JavaScript Program for the Minimum Index of a Repeating Element in an Array

Given an array of integers arr[], The task is to find the index of the first repeating element in it i.e. the element that occurs more than once and whose index of the first occurrence is the smallest. In this article, we will find the minimum index of a repeating element in an array in JavaScript.

Examples:

Input: arr[] = {10, 5, 3, 4, 3, 5, 6}
Output: 5 
Explanation: 5 is the first element that repeats

Input: arr[] = {6, 10, 5, 4, 9, 120, 4, 6, 10}
Output: 6 
Explanation: 6 is the first element that repeats

Approach 1: Brute force approach

Here we will use nested loops by using for to compare each element with every other element in the array to return duplicates.

Syntax:

for (let i = 0; i < arr.length; i++) {
        for (let j = i + 1; j < arr.length; j++) {
            if (arr[i] === arr[j]) {
                minIndex = Math.min(minIndex, i);
            }
        }
    }

Example: This example shows the use of the above-explained approach.

Javascript

function findMinIndexBruteForce(arr) {
    let minIndex = Infinity;

    for (let i = 0; i < arr.length; i++) {
        for (let j = i + 1;
                j < arr.length; j++) {
            if (arr[i] === arr[j]) {
                minIndex =
                    Math.min(minIndex, i);
            }
        }
    }

    return minIndex !== Infinity
        ? minIndex : -1;
}

const arr1 = [3, 2, 1, 2, 4, 3];
console.log(findMinIndexBruteForce(arr1));

Output

0

Approach 2: Using an Object (Hash Map)

Here we will use an object to store the elements you have seen so far as keys and their corresponding indices as values. When we encounter a repeating element, check if its index is smaller than the current minimum index.

Syntax:

const seen = {};

Javascript

function findMinIndexWithObject(arr) {
    const seen = {};

    let minIndex = Infinity;

    for (let i = 0; i < arr.length; i++) {
        if (seen[arr[i]] !== undefined) {
            minIndex = Math
                .min(minIndex, seen[arr[i]]);
        } else {
            seen[arr[i]] = i;
        }
    }

    return minIndex !== Infinity
        ? minIndex : -1;
}

const arr2 = [3, 2, 1, 2, 4, 3];
console.log(findMinIndexWithObject(arr2));

Output

0

Approach 3: Using a Set

Here we will use a set to keep track of the elements you have seen so far. When we encounter a repeating element, check if it exists in the set. If it does, compare its index with the current minimum index.

Syntax:

const seen = new Set();

Javascript

function findMinIndexWithSet(arr) {

    // Using a Map to store the indices of elements
    const seen = new Map();
    let minIndex = -1;

    for (let i = 0; i < arr.length; i++) {
        if (seen.has(arr[i])) {

            // Check if we found a repeating element
            const firstIndex = seen.get(arr[i]);
            if (minIndex === -1 || firstIndex < minIndex) {

                // If it's the first repeating element or
                //it's closer to the beginning of the array
                minIndex = firstIndex;
            }
        } else {
            seen.set(arr[i], i);
        }
    }

    return minIndex;
}

const arr3 = [3, 2, 1, 2, 4, 3];
console.log(findMinIndexWithSet(arr3));

Output

0