How do you solve Count Special Triplets on LeetCode?

Count Special Triplets (LeetCode #3583) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using frequency maps allows efficient counting of valid triplets.

What is the time complexity of Count Special Triplets?

The optimal solution for Count Special Triplets runs in O(n) time and uses O(n) space. We traverse the array a couple of times, and use maps to store frequencies, leading to linear complexity.

What is the brute force solution for Count Special Triplets?

The brute force approach for Count Special Triplets is Brute Force, with O(n²) time complexity and O(1) space complexity. Check all possible triplets in the array to see if they meet the special triplet condition. This is straightforward but inefficient for large arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Special Triplets?

Count Special Triplets uses the following concepts: Array, Hash Table, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Count Special Triplets?

Explain your thought process clearly. Consider edge cases, like empty arrays or arrays with all identical elements. Practice writing clean and efficient code.

What are common mistakes when solving Count Special Triplets?

Not handling the modulo operation correctly. Forgetting to check the conditions for valid indices.

#3583

Count Special Triplets

Medium

ArrayHash TableCountingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Use frequency maps to count occurrences of numbers before and after the current index, allowing efficient calculation of valid triplets.

⚙️

Algorithm

3 steps

1Step 1: Create two frequency arrays: freqPrev for counts before index j and freqNext for counts after index j.
2Step 2: For each index j, calculate the number of valid i's and k's using freqPrev and freqNext.
3Step 3: Sum the contributions for each j to get the total count of special triplets.

solution.py17 lines

1def countTriplets(nums):
2    MOD = 10**9 + 7
3    n = len(nums)
4    freqPrev = {}
5    freqNext = {}
6    for num in nums:
7        freqNext[num] = freqNext.get(num, 0) + 1
8    count = 0
9    for j in range(n):
10        freqNext[nums[j]] -= 1
11        if j > 0:
12            freqPrev[nums[j-1]] = freqPrev.get(nums[j-1], 0) + 1
13        if nums[j] % 2 == 0:
14            target = nums[j] // 2
15            count += freqPrev.get(target, 0) * freqNext.get(nums[j], 0)
16            count %= MOD
17    return count

ℹ

Complexity note: We traverse the array a couple of times, and use maps to store frequencies, leading to linear complexity.

1Using frequency maps allows efficient counting of valid triplets.
2The condition for special triplets relies on simple arithmetic relationships.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.