3583. Count Special Triplets

Problem Link

Description

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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)

  <ul>
  	<li><code>nums[0] = 8</code>, <code>nums[1] = 4</code>, <code>nums[3] = 8</code></li>
  	<li><code>nums[0] = nums[1] * 2 = 4 * 2 = 8</code></li>
  	<li><code>nums[3] = nums[1] * 2 = 4 * 2 = 8</code></li>
  </ul>
  </li>
  <li><code>(i, j, k) = (1, 2, 4)</code>
  <ul>
  	<li><code>nums[1] = 4</code>, <code>nums[2] = 2</code>, <code>nums[4] = 4</code></li>
  	<li><code>nums[1] = nums[2] * 2 = 2 * 2 = 4</code></li>
  	<li><code>nums[4] = nums[2] * 2 = 2 * 2 = 4</code></li>
  </ul>
  </li>
  

 

Constraints:

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

Solution

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Python3

class Solution:
    def specialTriplets(self, nums: List[int]) -> int:
        N = len(nums)
        MOD = 10 ** 9 + 7
        res = 0
        left = Counter()
        right = Counter(nums)
 
        for x in nums:
            right[x] -= 1
 
            t = x * 2
            res += (left[t] * right[t])
            res %= MOD
            
            left[x] += 1
            
        return res