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3885. Count Special Triplets

Array Hash Table Counting

Problem - 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

Solutions

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class Solution:
    def specialTriplets(self, nums: List[int]) -> int:
        left = Counter()
        right = Counter(nums)
        result = 0
        mod = 10**9 + 7

        for num in nums:
            right[num] -= 1
            result = (result + left[num * 2] * right[num * 2] % mod) % mod
            left[num] += 1

        return result

Submission Stats:

  • Runtime: 883 ms (18.91%)
  • Memory: 41.4 MB (77.93%)