LeetCode 2121: Intervals Between Identical Elements Solution

Master LeetCode problem 2121 (Intervals Between Identical Elements), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2121. Intervals Between Identical Elements

Problem Explanation

Explanation

To solve this problem, we can iterate through the array and keep track of the indices of each unique element in a HashMap. Then, for each element in the array, we calculate the sum of intervals between the current element and all other occurrences of the same element using the indices stored in the HashMap.

Algorithm

  1. Initialize a HashMap to store the indices of each unique element.
  2. Iterate through the array and store the indices of each element in the HashMap.
  3. For each element in the array, calculate the sum of intervals by iterating through the indices of the same element in the HashMap.
  4. Return the array containing the sum of intervals for each element.

Time Complexity

The time complexity of this solution is O(n), where n is the number of elements in the input array.

Space Complexity

The space complexity of this solution is O(n) to store the indices of each unique element in the HashMap.

Solution Code

import java.util.*;

class Solution {
    public int[] intervalBetweenIdenticalElements(int[] arr) {
        Map<Integer, List<Integer>> indices = new HashMap<>();
        
        for (int i = 0; i < arr.length; i++) {
            indices.computeIfAbsent(arr[i], k -> new ArrayList<>()).add(i);
        }
        
        int[] result = new int[arr.length];
        
        for (int i = 0; i < arr.length; i++) {
            List<Integer> idxList = indices.get(arr[i]);
            int sum = 0;
            
            for (int idx : idxList) {
                sum += Math.abs(i - idx);
            }
            
            result[i] = sum;
        }
        
        return result;
    }
}

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