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

You are given an unsorted array with both positive and negative elements. You have to find the smallest positive number missing from the array in O(n) time using constant extra space. You can modify the original array.

Examples 

 Input:  {2, 3, 7, 6, 8, -1, -10, 15}
 Output: 1

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

 Input: {1, 1, 0, -1, -2}
 Output: 2

A naive method to solve this problem is to search all positive integers, starting from 1 in the given array. We may have to search at most n+1 numbers in the given array. So this solution takes O(n^2) in worst case.

We can use sorting to solve it in lesser time complexity. We can sort the array in O(nLogn) time. Once the array is sorted, then all we need to do is a linear scan of the array. So this approach takes O(nLogn + n) time which is O(nLogn).

We can also use hashing. We can build a hash table of all positive elements in the given array. Once the hash table is built. We can look in the hash table for all positive integers, starting from 1. As soon as we find a number which is not there in hash table, we return it. This approach may take O(n) time on average, but it requires O(n) extra space.

A O(n) time and O(1) extra space solution:
The idea is similar to this post. We use array elements as index. To mark presence of an element x, we 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 we segregate positive from negative numbers as first step and then apply the approach.

Following is the two step algorithm. 
1) Segregate positive numbers from others i.e., move all non-positive numbers to left side. In the following code, segregate() function does this part. 
2) Now we can ignore non-positive elements and consider only the part of array which contains all positive elements. We traverse the array containing all positive numbers and to mark presence of an element x, we change the sign of value at index x to negative. We traverse the array again and print the first index which has positive value. In the following code, findMissingPositive() function does this part. Note that in findMissingPositive, we have subtracted 1 from the values as indexes start from 0 in C. 

C++

/* C++ program to find the
   smallest positive missing number */
#include <bits/stdc++.h>
using namespace std;

/* Utility to swap to integers */
void swap(int* a, int* b)
{
    int temp;
    temp = *a;
    *a = *b;
    *b = temp;
}

/* Utility function that puts all
non-positive (0 and negative) numbers on left
side of arr[] and return count of such numbers */
int segregate(int arr[], int size)
{
    int j = 0, i;
    for (i = 0; i < size; i++) {
        if (arr[i] <= 0) {
            swap(&arr[i], &arr[j]);

            // increment count of
            // non-positive integers
            j++;
        }
    }

    return j;
}

/* Find the smallest positive missing number
in an array that contains all positive integers */
int findMissingPositive(int arr[], int size)
{
    int i;

    // Mark arr[i] as visited by
    // making arr[arr[i] - 1] negative.
    // Note that 1 is subtracted
    // because index start
    // from 0 and positive numbers start from 1
    for (i = 0; i < size; i++) {
        if (abs(arr[i]) - 1 < size && arr[abs(arr[i]) - 1] > 0)
            arr[abs(arr[i]) - 1] = -arr[abs(arr[i]) - 1];
    }

    // Return the first index
    // value at which is positive
    for (i = 0; i < size; i++)
        if (arr[i] > 0)

            // 1 is added because
            // indexes start from 0
            return i + 1;

    return size + 1;
}

/* Find the smallest positive missing
number in an array that contains
both positive and negative integers */
int findMissing(int arr[], int size)
{

    // First separate positive
    // and negative numbers
    int shift = segregate(arr, size);

    // Shift the array and call
    // findMissingPositive for
    // positive part
    return findMissingPositive(arr + shift,
                               size - shift);
}

// Driver code
int main()
{
    int arr[] = { 0, 10, 2, -10, -20 };
    int arr_size = sizeof(arr) / sizeof(arr[0]);
    int missing = findMissing(arr, arr_size);
    cout << "The smallest positive missing number is " << missing;
    return 0;
}

// This is code is contributed by rathbhupendra

C

/* C program to find the smallest
   positive missing number */
#include <stdio.h>
#include <stdlib.h>

/* Utility to swap to integers */
void swap(int* a, int* b)
{
    int temp;
    temp = *a;
    *a = *b;
    *b = temp;
}

/* Utility function that puts all
non-positive (0 and negative) numbers on left
side of arr[] and return count of such numbers */
int segregate(int arr[], int size)
{
    int j = 0, i;
    for (i = 0; i < size; i++) {
        if (arr[i] <= 0) {
            swap(&arr[i], &arr[j]);
            j++; // increment count of non-positive integers
        }
    }

    return j;
}

/* Find the smallest positive missing number
in an array that contains all positive integers */
int findMissingPositive(int arr[], int size)
{
    int i;

