LeetCode in Kotlin

3583. Count Special Triplets

Medium

You are given an integer array nums.

A special triplet is defined as a triplet of indices (i, j, k) such that:

• 0 <= i < j < k < n, where n = nums.length
• nums[i] == nums[j] * 2
• nums[k] == nums[j] * 2

Return the total number of special triplets in the array.

Since the answer may be large, return it modulo 109 + 7.

Example 1:

Input: nums = [6,3,6]

Output: 1

Explanation:

The only special triplet is (i, j, k) = (0, 1, 2), where:

• nums[0] = 6, nums[1] = 3, nums[2] = 6
• nums[0] = nums[1] * 2 = 3 * 2 = 6
• nums[2] = nums[1] * 2 = 3 * 2 = 6

Example 2:

Input: nums = [0,1,0,0]

Output: 1

Explanation:

The only special triplet is (i, j, k) = (0, 2, 3), where:

• nums[0] = 0, nums[2] = 0, nums[3] = 0
• nums[0] = nums[2] * 2 = 0 * 2 = 0
• nums[3] = nums[2] * 2 = 0 * 2 = 0

Example 3:

Input: nums = [8,4,2,8,4]

Output: 2

Explanation:

There are exactly two special triplets:

• (i, j, k) = (0, 1, 3)
  • nums[0] = 8, nums[1] = 4, nums[3] = 8
  • nums[0] = nums[1] * 2 = 4 * 2 = 8
  • nums[3] = nums[1] * 2 = 4 * 2 = 8
• (i, j, k) = (1, 2, 4)
  • nums[1] = 4, nums[2] = 2, nums[4] = 4
  • nums[1] = nums[2] * 2 = 2 * 2 = 4
  • nums[4] = nums[2] * 2 = 2 * 2 = 4

Constraints:

• 3 <= n == nums.length <= 105
• 0 <= nums[i] <= 105

Solution

class Solution {
    fun specialTriplets(nums: IntArray): Int {
        val mod = 1_000_000_007
        var res = 0
        val left = mutableMapOf<Int, Int>()
        val right = mutableMapOf<Int, Int>()
        for (num in nums) {
            right[num] = right.getOrDefault(num, 0) + 1
        }
        for (num in nums) {
            right[num] = right[num]!! - 1
            val ci = left.getOrDefault(num * 2, 0)
            val ck = right.getOrDefault(num * 2, 0)
            res = ((res + 1L * ci * ck) % mod).toInt()
            left[num] = left.getOrDefault(num, 0) + 1
        }
        return res
    }
}

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