Formatted question description: https://leetcode.ca/all/1995.html

Easy

## Description

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

Loop over nums and count the number of quadruplets that satisfies the two conditions.

class Solution {
int count = 0;
int length = nums.length;
for (int i = 0; i < length - 3; i++) {
for (int j = i + 1; j < length - 2; j++) {
for (int k = j + 1; k < length - 1; k++) {
int sum = nums[i] + nums[j] + nums[k];
for (int m = k + 1; m < length; m++) {
if (sum == nums[m])
count++;
}
}
}
}
return count;
}
}