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Find the Missing Number

Last Updated : 19 Apr, 2025
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Given an array arr[] of size n-1 with distinct integers in the range of [1, n]. This array represents a permutation of the integers from 1 to n with one element missing. Find the missing element in the array.

Examples: 

Input: arr[] = [8, 2, 4, 5, 3, 7, 1]
Output: 6
Explanation: All the numbers from 1 to 8 are present except 6.

Input: arr[] = [1, 2, 3, 5]
Output: 4
Explanation: Here the size of the array is 4, so the range will be [1, 5]. The missing number between 1 to 5 is 4

Table of Content

[Naive Approach] Linear Search for Missing Number - O(n^2) Time and O(1) Space

This approach iterates through each number from 1 to n (where n is the size of the array + 1) and checks if the number is present in the array. For each number, it uses a nested loop to search the array. If a number is not found, it is returned as the missing number.

C++ 
#include <iostream>
#include <vector>
using namespace std;

int missingNum(vector<int>& arr) {
    int n = arr.size() + 1;

    // Iterate from 1 to n and check
    // if the current number is present
    for (int i = 1; i <= n; i++) {
        bool found = false;
        for (int j = 0; j < n - 1; j++) {
            if (arr[j] == i) {
                found = true;
                break;
            }
        }

        // If the current number is not present
        if (!found)
            return i;
    }
    return -1;
}

int main() {
    vector<int> arr = {8, 2, 4, 5, 3, 7, 1}; 
    cout << missingNum(arr) << endl; 
    return 0;
}
Java 
public class GfG {
    public static int missingNum(int[] arr) {
        int n = arr.length + 1;

        // Iterate from 1 to n and check
        // if the current number is present
        for (int i = 1; i <= n; i++) {
            boolean found = false;
            for (int j = 0; j < n - 1; j++) {
                if (arr[j] == i) {
                    found = true;
                    break;
                }
            }

            // If the current number is not present
            if (!found)
                return i;
        }
        return -1;
    }

    public static void main(String[] args) {
        int[] arr = {8, 2, 4, 5, 3, 7, 1};  
        System.out.println(missingNum(arr));
    }
}
Python 
def missingNum(arr):
    n = len(arr) + 1

    # Iterate from 1 to n and check
    # if the current number is present
    for i in range(1, n + 1):
        found = False
        for j in range(n - 1):
            if arr[j] == i:
                found = True
                break

        # If the current number is not present
        if not found:
            return i
    return -1

if __name__ == '__main__':
    arr = [8, 2, 4, 5, 3, 7, 1]
    print(missingNum(arr))
C# 
using System;

class GfG {
    static int missingNum(int[] arr) {
        int n = arr.Length + 1;

        // Iterate from 1 to n and check
        // if the current number is present
        for (int i = 1; i <= n; i++) {
            bool found = false;
            for (int j = 0; j < n - 1; j++) {
                if (arr[j] == i) {
                    found = true;
                    break;
                }
            }

            // If the current number is not present
            if (!found)
                return i;
        }
        return -1;
    }

    static void Main() {
        int[] arr = { 8, 2, 4, 5, 3, 7, 1 };  
        Console.WriteLine(missingNum(arr));
    }
}
JavaScript 
function missingNum(arr) {
    const n = arr.length + 1;

    // Iterate from 1 to n and check
    // if the current number is present
    for (let i = 1; i <= n; i++) {
        let found = false;
        for (let j = 0; j < n - 1; j++) {
            if (arr[j] === i) {
                found = true;
                break;
            }
        }

        // If the current number is not present
        if (!found)
            return i;
    }
    return -1;
}

// drvier code 
const arr = [8, 2, 4, 5, 3, 7, 1];
console.log(missingNum(arr));

Output
6

[Better Approach] Using Hashing - O(n) Time and O(n) Space

This approach uses a hash array (or frequency array) to track the presence of each number from 1 to n in the input array. It first initializes a hash array to store the frequency of each element. Then, it iterates through the hash array to find the number that is missing (i.e., the one with a frequency of 0).

