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Leetcode 1995. Count Special Quadruplets

Problem Statement

Given an array nums of n integers, return the number of special quadruplets (a, b, c, d) that satisfy the following conditions:

  1. 0 <= a < b < c < d < n
  2. nums[a] + nums[b] + nums[c] == nums[d]

Clarifying Questions

  1. Input Range: What is the range of n and the values in nums?
    • Typically for competitive programming, n could go up to a few thousand. The values in nums are generally within the range of standard integer limits.
  2. Distinct Elements: Are all elements in nums distinct?
    • The problem doesn’t specify that the numbers are distinct, so we assume they can have duplicates.
  3. Target Complexity: Is there a preferred time complexity we should aim for?
    • Because of the nature of the problem, a brute-force approach would have a time complexity of O(n^4). However, optimizations should be attempted to reduce it if possible.

Strategy

A brute-force approach would involve iterating over all possible quadruplets, but that results in O(n^4) complexity, which is not efficient for large n.

Optimized Approach

  1. Dictionary Based Storage: Utilize a HashMap to store sums of pairs of indices as we iterate.
  2. Iterate Efficiently: As we choose the fourth element, leverage stored pairs sums to find the required combination efficiently.

We will approach this problem in two main steps:

  1. Use a nested loop to generate sums and store them in a dictionary.
  2. Use another set of loops to look for quadruplets based on precomputed sums.

Pseudo Code Outline

  1. Iterate over c and d in nested loops.
  2. For each c and d, iterate over pairs of indices for sums before c.
  3. Utilize a HashMap to check if the required sum exists.

Code

import java.util.HashMap;
import java.util.Map;

public class Solution {
    public int countQuadruplets(int[] nums) {
        int n = nums.length;
        int count = 0;
        Map<Integer, Integer> sumPairs = new HashMap<>();
        
        // Loop through each element considering it as d where 3 <= d < n
        for (int d = n - 1; d >= 3; d--) {
            // Update the sumPairs map by considering pairs nums[b] + nums[c]
            for (int c = d - 1; c >= 2; c--) {
                int target = nums[d] - nums[c];
                if (sumPairs.containsKey(target)) {
                    count += sumPairs.get(target);
                }
            }
            
            // Update the sumPairs map by considering pairs nums[a] + nums[b] before c
            for (int b = 0; b < d - 2; b++) {
                for (int a = 0; a < b; a++) {
                    int sum = nums[a] + nums[b];
                    sumPairs.put(sum, sumPairs.getOrDefault(sum, 0) + 1);
                }
            }
        }
        
        return count;
    }
}

Time Complexity

Thus, our solution has an overall time complexity of O(n^3), which is a significant improvement over bruteforce methods for large n.

With this approach, large arrays up to the typical competitive programming limits (like 1000) can be handled efficiently within time constraints.

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