count-special-triplets

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