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LeetCode solutions

3885. Count Special Triplets

My accepted Python solution to LeetCode problem 3885, Count Special Triplets, running in 557ms.

  • Difficulty: Medium
  • Python
  • Runtime 557ms
  • Memory 39.3MB

Read the problem on LeetCode View on GitHub

The problem statement is LeetCode’s and stays on their site. What follows is my accepted solution.

Python

Accepted on LeetCode — runtime 557ms, memory 39.3MB, accepted 2025-12-30.

python
class Solution:
    def specialTriplets(self, nums: List[int]) -> int:
        MOD = 10**9 + 7
        n = len(nums)
        
        # For each j, count i's before it where nums[i] = nums[j]*2
        # and k's after it where nums[k] = nums[j]*2
        
        # First pass: count occurrences of each value for suffix
        suffix_count = {}
        for num in nums:
            suffix_count[num] = suffix_count.get(num, 0) + 1
        
        prefix_count = {}
        result = 0
        
        for j in range(n):
            target = nums[j] * 2
            
            # Remove current element from suffix
            suffix_count[nums[j]] -= 1
            
            # Count triplets with j as middle element
            left_count = prefix_count.get(target, 0)
            right_count = suffix_count.get(target, 0)
            
            result = (result + left_count * right_count) % MOD
            
            # Add current element to prefix
            prefix_count[nums[j]] = prefix_count.get(nums[j], 0) + 1
        
        return result

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