How do you solve Count Special Subsequences on LeetCode?

Count Special Subsequences (LeetCode #3404) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n) space complexity. Key insight: Understanding the conditions for valid subsequences is crucial.

What is the brute force solution for Count Special Subsequences?

The brute force approach for Count Special Subsequences is Brute Force, with O(n^4) time complexity and O(1) space complexity. The brute force approach involves checking all possible combinations of indices (p, q, r, s) to see if they satisfy the special subsequence condition. This is straightforward but inefficient due to the nested loops. The optimal Optimal Solution approach improves this to O(n²).

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

Count Special Subsequences uses the following concepts: Array, Hash Table, Math, Enumeration. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Count Special Subsequences?

Always clarify the problem constraints and requirements before diving into coding. Think about edge cases, such as the minimum size of the input array. Discuss your thought process with the interviewer to showcase your problem-solving approach.

What are common mistakes when solving Count Special Subsequences?

Not considering the index constraints properly (p < q < r < s). Overlooking the need for at least one element between indices.

#3404

Count Special Subsequences

Medium

ArrayHash TableMathEnumerationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n^4)	O(n²)
Space	O(1)	O(n)

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Intuition

Time O(n²)Space O(n)

Instead of checking all combinations, we can use a hashmap to store the ratios of products formed by pairs (p, r) and (q, s). This allows us to efficiently count valid pairs that satisfy the condition.

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Algorithm

3 steps

1Step 1: Create a hashmap to store counts of products formed by pairs (p, r).
2Step 2: Iterate through all pairs (p, r) and store the product in the hashmap with the ratio as the key.
3Step 3: For each valid (q, s) pair, check if the product exists in the hashmap and add to the count.

solution.py16 lines

1from collections import defaultdict
2
3def countSpecialSubsequences(nums):
4    count = 0
5    n = len(nums)
6    product_map = defaultdict(int)
7
8    for p in range(n - 3):
9        for r in range(p + 2, n):
10            product_map[nums[p] * nums[r]] += 1
11
12    for q in range(1, n - 2):
13        for s in range(q + 2, n):
14            count += product_map[nums[q] * nums[s]]
15
16    return count

ℹ

Complexity note: The time complexity is O(n²) because we iterate through pairs for both (p, r) and (q, s), but we use a hashmap to store intermediate results, which reduces redundant calculations.

1Understanding the conditions for valid subsequences is crucial.
2Using a hashmap can significantly reduce the number of checks needed.

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