    // Mark arr[i] as visited by
    // making arr[arr[i] - 1] negative.
    // Note that 1 is subtracted
    // because index start
    // from 0 and positive numbers start from 1
    for (i = 0; i < size; i++) {
        if (abs(arr[i]) - 1 < size && arr[abs(arr[i]) - 1] > 0)
            arr[abs(arr[i]) - 1] = -arr[abs(arr[i]) - 1];
    }

    // Return the first index value at which is positive
    for (i = 0; i < size; i++)
        if (arr[i] > 0)
            // 1 is added because indexes start from 0
            return i + 1;

    return size + 1;
}

/* Find the smallest positive missing
number in an array that contains
both positive and negative integers */
int findMissing(int arr[], int size)
{
    // First separate positive and negative numbers
    int shift = segregate(arr, size);

    // Shift the array and call findMissingPositive for
    // positive part
    return findMissingPositive(arr + shift, size - shift);
}

// Driver code
int main()
{
    int arr[] = { 0, 10, 2, -10, -20 };
    int arr_size = sizeof(arr) / sizeof(arr[0]);
    int missing = findMissing(arr, arr_size);
    printf("The smallest positive missing number is %d ", missing);
    getchar();
    return 0;
}

Java

// Java program to find the smallest
// positive missing number
import java.util.*;

class Main {

    /* Utility function that puts all non-positive
       (0 and negative) numbers on left side of
       arr[] and return count of such numbers */
    static int segregate(int arr[], int size)
    {
        int j = 0, i;
        for (i = 0; i < size; i++) {
            if (arr[i] <= 0) {
                int temp;
                temp = arr[i];
                arr[i] = arr[j];
                arr[j] = temp;
                // increment count of non-positive
                // integers
                j++;
            }
        }

        return j;
    }

    /* Find the smallest positive missing
       number in an array that contains
       all positive integers */
    static int findMissingPositive(int arr[], int size)
    {
        int i;

        // Mark arr[i] as visited by making
        // arr[arr[i] - 1] negative. Note that
        // 1 is subtracted because index start
        // from 0 and positive numbers start from 1
        for (i = 0; i < size; i++) {
            int x = Math.abs(arr[i]);
            if (x - 1 < size && arr[x - 1] > 0)
                arr[x - 1] = -arr[x - 1];
        }

        // Return the first index value at which
        // is positive
        for (i = 0; i < size; i++)
            if (arr[i] > 0)
                return i + 1; // 1 is added because indexes
        // start from 0

        return size + 1;
    }

    /* Find the smallest positive missing
       number in an array that contains
       both positive and negative integers */
    static int findMissing(int arr[], int size)
    {
        // First separate positive and
        // negative numbers
        int shift = segregate(arr, size);
        int arr2[] = new int[size - shift];
        int j = 0;
        for (int i = shift; i < size; i++) {
            arr2[j] = arr[i];
            j++;
        }
        // Shift the array and call
        // findMissingPositive for
        // positive part
        return findMissingPositive(arr2, j);
    }

    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 0, 10, 2, -10, -20 };
        int arr_size = arr.length;
        int missing = findMissing(arr, arr_size);
        System.out.println("The smallest positive missing number is " + missing);
    }
}

Python3

''' Python3 program to find the
smallest positive missing number '''

''' Utility function that puts all
non-positive (0 and negative) numbers on left
side of arr[] and return count of such numbers '''
def segregate(arr, size):
    j = 0
    for i in range(size):
        if (arr[i] <= 0):
            arr[i], arr[j] = arr[j], arr[i]
            j += 1 # increment count of non-positive integers
    return j


''' Find the smallest positive missing number
in an array that contains all positive integers '''
def findMissingPositive(arr, size):

    # Mark arr[i] as visited by
    # making arr[arr[i] - 1] negative.
    # Note that 1 is subtracted
    # because index start
    # from 0 and positive numbers start from 1
    for i in range(size):
        if (abs(arr[i]) - 1 < size and arr[abs(arr[i]) - 1] > 0):
            arr[abs(arr[i]) - 1] = -arr[abs(arr[i]) - 1]

