1995. Count Special Quadruplets

Problem

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

• nums[a] + nums[b] + nums[c] == nums[d], and

• a < b < c < d

  Example 1:

Input: nums = [1,2,3,6]
Output: 1
Explanation: The only quadruplet that satisfies the requirement is (0, 1, 2, 3) because 1 + 2 + 3 == 6.

Example 2:

Input: nums = [3,3,6,4,5]
Output: 0
Explanation: There are no such quadruplets in [3,3,6,4,5].

Example 3:

Input: nums = [1,1,1,3,5]
Output: 4
Explanation: The 4 quadruplets that satisfy the requirement are:
- (0, 1, 2, 3): 1 + 1 + 1 == 3
- (0, 1, 3, 4): 1 + 1 + 3 == 5
- (0, 2, 3, 4): 1 + 1 + 3 == 5
- (1, 2, 3, 4): 1 + 1 + 3 == 5

  Constraints:

• 4 <= nums.length <= 50

• 1 <= nums[i] <= 100

Solution (Java)

class Solution {
    public int countQuadruplets(int[] nums) {
        int count = 0;
        // max nums value is 100 so two elements sum can be max 200
        int[] m = new int[201];
        for (int i = 1; i < nums.length - 2; i++) {
            for (int j = 0; j < i; j++) {
                // update all possible 2 sums
                m[nums[j] + nums[i]]++;
            }
            for (int j = i + 2; j < nums.length; j++) {
                // fix third element and search for fourth - third in 2 sums as a  + b + c = d == a
                // +  b = d - c
                int diff = nums[j] - nums[i + 1];
                if (diff >= 0) {
                    count += m[diff];
                }
            }
        }
        return count;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).