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Leetcode 1995. Count Special Quadruplets

Problem Statement

Given a 0-indexed integer array nums, return the number of distinct quadruplets (a, b, c, d) such that:

Clarifying Questions

  1. What is the length range of the nums array?
    • The length of nums can go up to 50 as per the constraints given on LeetCode.
  2. Can the elements of nums be negative?
    • Yes, the elements of nums can be negative, zero, or positive.
  3. Should we consider quadruplets with a, b, c, and d in any order?
    • No, we need to count only distinct combinations where a < b < c < d as specified by the constraints of indices being distinct and having a specific order.

Strategy

To solve the problem, we need an effective way to identify quadruplets that meet the given conditions. Given the maximum size of the array (50), we can use a brute force approach to check all possible quadruplets.

  1. Brute-force Solution:
    • Use four nested loops to iterate over all possible combinations of indices a, b, c, and d.
    • Check if the condition nums[a] + nums[b] + nums[c] == nums[d] is met for every combination.
    • Count the valid combinations.
#include <vector>
using namespace std;

class Solution {
public:
    int countQuadruplets(vector<int>& nums) {
        int n = nums.size();
        int count = 0;
        
        // Brute-force solution with four nested loops
        for (int a = 0; a < n - 3; ++a) {
            for (int b = a + 1; b < n - 2; ++b) {
                for (int c = b + 1; c < n - 1; ++c) {
                    for (int d = c + 1; d < n; ++d) {
                        if (nums[a] + nums[b] + nums[c] == nums[d]) {
                            count++;
                        }
                    }
                }
            }
        }
        
        return count;
    }
};

Time Complexity

The time complexity of the brute-force solution is O(n^4) where n is the length of the array. Given the constraints (n <= 50), this approach is feasible as it will have a maximum of 50^4 = 6,250,000 iterations which is manageable in most scenarios.

Final Thoughts

While this approach works given the constraints, in the event of larger input sizes, the solution would need optimization to reduce the time complexity. For instance, using hash maps to store previously computed sums and improving lookup times could be a next step for optimization.

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