    # Return the first index value at which is positive
    for i in range(size):
        if (arr[i] > 0):

            # 1 is added because indexes start from 0
            return i + 1
    return size + 1

''' Find the smallest positive missing
number in an array that contains
both positive and negative integers '''
def findMissing(arr, size):

    # First separate positive and negative numbers
    shift = segregate(arr, size)

    # Shift the array and call findMissingPositive for
    # positive part
    return findMissingPositive(arr[shift:], size - shift)

# Driver code
arr = [ 0, 10, 2, -10, -20 ]
arr_size = len(arr)
missing = findMissing(arr, arr_size)
print("The smallest positive missing number is ", missing)

# This code is contributed by Shubhamsingh10

C#

// C# program to find the smallest
// positive missing number
using System;

class main {

    // Utility function that puts all
    // non-positive (0 and negative)
    // numbers on left side of arr[]
    // and return count of such numbers
    static int segregate(int[] arr, int size)
    {
        int j = 0, i;
        for (i = 0; i < size; i++) {
            if (arr[i] <= 0) {
                int temp;
                temp = arr[i];
                arr[i] = arr[j];
                arr[j] = temp;

                // increment count of non-positive
                // integers
                j++;
            }
        }

        return j;
    }

    // Find the smallest positive missing
    // number in an array that contains
    // all positive integers
    static int findMissingPositive(int[] arr, int size)
    {
        int i;

        // Mark arr[i] as visited by making
        // arr[arr[i] - 1] negative. Note that
        // 1 is subtracted as index start from
        // 0 and positive numbers start from 1
        for (i = 0; i < size; i++) {
            if (Math.Abs(arr[i]) - 1 < size && arr[ Math.Abs(arr[i]) - 1] > 0)
                arr[ Math.Abs(arr[i]) - 1] = -arr[ Math.Abs(arr[i]) - 1];
        }

        // Return the first index value at
        // which is positive
        for (i = 0; i < size; i++)
            if (arr[i] > 0)
                return i + 1;

        // 1 is added because indexes
        // start from 0
        return size + 1;
    }

    // Find the smallest positive
    // missing number in array that
    // contains both positive and
    // negative integers
    static int findMissing(int[] arr, int size)
    {

        // First separate positive and
        // negative numbers
        int shift = segregate(arr, size);
        int[] arr2 = new int[size - shift];
        int j = 0;

        for (int i = shift; i < size; i++) {
            arr2[j] = arr[i];
            j++;
        }

        // Shift the array and call
        // findMissingPositive for
        // positive part
        return findMissingPositive(arr2, j);
    }

    // Driver code
    public static void Main()
    {
        int[] arr = { 0, 10, 2, -10, -20 };
        int arr_size = arr.Length;
        int missing = findMissing(arr, arr_size);
        Console.WriteLine("The smallest positive missing number is " + missing);
    }
}

// This code is contributed by Anant Agarwal.

Javascript

<script>
// Javascript program to find the smallest
// positive missing number

 /* Utility function that puts all non-positive
       (0 and negative) numbers on left side of
       arr[] and return count of such numbers */
function segregate(arr,size)
{
     let j = 0, i;
        for (i = 0; i < size; i++) {
            if (arr[i] <= 0) {
                let temp;
                temp = arr[i];
                arr[i] = arr[j];
                arr[j] = temp;
                // increment count of non-positive
                // integers
                j++;
            }
        }

        return j;
}

 /* Find the smallest positive missing
       number in an array that contains
       all positive integers */
function findMissingPositive(arr,size)
{
    let i;

        // Mark arr[i] as visited by making
        // arr[arr[i] - 1] negative. Note that
        // 1 is subtracted because index start
        // from 0 and positive numbers start from 1
        for (i = 0; i < size; i++) {
            let x = Math.abs(arr[i]);
            if (x - 1 < size && arr[x - 1] > 0)
                arr[x - 1] = -arr[x - 1];
        }

        // Return the first index value at which
        // is positive
        for (i = 0; i < size; i++)
            if (arr[i] > 0)
                return i + 1; // 1 is added because indexes
        // start from 0

        return size + 1;
}

/* Find the smallest positive missing
       number in an array that contains
       both positive and negative integers */
function findMissing(arr,size)
{

