# Find the missing number

Find the missing number: In this tutorial, we will learn an advance algorithm to find a missing number in an array of length n range from 1 to n. By Radib Kar Last updated : August 21, 2023

## Table of Contents

## Problem statement

We are given a list of n-1 integers and the integers range from 1 to n. There are no duplicates in the list. Only one of the integer from 1 to n is missing. We need to find the missing number in O(n) time complexity.

## Solution

The simplest solution is to take each number from **1** to **n** and to check whether it exists in the array or not. But such approach has **O(n ^{2}) time complexity**. Thus we need to find some advance algorithm to reduce the time complexity.

## Algorithm

The algorithm is illustrated below:

**XOR**all the elements presented in the array. Let the result be**x**.**XOR**all numbers from**1**to**n**. Let the result be**y**.**XORing****x**&**y**gives the missing number.

## C Implementation: Program to find the missing number

#include <stdio.h> #include <stdlib.h> int missingNo(int* a, int n) { int x = 0, y = 0; for (int i = 0; i < n - 1; i++) { //xoring all elements x ^= a[i]; } for (int i = 1; i <= n; i++) { //xoring 1 to n y ^= i; } //xoring x & y outputs missing number return x ^ y; } int main() { int n, x, count = 0; printf("enter your range\n"); scanf("%d", &n); printf("enter elements leaving one nummber in the range 1 to n\n"); // dynamic array created for n-1 elements int* a = (int*)(malloc(sizeof(int) * (n - 1))); for (int i = 0; i < n - 1; i++) { scanf("%d", &a[i]); } // function to check duplicate exists or not printf("\nthe missing number is %d\n", missingNo(a, n)); return 0; }

## Output

enter your range 5 enter elements leaving one nummber in the range 1 to n 2 4 3 5 the missing number is 1

## Time and Space Complexity

**Time complexity:**O(n) (for scanning the array)**Space complexity:**O(1)

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