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

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

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

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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)

// 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);
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);
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 loose 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
// 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);
cout << firstMissingPos(A, size);
}

// This code is contributed by gp6

Java

// Java Program to find the smallest
// positive missing number
import java.util.Arrays;
public class GFG {

static int solution(int[] A)
{
int n = A.length;

// To mark the occurrence of elements
boolean[] present = new boolean[n + 1];

// 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
// 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
public static void main(String[] args)
{

int A[] = { 0, 10, 2, -10, -20 };
System.out.println(solution(A));
}
}
// This code is contributed by 29AjayKumar

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 == 1 else 1
l =  * 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 == 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 == 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
// 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

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);

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);
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)
Space Complexity : 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);

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[arr[i] - 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

Time Complexity: O(n)

Space Complexity: O(1)

Another simple approach with small and crisp code.

Intutuion: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 wheather nums[i]==ans (means we are cheacking for value from 1 upto missing number).

By iterating if that condition mate where nums[i]==ans then increamnet ans by 1 and again check for the same condition uptill 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 intution 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;
}

Time complexity:O(nlogn)+O(n)[O(nlogn for sorting and O(n)for iteration)].

Space complexity:O(n) [As we are not using any extra space].

https://www.yotutube.com/watch?v=_OAFHDVihjo