C++ 
#include <iostream>
#include <vector>
using namespace std;

int missingNum(vector<int> &arr) {

    int n = arr.size() + 1;

    // Create hash array of size n+1
    vector<int> hash(n + 1, 0);

    // Store frequencies of elements
    for (int i = 0; i < n - 1; i++) {
        hash[arr[i]]++;
    }

    // Find the missing number
    for (int i = 1; i <= n; i++) {
        if (hash[i] == 0) {
            return i;
        }
    }
    return -1;
}

int main() {
    vector<int> arr = {8, 2, 4, 5, 3, 7, 1};
    int res = missingNum(arr);
    cout << res << endl;
    return 0;
}
Java 
import java.util.Arrays;

public class GfG {
    public static int missingNum(int[] arr) {
        int n = arr.length + 1;

        // Create hash array of size n+1
        int[] hash = new int[n + 1];

        // Store frequencies of elements
        for (int i = 0; i < n - 1; i++) {
            hash[arr[i]]++;
        }

        // Find the missing number
        for (int i = 1; i <= n; i++) {
            if (hash[i] == 0) {
                return i;
            }
        }
        return -1;
    }

    public static void main(String[] args) {
        int[] arr = {8, 2, 4, 5, 3, 7, 1};
        int res = missingNum(arr);
        System.out.println(res);
    }
}
Python 
def missingNum(arr):
    n = len(arr) + 1

    # Create hash array of size n+1
    hash = [0] * (n + 1)

    # Store frequencies of elements
    for i in range(n - 1):
        hash[arr[i]] += 1

    # Find the missing number
    for i in range(1, n + 1):
        if hash[i] == 0:
            return i
    return -1

if __name__ == '__main__':
    arr = [8, 2, 4, 5, 3, 7, 1]
    res = missingNum(arr)
    print(res)
C# 
using System;

class GfG {
    public static int missingNum(int[] arr) {
        int n = arr.Length + 1;

        // Create hash array of size n+1
        int[] hash = new int[n + 1];

        // Store frequencies of elements
        for (int i = 0; i < n - 1; i++) {
            hash[arr[i]]++;
        }

        // Find the missing number
        for (int i = 1; i <= n; i++) {
            if (hash[i] == 0) {
                return i;
            }
        }
        return -1;
    }

    static void Main() {
        int[] arr = {8, 2, 4, 5, 3, 7, 1}; 
        int res = missingNum(arr);
        Console.WriteLine(res);
    }
}
JavaScript 
function missingNum(arr) {
    let n = arr.length + 1;

    // Create hash array of size n+1
    let hash = new Array(n + 1).fill(0);

    // Store frequencies of elements
    for (let i = 0; i < n - 1; i++) {
        hash[arr[i]]++;
    }

    // Find the missing number
    for (let i = 1; i <= n; i++) {
        if (hash[i] === 0) {
            return i;
        }
    }
    return -1;
}

// driver code
const arr = [8, 2, 4, 5, 3, 7, 1];
const res = missingNum(arr);
console.log(res);

Output
6

[Expected Approach 1] Using Sum of n terms Formula - O(n) Time and O(1) Space

The sum of the first n natural numbers is given by the formula (n * (n + 1)) / 2. The idea is to compute this sum and subtract the sum of all elements in the array from it to get the missing number.

C++ 
#include <iostream>
#include <vector>
using namespace std;

int missingNum(vector<int> &arr) {
    int n = arr.size() + 1;
  
    // Calculate the sum of array elements
    int sum = 0;
    for (int i = 0; i < n - 1; i++) {
        sum += arr[i];
    }

    // Calculate the expected sum
    long long expSum = (n *1LL* (n + 1)) / 2;  

    // Return the missing number
    return expSum - sum;
}

int main() {
    vector<int> arr = {8, 2, 4, 5, 3, 7, 1};  
    cout << missingNum(arr);  
    return 0;
}
Java 
import java.util.*;

public class GfG {
   public static int missingNum(int[] arr) {
        long n = arr.length + 1;
    
        // Calculate the sum of array elements
        long sum = 0;
        for (int i = 0; i < arr.length; i++) {
            sum += arr[i];
        }
    
        // Use long for expected sum to avoid overflow
        long expSum = n * (n + 1) / 2;
    
        // Return the missing number
        return (int)(expSum - sum);
    }

    public static void main(String[] args) {
        int[] arr = {8, 2, 4, 5, 3, 7, 1};
        System.out.println(missingNum(arr));
    }
}
Python 
def missingNum(arr):
    n = len(arr) + 1

    # Calculate the sum of array elements
    totalSum = sum(arr)

    # Calculate the expected sum
    expSum = n * (n + 1) // 2

    # Return the missing number
    return expSum - totalSum

if __name__ == '__main__':
    arr = [8, 2, 4, 5, 3, 7, 1]
    print(missingNum(arr))
C# 
using System;
using System.Linq;

class GfG {
    static int missingNum(int[] arr) {
      int n = arr.Length + 1;

