LeetCode 1995: Count Special Quadruplets Solution

Master LeetCode problem 1995 (Count Special Quadruplets), 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.

1995. Count Special Quadruplets

Easy

Array Hash Table Enumeration

Problem Explanation

Explanation

To solve this problem, we can use a brute force approach where we consider all possible quadruplets of indices (a, b, c, d) and check if the sum of corresponding elements in the array equals the value at index d. We need to ensure that a < b < c < d for each quadruplet. By keeping track of the count of valid quadruplets, we can return the final answer.

Algorithm:
1. Initialize a variable count to store the number of valid quadruplets.
2. Iterate over all possible quadruplets of indices (a, b, c, d) where a < b < c < d.
3. Check if nums[a] + nums[b] + nums[c] == nums[d]. If true, increment count.
4. Return the final value of count.
Time Complexity: O(n^4) where n is the length of the input array nums.
Space Complexity: O(1)

Solution Code

class Solution {
    public int countQuadruplets(int[] nums) {
        int count = 0;
        int n = nums.length;
        
        for (int a = 0; a < n; a++) {
            for (int b = a + 1; b < n; b++) {
                for (int c = b + 1; c < n; c++) {
                    for (int d = c + 1; d < n; d++) {
                        if (nums[a] + nums[b] + nums[c] == nums[d]) {
                            count++;
                        }
                    }
                }
            }
        }
        
        return count;
    }
}

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