(key ``=` `cmp_to_key(``cmp``))``    ``ans ``=` `1``    ``for` `i ``in` `range``(``len``(nums)):` `        ``if``(nums[i] ``=``=` `ans):``            ``ans ``+``=` `1` `    ``return` `ans`  `# driver code``if` `__name__ ``=``=` `'__main__'``:``    ``arr ``=` `[``0``, ``10``, ``2``, ``-``10``, ``-``20``]``    ``print``(firstMissingPositive(arr))` `# This code is contributed by shinjanpatra`

## C#

 `// C# program for the above approach``using` `System;` `public` `class` `GFG {``    ``static` `public` `int` `firstMissingPositive(``int``[] nums,``                                           ``int` `n)``    ``{``        ``Array.Sort(nums);``        ``int` `ans = 1;``        ``for` `(``int` `i = 0; i < n; i++) {``            ``if` `(nums[i] == ans)``                ``ans++;``        ``}``        ``return` `ans;``    ``}` `    ``// Driver Code``    ``static` `public` `void` `Main()``    ``{` `        ``int``[] arr = { 0, 10, 2, -10, -20 };``        ``int` `n = arr.Length;``        ``int` `ans = firstMissingPositive(arr, n);` `        ``Console.WriteLine(ans);``    ``}``}` `// This code is contributed by kothavvsaakash`

## Javascript

 ``

Output

`1`

Time Complexity: O(N*log(N)), Time required to sort the array
Auxiliary Space: O(1)