JavaScript Program to Find the Missing Number

Given an array of size N-1 with integers in the range of [1, N]. The task is to find the missing number from the first N integers. There are no duplicate elements allowed in the array.

Examples:

Input :  arr = [1, 2, 3, 5]
Output : 4
Input : arr = [1, 4, 3, 2, 6, 5, 7, 10, 9]
Output : 8

Approach 1: Using the Mathematical Approach (Summation of first N natural Numbers)

The sum of the first N natural Numbers in a Range [1, N] is given by N * (N + 1) / 2.

To find the missing number, first find the sum of the first N natural number using the formula. And then use array traversal using a loop (for / while, …) and find the sum of array elements. At last, subtract the sum of array elements from the sum of natural numbers to find the missing element of an array.

Syntax:

missingElement = (N * (N + 1) / 1) - Sum of Array Elements

Example: The function findMissingNumber calculates the missing number in an array by finding the difference between the sum of the expected sequence and the sum of the array elements.

Javascript

function findMissingNumber(arr) {
    const n = arr.length + 1;
    const sumOfFirstN = (n * (n + 1)) / 2;

    let sumOfArray = 0;
    for (let i = 0; i < n - 1; i++) {
        sumOfArray = sumOfArray + arr[i];
    }

    let missingNumber = sumOfFirstN - sumOfArray;

    return missingNumber;
}

const arr = [1, 2, 5, 4, 6, 8, 7];
const missingNumber = findMissingNumber(arr);
console.log("Missing Number: ", missingNumber);

Output

Missing Number:  3

Approach 2: Using Hashing

• Create an array temp[] of size N + 1 (where N is the length of array) with all initial values as 0.
• Traverse the input array arr[], and set the temp index frequency to 1, i.e. if(temp[arr[i]] == 0) temp[arr[i]] = 1
• Traverse temp[] and output the array element having value as 0 (This is the missing element).

Example:

Javascript

// Function to find the missing number 
function findMissing(arr, N) {
    let i;

    // Create an Array of size N 
    // and filled with 0 
    let temp = new Array(N).fill(0);

    // If array element exist then 
    // set the frequency to 1 
    for (i = 0; i < N; i++) {
        temp[arr[i] - 1] = 1;
    }

    let ans = 0;
    for (i = 0; i <= N; i++) {
        if (temp[i] === 0)
            ans = i + 1;
    }
    console.log(ans);
}

// Driver code 
let arr = [1, 3, 7, 5, 6, 2];
let n = arr.length;

// Function call 
findMissing(arr, n);

Output

4

Approach 3: Using Sorting

• First we will sort the array in ascending order. Sorting will helps us identify the missing number because it allows us to easily find where the sequence breaks.
• Next we will iterate over the sorted array. For each element at index i, we compare it with i + 1. If they are not equal then i + 1 is the missing numbe.

JavaScript

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

const numbers = [1, 2, 3, 4, 5, 6, 8, 9, 10];
const missingNumber = findMissingNumber(numbers);
console.log("The missing number is:", missingNumber);

Output

The missing number is: 7