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], anda < 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 <= 501 <= 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
}
}