#3404

Medium

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Count Special Subsequences

Count the number of special subsequences in an array of positive integers, focusing on efficient array scanning and hash lookup.

array hash table math enumeration

Problem Statement

You are given an array of positive integers nums, and you need to count how many special subsequences of length 4 can be formed. A special subsequence is defined by four indices (p, q, r, s), where p < q < r < s, and the values in the subsequence must satisfy the condition nums[p] / nums[q] == nums[s] / nums[r].

Return the number of different special subsequences in nums, with the constraint that the length of nums is between 7 and 1000 and each element is a positive integer between 1 and 1000.

Examples

Example 1

Input: nums = [1,2,3,4,3,6,1]

Output: 1

There is one special subsequence in nums .

Example 2

Input: nums = [3,4,3,4,3,4,3,4]

Output: 3

There are three special subsequences in nums .

Constraints

7 <= nums.length <= 1000
1 <= nums[i] <= 1000

Solution Approach

Array Scanning with Hash Lookup

This solution involves scanning the array and leveraging a hash table to store counts of possible ratios formed by pairs of elements. By using GCD (Greatest Common Divisor), we can efficiently check for pairs where nums[p] / nums[q] equals nums[s] / nums[r].

Optimized Counting with GCD

Since calculating ratios directly could be inefficient, we use GCD to simplify fractions and store these ratios in a hash table. This allows us to quickly identify matching ratios between different subsequences.

Efficient Enumeration

The solution involves enumerating all possible subsequences of length 4 by iterating through pairs of indices (p, q) and (r, s). By keeping track of ratios during this process, we can count valid subsequences efficiently without brute-force checking.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time and space complexities depend on the final approach. Using GCD for ratio comparison reduces the overhead of direct fraction comparison, while hash tables allow for efficient lookups. The solution may involve O(n^2) for scanning pairs and efficient GCD ratio comparisons, where n is the length of the array.

What Interviewers Usually Probe

Evaluating the candidate's ability to optimize ratio calculations using GCD.
Assessing the approach to using hash tables for efficient subsequence counting.
Looking for a clean solution that balances time complexity with correctness.

Common Pitfalls or Variants

Common pitfalls

Overcomplicating the ratio calculation, missing the optimization using GCD.
Inefficient brute-force enumeration of all subsequences.
Incorrectly handling the array bounds and missing subsequences due to index errors.

Follow-up variants

What if the array length was increased significantly? The approach should scale accordingly.
How would the solution change if the array contained zeros or negative numbers?
What if subsequences of length other than 4 were required?

FAQ

What is the main pattern in 'Count Special Subsequences'?

The main pattern is array scanning with hash lookups, utilizing GCD to simplify ratio calculations and efficiently count valid subsequences.

What does a 'special subsequence' mean in this problem?

A special subsequence consists of four indices (p, q, r, s) such that nums[p] / nums[q] == nums[s] / nums[r], where p < q < r < s.

What is the time complexity of this problem?

The time complexity depends on the final approach, but it typically involves O(n^2) due to scanning pairs and efficient ratio checking using GCD.

How can I improve the efficiency of my solution?

Using GCD to reduce fraction complexity and hash tables for efficient lookup of ratios will significantly improve the efficiency of your solution.

Are there any edge cases I should be aware of?

Yes, make sure to handle cases where no valid subsequences exist and ensure that you do not miss any subsequences due to incorrect index handling.

terminal

Solution

Solution 1

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class Solution:
    def numberOfSubsequences(self, nums: List[int]) -> int:
        n = len(nums)
        cnt = defaultdict(int)
        for r in range(4, n - 2):
            c = nums[r]
            for s in range(r + 2, n):
                d = nums[s]
                g = gcd(c, d)
                cnt[(d // g, c // g)] += 1
        ans = 0
        for q in range(2, n - 4):
            b = nums[q]
            for p in range(q - 1):
                a = nums[p]
                g = gcd(a, b)
                ans += cnt[(a // g, b // g)]
            c = nums[q + 2]
            for s in range(q + 4, n):
                d = nums[s]
                g = gcd(c, d)
                cnt[(d // g, c // g)] -= 1
        return ans

Continue Practicing

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