Most Efficient Way to Find a Missing Number in an array

Last Updated : 24 May, 2024

Have you wondered which is the most efficient way for a popular problem to find a missing Number? well in this post we are going to discuss this phenomenon in brief.

First, let's discuss the problem statement:

Given an array of numbers from 1 to N (both inclusive). The size of the array is N-1. The numbers are randomly added to the array, there is one missing number in the array from 1 to N. What is the quickest way to find that missing number?

An approach using the XOR operator:

We can use the XOR operation which is safer than summation because in programming languages if the given input is large it may overflow and may give wrong answers.
Before going to the solution, know that A xor A = 0. So if we XOR two identical numbers the value is 0.
Now, XORing [1..n] with the elements present in the array cancels the identical numbers. So in the end we will get the missing number.

How the XOR operator makes this algorithm efficient:

XOR has certain properties

Assume a₁ ⊕ a₂ ⊕ a₃ ⊕ . . . ⊕ a_n = a and a₁ ⊕ a₂ ⊕ a₃ ⊕ . . . ⊕ a_n-1 = b
Then a ⊕ b = a_n
X ^ X = 0, i.e. Xor of the same values is zero.

Below is the implementation of the above idea:

C++

#include <iostream>
using namespace std;

// Function to find the missing number
int findMissingNumbers(int arr[], int n)
{
    int XOR = 0;
    // XORing [1..n] with the elements present in the array
    // cancels the identical numbers. So at the end we will
    // get the missing number.
    for (int i = 0; i < n - 1; i++) {
        XOR ^= arr[i];
    }
    for (int i = 1; i < n + 1; i++) {
        XOR ^= i;
    }
    return XOR;
}

// Driver code
int main()
{
    int arr[] = { 1, 2, 4, 5 };
    int n = 5;
    cout << findMissingNumbers(arr, n) << endl;
    return 0;
}

Java

/*code by flutterfly */
public class MissingNumberFinder {

    // Function to find the missing number
    public static int findMissingNumbers(int[] arr, int n) {
        int XOR = 0;
        
        // XORing [1..n] with the elements present in the array
        // cancels the identical numbers. So at the end, we will
        // get the missing number.
        for (int i = 0; i < n - 1; i++) {
            XOR ^= arr[i];
        }
        for (int i = 1; i < n + 1; i++) {
            XOR ^= i;
        }
        return XOR;
    }

    // Driver code
    public static void main(String[] args) {
        int[] arr = {1, 2, 4, 5};
        int n = 5;
        System.out.println(findMissingNumbers(arr, n));
    }
}

Python

# code by flutterfly
# Function to find the missing number
def find_missing_numbers(arr, n):
    XOR = 0
    
    # XORing [1..n] with the elements present in the array
    # cancels the identical numbers. So at the end, we will
    # get the missing number.
    for i in range(n - 1):
        XOR ^= arr[i]
    
    for i in range(1, n + 1):
        XOR ^= i
    
    return XOR

# Driver code
if __name__ == "__main__":
    arr = [1, 2, 4, 5]
    n = 5
    print(find_missing_numbers(arr, n))

//code by flutterfly
using System;

public class MissingNumberFinder
{
    // Function to find the missing number
    public static int FindMissingNumbers(int[] arr, int n)
    {
        int XOR = 0;
        
        // XORing [1..n] with the elements present in the array
        // cancels the identical numbers. So at the end, we will
        // get the missing number.
        for (int i = 0; i < n - 1; i++)
        {
            XOR ^= arr[i];
        }

        for (int i = 1; i < n + 1; i++)
        {
            XOR ^= i;
        }

        return XOR;
    }

    // Driver code
    public static void Main()
    {
        int[] arr = {1, 2, 4, 5};
        int n = 5;
        Console.WriteLine(FindMissingNumbers(arr, n));
    }
}

JavaScript

// Function to find the missing number
function findMissingNumbers(arr, n) {
    let XOR = 0;

    // XORing [1..n] with the elements present in the array
    // cancels the identical numbers. So at the end, we will
    // get the missing number.
    for (let i = 0; i < n - 1; i++) {
        XOR ^= arr[i];
    }

    for (let i = 1; i < n + 1; i++) {
        XOR ^= i;
    }

    return XOR;
}

// Driver code
const arr = [1, 2, 4, 5];
const n = 5;
console.log(findMissingNumbers(arr, n));

Output

Time Complexity : O(N)
Auxillary space: O(1)