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Leetcode 2670. Find the Distinct Difference Array

Problem Statement

You are given a 0-indexed array nums of length n. The distinct difference array of nums is an array diff of length n such that diff[i] is equal to the number of distinct elements in the suffix nums[i+1, n-1] minus the number of distinct elements in the prefix nums[0, i].

Return the distinct difference array of nums.

Clarifying Questions

  1. Input Size: What is the maximum length of the input array? (This helps in understanding the potential time complexity limitations.)
  2. Element Range: Are there any constraints on the values within the array? (This could affect data structures we might choose, like if values are within a certain range.)
  3. Examples: Can you provide some example input and output for better understanding?

Strategy

To solve this problem, we’ll break it down into several steps:

  1. Identify Prefix Distinct Counts: Use a set to track distinct elements from left (prefix).
  2. Identify Suffix Distinct Counts: Use another set to track distinct elements from right (suffix).
  3. Construct Result Array: Use the information gathered in the previous steps to compute the result for each index.

Detailed Steps and Code

  1. Prefix Distinct Count: Traverse the array from left to right updating a set and storing the size at each index.
  2. Suffix Distinct Count: Traverse the array from right to left in a similar manner.
  3. Compute Distinct Difference Array: Use the prefix and suffix arrays to compute the result array where each element is the difference between distinct counts in the suffix and prefix up to that index.

Java Code Implementation

import java.util.*;

public class DistinctDifferenceArray {
    public static int[] distinctDifferenceArray(int[] nums) {
        int n = nums.length;
        int[] prefixDistinctCount = new int[n];
        int[] suffixDistinctCount = new int[n];
        int[] diff = new int[n];
        
        // Set to track distinct elements in prefix
        Set<Integer> prefixSet = new HashSet<>();
        for (int i = 0; i < n; i++) {
            prefixSet.add(nums[i]);
            prefixDistinctCount[i] = prefixSet.size();
        }
        
        // Set to track distinct elements in suffix
        Set<Integer> suffixSet = new HashSet<>();
        for (int i = n - 1; i >= 0; i--) {
            suffixSet.add(nums[i]);
            suffixDistinctCount[i] = suffixSet.size();
        }
        
        // Compute the diff array
        for (int i = 0; i < n - 1; i++) {
            diff[i] = suffixDistinctCount[i + 1] - prefixDistinctCount[i];
        }
        // The last element remains as -prefixDistinctCount[n-1] as suffix beyond this is EMPTY
        diff[n - 1] = -prefixDistinctCount[n - 1];
        
        return diff;
    }
    
    public static void main(String[] args) {
        int[] nums = {5, 2, 5, 2, 1};
        System.out.println(Arrays.toString(distinctDifferenceArray(nums))); 
        // Example output for given input []
    }
}

Time Complexity

This approach ensures that we efficiently calculate the distinct differences with minimal passes through the array. The use of sets ensures that we are accurately tracking distinct elements.

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