JavaScript Program to find Maximum Distance Between two Occurrences of Same Element in Array

Last Updated : 26 Aug, 2024
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Given an array of repeated elements, the task is to find the maximum distance between two occurrences of an element.

Example:

Input : arr[] = {3, 2, 1, 2, 1, 4, 5, 8, 6, 7, 4, 2}
Output: 10
// maximum distance for 2 is 11-1 = 10 
// maximum distance for 1 is 4-2 = 2 
// maximum distance for 4 is 10-5 = 5

Table of Content

Brute Force

  • Use a dictionary to store element indices.
  • Initialize max distance as -1.
  • Iterate through the array.
  • Update max distance when encountering the same element.
  • Return the maximum distance found.

Example: Below are implementations of the idea.

JavaScript 
function maxDistance(arr) {
    let dict = {};
    let maxD = -1;

    for (let i = 0; i < arr.length; i++) {
        if (dict[arr[i]] !== undefined) {
            maxD = Math.max(maxD, i - dict[arr[i]]);
        } else {
            dict[arr[i]] = i;
        }
    }

    return maxD;
}

let arr = [1, 2, 4, 1, 3, 4, 2, 5, 6, 5];
console.log(
    "Max distance between occurrences of same element: " 
        + maxDistance(arr));

Output
Max distance between occurrences of same element: 5

Time Complexity: O(N^2)

Space Complexity: O(1)

Optimal Solution

  • Use a Map to store the first occurrence index of each element.
  • Iterates through the array, updating the maximum distance for repeated elements.
  • Returns the maximum distance found between occurrences of the same element in the array.

Example: Below are implementations of the idea.

JavaScript 
function maxDistance(arr, n) {
    let map = new Map();
    let maxDist = 0;

    for (let i = 0; i < n; i++) {
        if (!map.has(arr[i]))
            map.set(arr[i], i);
        else
            maxDist = Math.max(maxDist, i - map.get(arr[i]));
    }

    return maxDist;
}


let arr = [1, 2, 4, 1, 3, 4, 2, 5, 6, 5];
console.log(
    "Max distance between occurrences of same element: " 
        + maxDistance(arr,10));

Output
Max distance between occurrences of same element: 5

Time Complexity: O(N) under the assumption that unordered_map’s search and insert operations take O(1) time.

Space Complexity: O(N)

Sorting and Two-Pointer Approach

In this method, we first store the indices of each element in a dictionary. Then, we sort the indices of each element and use a two-pointer approach to find the maximum distance for each element.

Example:

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

Explanation:

  • Maximum distance for 2 is 11 - 1 = 10
  • Maximum distance for 1 is 4 - 2 = 2
  • Maximum distance for 4 is 10 - 5 = 5

Example:

JavaScript 
function maxDistance(arr) {
    const elementIndices = {};

    // Store indices of each element
    for (let i = 0; i < arr.length; i++) {
        if (elementIndices[arr[i]] === undefined) {
            elementIndices[arr[i]] = [];
        }
        elementIndices[arr[i]].push(i);
    }

    let maxDist = -1;

    // Iterate through each element in the dictionary
    for (const key in elementIndices) {
        const indices = elementIndices[key];
        if (indices.length > 1) {
            // Find the max distance using first and last index
            const dist = indices[indices.length - 1] - indices[0];
            maxDist = Math.max(maxDist, dist);
        }
    }

    return maxDist;
}

// Examples
const arr1 = [3, 2, 1, 2, 1, 4, 5, 8, 6, 7, 4, 2];
console.log(maxDistance(arr1)); // Output: 10

const arr2 = [1, 1, 1, 1];
console.log(maxDistance(arr2)); // Output: 3

const arr3 = [1, 2, 3, 4, 5];
console.log(maxDistance(arr3)); // Output: -1 (no repeated elements)

Output
10
3
-1

Prefix Sum with Hash Map

Idea:

  1. Use a Hash Map to store the first occurrence index of each element.
  2. Compute Distances dynamically while iterating through the array.
  3. Update Maximum Distance if a larger distance is found.

Steps:

  1. Initialize a Hash Map: This map will store the first occurrence index of each element.
  2. Iterate Through the Array: For each element:
    • If the element is not in the map, store its index as its first occurrence.
    • If the element is already in the map, calculate the distance from its first occurrence to the current index.
    • Update the maximum distance if the calculated distance is larger.
  3. Return the Maximum Distance.

Example:

JavaScript 
function maxDistance(arr) {
    let map = new Map();
    let maxDist = -1; 

    for (let i = 0; i < arr.length; i++) {
        if (!map.has(arr[i])) {
            map.set(arr[i], i);
        } else {
            const firstIndex = map.get(arr[i]);
            maxDist = Math.max(maxDist, i - firstIndex);
        }
    }

    return maxDist;
}
const arr1 = [3, 2, 1, 2, 1, 4, 5, 8, 6, 7, 4, 2];
console.log(maxDistance(arr1));

const arr2 = [1, 1, 1, 1];
console.log(maxDistance(arr2)); 

const arr3 = [1, 2, 3, 4, 5];
console.log(maxDistance(arr3));

Output
10
3
-1

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