Is Count Special Triplets easy or hard?

Count Special Triplets is generally classified as a Medium problem. The difficulty comes from recognizing that fixing the middle element allows you to convert a triple nested search into a counting problem with hash tables. Once that insight is clear, the implementation becomes straightforward.

Count Special Triplets Python/Java solution

Implement the optimized approach by iterating through the array and maintaining two hash maps: one for elements already seen and one for elements remaining. For each middle index, compute the required partner values and use map lookups to count valid combinations. This logic translates directly to Python dictionaries, Java HashMap, C++ unordered_map, and similar structures.

How to solve Count Special Triplets in O(n)?

Iterate through the array and treat each element as the middle index j. Maintain a hash map for values to the left and another for values to the right. For the current middle value, determine which values must appear on both sides to form a valid triplet and multiply their frequencies. Update the maps as you move forward to maintain correct counts.

What is the best approach for Count Special Triplets?

The most efficient approach enumerates the middle index j and uses hash tables to track frequencies of numbers on the left and right of j. For each middle value, compute the required partner values that satisfy the special relation and use constant-time hash lookups to count matches. This reduces the complexity from O(n^3) brute force to O(n) time with O(n) extra space.

Is Count Special Triplets asked at Google/Amazon/Meta?

Problems based on counting triplets with hash maps and index ordering frequently appear in interviews at companies like Google, Amazon, and Meta. Variations typically test array traversal, frequency maps, and recognizing patterns where fixing one element simplifies the search space.

What data structure is used in Count Special Triplets?

The optimized solution relies on a hash table (hash map) to maintain frequency counts of numbers on the left and right sides of the current middle index. Arrays are used for traversal, while the hash table enables constant-time lookups for matching values required to form valid triplets.

What is the time complexity of Count Special Triplets?

The optimal solution runs in O(n) time using a middle enumeration strategy with hash maps for frequency counting. Each element is processed once as the middle, while lookups for matching left and right values take O(1) on average. Space complexity is O(n) for storing frequency counts.

Description Hints 2 Solution Video 15:16

3583. Count Special Triplets

Medium47.1% acceptance3 companies

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

Asked at 3 companies

Bloomberg

Google Meta

Topics

Hints open one at a time

Hint 1

Use frequency arrays or maps, e.g. freqPrev and freqNext—to track how many times each value appears before and after the current index.

Hint 2

For each index j in the triplet (i,j,k), compute its contribution to the answer using your freqPrev and freqNext counts.

15:16

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1class Solution:
2    def specialTriplets(self, nums: List[int]) -> int:
3        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2Case 3

Input

[6,3,6]

Input

[0,1,0,0]

Input

[8,4,2,8,4]

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