    // Calculate the sum of array elements
    int sum = arr.Sum();
    
    // Calculate the expected sum using long to avoid overflow
    long expSum = (long)n * (n + 1) / 2;
    
    // Return the missing number
    return (int)(expSum - sum);

    }

    static void Main() {
        int[] arr = {8, 2, 4, 5, 3, 7, 1};
        Console.WriteLine(missingNum(arr));
    }
}
JavaScript 
function missingNum(arr) {
    let n = arr.length + 1;

    // Calculate the sum of array elements
    let sum = 0;
    for (let i = 0; i < n - 1; i++) {
        sum += arr[i];
    }

    // Calculate the expected sum
    let expSum = (n * (n + 1)) / 2;

    // Return the missing number
    return expSum - sum;
}

// driver code 
let arr = [8, 2, 4, 5, 3, 7, 1];
console.log(missingNum(arr));

Output
6

[Expected Approach 2] Using XOR Operation - O(n) Time and O(1) Space

XOR of a number with itself is 0 i.e. x ^ x = 0 and the given array arr[] has numbers in range [1, n]. This means that the result of XOR of first n natural numbers with the XOR of all the array elements will be the missing number. To do so, calculate XOR of first n natural numbers and XOR of all the array arr[] elements, and then our result will be the XOR of both the resultant values.

C++ 
#include <iostream>
#include <vector>
using namespace std;

int missingNum(vector<int>& arr) {
    int n = arr.size() + 1;
    int xor1 = 0, xor2 = 0;

    // XOR all array elements
    for (int i = 0; i < n - 1; i++) {
        xor2 ^= arr[i];
    }

    // XOR all numbers from 1 to n
    for (int i = 1; i <= n; i++) {
        xor1 ^= i;
    }

    // Missing number is the XOR of xor1 and xor2
    return xor1 ^ xor2;
}

int main() {
    vector<int> arr = {8, 2, 4, 5, 3, 7, 1}; 
    int res = missingNum(arr);  
    cout << res << endl;  
    return 0;
}
Java 
import java.util.Arrays;

public class GfG {
    public static int missingNum(int[] arr) {
        int n = arr.length + 1;
        int xor1 = 0, xor2 = 0;

        // XOR all array elements
        for (int i = 0; i < n - 1; i++) {
            xor2 ^= arr[i];
        }

        // XOR all numbers from 1 to n
        for (int i = 1; i <= n; i++) {
            xor1 ^= i;
        }

        // Missing number is the XOR of xor1 and xor2
        return xor1 ^ xor2;
    }

    public static void main(String[] args) {
        int[] arr = {8, 2, 4, 5, 3, 7, 1};
        int res = missingNum(arr);
        System.out.println(res);
    }
}
Python 
def missingNum(arr):
    n = len(arr) + 1
    xor1 = 0
    xor2 = 0

    # XOR all array elements
    for i in range(n - 1):
        xor2 ^= arr[i]

    # XOR all numbers from 1 to n
    for i in range(1, n + 1):
        xor1 ^= i

    # Missing number is the XOR of xor1 and xor2
    return xor1 ^ xor2

if __name__ == '__main__':
    arr = [8, 2, 4, 5, 3, 7, 1]
    res = missingNum(arr)
    print(res)
C# 
using System;

class GfG {
    static int missingNum(int[] arr) {
        int n = arr.Length + 1;
        int xor1 = 0, xor2 = 0;

        // XOR all array elements
        for (int i = 0; i < n - 1; i++) {
            xor2 ^= arr[i];
        }

        // XOR all numbers from 1 to n
        for (int i = 1; i <= n; i++) {
            xor1 ^= i;
        }

        // Missing number is the XOR of xor1 and xor2
        return xor1 ^ xor2;
    }

    static void Main() {
        int[] arr = { 8, 2, 4, 5, 3, 7, 1 };
        int res = missingNum(arr);
        Console.WriteLine(res);
    }
}
JavaScript 
function missingNum(arr) {
    const n = arr.length + 1;
    let xor1 = 0, xor2 = 0;

    // XOR all array elements
    for (let i = 0; i < n - 1; i++) {
        xor2 ^= arr[i];
    }

    // XOR all numbers from 1 to n
    for (let i = 1; i <= n; i++) {
        xor1 ^= i;
    }

    // Missing number is the XOR of xor1 and xor2
    return xor1 ^ xor2;
}

// driver code
const arr = [8, 2, 4, 5, 3, 7, 1];
const res = missingNum(arr);
console.log(res);

Output
6

Next Article
Find the first repeating element in an array of integers

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