LeetCode 2670: Find the Distinct Difference Array Solution

Master LeetCode problem 2670 (Find the Distinct Difference Array), a easy 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.

2670. Find the Distinct Difference Array

Easy

Array Hash Table

Problem Explanation

Explanation

To solve this problem, we need to iterate through the array from left to right and calculate the distinct difference for each index. For each index i, we count the number of distinct elements in the prefix nums[0, ..., i] and the number of distinct elements in the suffix nums[i + 1, ..., n - 1]. Then, we subtract the count of distinct elements in the suffix from the count of distinct elements in the prefix to get the distinct difference for that index.

We can use a set data structure to keep track of distinct elements in the prefix and suffix as we iterate through the array. By adding elements to the set as we move forward in the array and removing elements as we move backward, we can efficiently calculate the distinct elements in the prefix and suffix.

Time complexity: O(n) where n is the length of the input array nums. Space complexity: O(n) for the sets used to store distinct elements in the prefix and suffix.

Solution Code

import java.util.*;

class Solution {
    public int[] distinctDifferenceArray(int[] nums) {
        int n = nums.length;
        int[] result = new int[n];
        
        for (int i = 0; i < n; i++) {
            Set<Integer> prefixSet = new HashSet<>();
            Set<Integer> suffixSet = new HashSet<>();
            int prefixCount = 0, suffixCount = 0;
            
            for (int j = 0; j < i; j++) {
                if (!prefixSet.contains(nums[j])) {
                    prefixSet.add(nums[j]);
                    prefixCount++;
                }
            }
            
            for (int j = i + 1; j < n; j++) {
                if (!suffixSet.contains(nums[j])) {
                    suffixSet.add(nums[j]);
                    suffixCount++;
                }
            }
            
            result[i] = prefixCount - suffixCount;
        }
        
        return result;
    }
}

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