LeetCode in Kotlin

1995. Count Special Quadruplets

Easy

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

class Solution {
    fun countQuadruplets(nums: IntArray): Int {
        var count = 0
        // max nums value is 100 so two elements sum can be max 200
        val m = IntArray(201)
        for (i in 1 until nums.size - 2) {
            for (j in 0 until i) {
                // update all possible 2 sums
                m[nums[j] + nums[i]]++
            }
            for (j in i + 2 until nums.size) {
                // fix third element and search for fourth - third in 2 sums as a  + b + c = d == a
                // +  b = d - c
                val diff = nums[j] - nums[i + 1]
                if (diff >= 0) {
                    count += m[diff]
                }
            }
        }
        return count
    }
}

This site is open source. Improve this page.