Find the Repeating & Missing Numbers in JavaScript Array ?

JavaScript allows us to find the repeating and missing number in a given array which includes numbers from 1 to N range. We are given an array that has numbers present from 1 to N range, where N is any natural number. One number is present two times and one number is missing from the range.

We have to return these numbers which can be achieved by several methods using JavaScript which are as follows:

Finding repeating and missing numbers Using array manipulation

In this approach, we will use a nested loop to return repeating and missing numbers. We will use a loop from 1 to N range. We use another loop inside it to store the count of each number. We will return the repeating number which has a count of 2 and return the missing number which has a count of 0 (not present).

Example: This JavaScript code efficiently identifies a repeating and a missing number within an array using array manipulation and negation, ensuring a concise and effective solution.

Javascript

function MissingandRepeating(vec) {
    const n = vec.length; // size of the array
    let repeating = -1, missing = -1;

    // Find the repeating and missing number in the array
    for (let i = 0; i < n; i++) {
        const index = Math.abs(vec[i]) - 1;
        if (vec[index] < 0)
            repeating = Math.abs(vec[i]);
        else
            vec[index] = -vec[index];
    }
    for (let i = 0; i < n; i++) {
        if (vec[i] > 0) {
            missing = i + 1;
            break;
        }
    }
    return [repeating, missing];
}

const vec = [1 , 2, 4 , 5, 5, 6 , 7];
const ans = MissingandRepeating(vec);
console.log(
    "The repeating and missing numbers are: {" + ans[0] + ", " + ans[1] + "}");

Output

The repeating and missing numbers are: {5, 3}

Time Complexity: O(n) , because we are using a nested loop to store the count of elements.

Space Complexity: O(1) , as it is taking constant time.

Finding repeating and missing numbers Using Hashing

We use hashstore the count every element from the 1 to N range. We store the count of every element by traversing the whole array. We will return the repeating number which has a count of 2 in the hash array, and return the missing number which has a count of 0.

Example: This JavaScript code employs a hash array to efficiently find the repeating and missing numbers within an array, providing a concise and straightforward solution.

Javascript

function MissingandRepeatingNumbers(vec) {
    const n = vec.length;
    const hash = new Array(n + 1).fill(0);

    // store count of every element in hash array
    for (let i = 0; i < n; i++) {
        hash[vec[i]]++;
    }

    // Find the repeating and missing number:
    let repeating = -1, missing = -1;
    for (let i = 1; i <= n; i++) {
        if (hash[i] === 2) repeating = i;
        else if (hash[i] === 0) missing = i;

        if (repeating !== -1 && missing !== -1)
            break;
    }
    return [repeating, missing];
}

const vec = [3, 1, 2, 5, 4, 6, 7, 5];
const ans = MissingandRepeatingNumbers(vec);
console.log(
    "The repeating and missing numbers are {" + ans[0] + ", " + ans[1] + "}");

Output

The repeating and missing numbers are {5, 8}

Time Complexity: O(2n), as we are using two loops iterations.

Space Complexity: O(n), as we are using hash array.