    // First separate positive and
        // negative numbers
        let shift = segregate(arr, size);
        let arr2 = new Array(size - shift);
        let j = 0;
        for (let i = shift; i < size; i++) {
            arr2[j] = arr[i];
            j++;
        }

        // Shift the array and call
        // findMissingPositive for
        // positive part
        return findMissingPositive(arr2, j);
}

// Driver code
let arr = [0, 10, 2, -10, -20];
let arr_size = arr.length;
let  missing = findMissing(arr, arr_size);
document.write("The smallest positive missing number is " + missing);

// This code is contributed by rag2127
</script>

Output

The smallest positive missing number is 1

Note that this method modifies the original array. We can change the sign of elements in the segregated array to get the same set of elements back. But we still lose the order of elements. If we want to keep the original array as it was, then we can create a copy of the array and run this approach on the temp array.

Another approach: In this problem, we have created a list full of 0’s with size of the max() value of our given array. Now, whenever we encounter any positive value in our original array, we change the index value of our list to 1. So, after we are done, we simply iterate through our modified list, the first 0 we encounter, its (index value + 1) should be our answer since index in python starts from 0.

Below is the implementation of above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
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 A[] = { 0, 10, 2, -10, -20 };
    int size = sizeof(A) / sizeof(A[0]);
    cout << firstMissingPos(A, size);
}

// This code is contributed by Aditya Kumar (adityakumar129)

C

// C implementation of the approach
#include <stdbool.h>
#include <stdio.h>

// 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 A[] = { 0, 10, 2, -10, -20 };
    int size = sizeof(A) / sizeof(A[0]);
    printf("%d", firstMissingPos(A, 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 A[] = { 0, 10, 2, -10, -20 };
        System.out.println(solution(A));
        int arr[] = { -2, -1, 0, 1, 2, 3, 4 };
        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
A = [0, 10, 2, -10, -20]
print(solution(A))

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[0] == 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[] A = { 0, 10, 2, -10, -20 };
        Console.WriteLine(solution(A));
    }
}

// This code is contributed by PrinciRaj1992

PHP

<?php
// PHP Program to find the smallest
// positive missing number

function solution($A){//Our original array

    $m = max($A); //Storing maximum value
    if ($m < 1)
    {
        // In case all values in our array are negative
        return 1;
    }
    if (sizeof($A) == 1)
    {
        //If it contains only one element
        if ($A[0] == 1)
            return 2 ;
        else
            return 1 ;
    }
    $l = array_fill(0, $m, NULL);
    for($i = 0; $i < sizeof($A); $i++)
    {
        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 = 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
$A = array(0, 10, 2, -10, -20);
echo solution($A);
return 0;
?>

Javascript

<script>
// Javascript Program to find the smallest
// positive missing number

    function solution(A)
    {
        let n = A.length;
        // To mark the occurrence of elements
        let present = new Array(n+1);


        for(let i=0;i<n+1;i++)
        {
            present[i]=false;
        }
        // Mark the occurrences
        for (let 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 (let 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
    let A=[0, 10, 2, -10, -20]
    document.write(solution(A));

</script>

Output

1

Complexity Analysis:

• Time Complexity: O(n). 
  Only two traversals are needed. So the time complexity is O(n).
• Auxiliary Space: O(n). 
  using the list will require extra space

Another Approach:

• 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 array. This will happen when array have all the elements from 1 to N. When we are traversing the array, if we find any number less than 1 or greater than N, then we will change it to 1. This will not change anything as answer will always 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]. What we have done is for each value we have increased value at that index by N.
• We will find now which index has 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 <bits/stdc++.h>
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[] = { 2, 3, -7, 6, 8, 1, -10, 15 };
    int n = sizeof(arr) / sizeof(arr[0]);

    int ans = firstMissingPositive(arr, n);

    cout << ans;

    return 0;
}

C

// C program for the above approach
#include <stdio.h>
#include <stdlib.h>

// 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[] = { 2, 3, -7, 6, 8, 1, -10, 15 };
    int n = sizeof(arr) / sizeof(arr[0]);
    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[] = { 2, 3, -7, 6, 8, 1, -10, 15 };
    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

# Given array
A = [ 2, 3, -7, 6, 8, 1, -10, 15 ]

# 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 = { 2, 3, -7, 6, 8, 1, -10, 15 };
    int n = A.Length;
    int ans = firstMissingPositive(A, n);

