leetcode/src/notes/3583_count_special_triplets.md

Count Special Triplets

Problem Number: 3583 Difficulty: Medium Category: Array, Hash Table, Two Pointers LeetCode Link: https://leetcode.com/problems/count-special-triplets/

Problem Description

You are given a 0-indexed integer array nums. A triplet of indices (i, j, k) is a special triplet if the following conditions are met:

• 0 <= i < j < k < nums.length
• nums[i] * 2 == nums[j]
• nums[j] * 2 == nums[k]

Return the number of special triplets.

Example 1:

Input: nums = [1,2,4,8]
Output: 2
Explanation: The special triplets are:
- (0,1,2): nums[0] * 2 == nums[1] and nums[1] * 2 == nums[2]
- (1,2,3): nums[1] * 2 == nums[2] and nums[2] * 2 == nums[3]

Example 2:

Input: nums = [1,2,4,8,16]
Output: 4
Explanation: The special triplets are:
- (0,1,2): nums[0] * 2 == nums[1] and nums[1] * 2 == nums[2]
- (1,2,3): nums[1] * 2 == nums[2] and nums[2] * 2 == nums[3]
- (2,3,4): nums[2] * 2 == nums[3] and nums[3] * 2 == nums[4]
- (0,1,2,3): nums[0] * 2 == nums[1] and nums[1] * 2 == nums[2] and nums[2] * 2 == nums[3]

Constraints:

• 3 <= nums.length <= 10^5
• 1 <= nums[i] <= 10^9

My Approach

I used a Hash Table approach with sliding window to count special triplets efficiently. The key insight is to track left and right counts for each middle element.

Algorithm:

1. Initialize hash tables for left and right counts
2. Pre-populate right count hash table
3. For each middle element (j):
   • Calculate target as nums[j] * 2
   • Get left count (elements before j that equal target)
   • Get right count (elements after j that equal target)
   • Add left_count * right_count to total
   • Update left count hash table
   • Decrement right count hash table

Solution

The solution uses hash tables to efficiently count special triplets. See the implementation in the solution file.

Key Points:

• Uses hash tables for O(1) lookup
• Tracks left and right counts
• Updates counts dynamically
• Handles large numbers with modulo

Time & Space Complexity

Time Complexity: O(n)

• Single pass through array: O(n)
• Hash table operations: O(1) average
• Total: O(n)

Space Complexity: O(n)

• Hash tables for left and right counts: O(n)
• Total: O(n)

Key Insights

1. Hash Table Tracking: Use hash tables to track element frequencies.

2. Sliding Window: Update counts as we move through the array.

3. Triplet Counting: For each middle element, count valid left and right pairs.

4. Dynamic Updates: Update left and right counts efficiently.

5. Modulo Operation: Handle large numbers with modulo arithmetic.

6. Efficient Lookup: O(1) lookup for element frequencies.

Mistakes Made

1. Wrong Counting: Initially might count triplets incorrectly.

2. Complex Logic: Overcomplicating the triplet counting.

3. Update Issues: Not properly updating left and right counts.

4. Inefficient Approach: Using O(n³) approach when hash table suffices.

Related Problems

• 3Sum (Problem 15): Find triplets with target sum
• Two Sum (Problem 1): Find pair with target sum
• Contains Duplicate (Problem 217): Check for duplicates
• Majority Element (Problem 169): Find majority element

Alternative Approaches

1. Brute Force: Check all triplets - O(n³) time
2. Two Pointers: Use two pointers for each middle element - O(n²) time
3. Binary Search: Use binary search for target elements - O(n log n) time

Common Pitfalls

1. Wrong Counting: Not counting triplets correctly.
2. Complex Logic: Overcomplicating the triplet counting.
3. Update Issues: Not properly updating left and right counts.
4. Inefficient Approach: Using O(n³) approach when hash table suffices.
5. Overflow: Not handling large numbers properly.

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Note: This is a hash table problem that demonstrates efficient triplet counting with sliding window technique.