    Console.WriteLine(ans);
}
}

// This code is contributed by shailjapriya

Javascript

<script>

// Javascript program for the above approach

// Function for finding the first
// missing positive number
function firstMissingPositive(arr, n)
{
    let ptr = 0;

    // Check if 1 is present in array or not
    for(let 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(let i = 0; i < n; i++)
        if (arr[i] <= 0 || arr[i] > n)
            arr[i] = 1;

    // Updating indices according to values
    for(let i = 0; i < n; i++)
        arr[(arr[i] - 1) % n] += n;

    // Finding which index has value less than n
    for(let 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
let arr = [ 2, 3, -7, 6, 8, 1, -10, 15 ];
let n = arr.length;
let ans = firstMissingPositive(arr, n);

document.write(ans);

// This code is contributed by telimayur

</script>

Output:

4

Time Complexity : O(n)
Auxiliary Space:  O(1) 

Another Approach:

1. Traverse the array, Ignore the elements which are greater than n and less than 1.
2. While traversing check a[i]!=a[a[i]-1] if this condition hold true or not .
3. If the above condition is true then swap a[i], a[a[i] – 1]  && swap until a[i] != a[a[i] – 1]  condition will fail .
4. Traverse the array and check whether a[i] != i + 1 then return i + 1.
5. 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 <bits/stdc++.h>
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[] = { 2, 3, -7, 6, 8, 1, -10, 15 };
    int n = sizeof(arr) / sizeof(arr[0]);

    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[] = {2, 3, -7, 6, 8, 1, -10, 15 };
    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
arr = [ 2, 3, -7, 6, 8, 1, -10, 15 ];
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 = {2, 3, -7, 6, 8, 1, -10, 15 };
    int n = arr.Length;
    int ans = firstMissingPositive(arr, n);

    Console.WriteLine(ans);
  }
}

// This code is contributed by ab2127

Javascript

<script>
// Javascript program for the above approach

// Function for finding the first
// missing positive number
function firstMissingPositive(arr,n)
{
    // Check if 1 is present in array or not
    for(let 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]) {

        let temp=arr[arr[i]-1];
            arr[arr[i]-1]=arr[i];
            arr[i]=temp;
      }
    }

    // Finding which index has value less than n
    for(let 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
let arr=[2, 3, -7, 6, 8, 1, -10, 15 ];
let  n = arr.length;
let  ans = firstMissingPositive(arr, n);
document.write(ans);



// This code is contributed by patel2127
</script>

Output

4

Complexity Analysis:

• Time Complexity: O(n). 
  Only two traversals are needed. So the time complexity is O(n).
• Auxiliary Space: O(1). 
  No extra space is needed, so the space complexity is constant.

Another simple approach with small and crisp code. 

Intuition: first we will sort the array now as we have to calculate the first missing positive integer,and the smallest positive integer is 1.

So, take ans=1 and iterate over the array once and check whether nums[i]==ans (means we are checking for value from 1 upto missing number).

By iterating if that condition meet where nums[i]==ans then increment ans by 1 and again check for the same condition until size of the array.

After one scan of array we got the missing number stored in ans variable.

Now return that ans to the function as return type in int.

Code implementation of above intuition is as below:

C++

#include <bits/stdc++.h>
using namespace std;
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;
}
int main() {
vector<int>arr={-1,0,8,1};
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[] = { 2, 3, -7, 6, 8, 1, -10, 15 };
    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
arr = [-1, 0, 8, 1]
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 = { 2, 3, -7, 6, 8, 1, -10, 15 };
    int n = arr.Length;
    int ans = firstMissingPositive(arr, n);

    Console.WriteLine(ans);
  }
}

// This code is contributed by kothavvsaakash

Javascript

<script>

function firstMissingPositive(nums)
{
    nums.sort((a,b)=>a-b);
    let ans = 1;
    for(let i = 0; i < nums.length; i++)
    {
        if(nums[i] == ans)
            ans++;
    }
    return ans;
}

// driver code
let arr = [-1,0,8,1];
document.write(firstMissingPositive(arr));

// This code is contributed by shinjanpatra

</script>

Output

2

Time Complexity: O(n*log(n))

Auxiliary Space: O(1) 

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