We use cookies (including Google cookies) to personalize ads and analyze traffic. By continuing to use our site, you accept our Privacy Policy.

SimplyLeet
TagsDisclaimersPrivacyTermsContactAbout Us

Count Special Subsequences

Difficulty: Medium

Problem Description

You are given an array nums consisting of positive integers. A special subsequence is defined as a subsequence of length 4, represented by indices (p, q, r, s), where p < q < r < s. This subsequence must satisfy the following conditions:

  • nums[p] * nums[r] == nums[q] * nums[s]
  • There must be at least one element between each pair of indices. In other words, q - p > 1, r - q > 1 and s - r > 1.

Return the number of different special subsequences in nums.

Key Insights

  • The problem requires finding subsequences of length 4 that meet specific multiplication criteria.
  • The indices must be spaced apart, enforcing a minimum distance between them.
  • A brute force approach would be inefficient, especially with the maximum input size; thus, an optimized approach using hashing or indexed storage may be necessary.
  • We need to keep track of potential pairs that may form valid subsequences to efficiently count them.

Space and Time Complexity

Time Complexity: O(n^3) in the worst case, where n is the length of the array, due to three nested loops. Space Complexity: O(n) for storing intermediate results if using hash maps or lists to track pairs.

Solution

To solve the problem, we can use a nested loop approach where we iterate through the array. For each potential pair of indices (p, r), we calculate the product of nums[p] and nums[r]. We then look for valid pairs (q, s) that satisfy the condition nums[q] * nums[s] = nums[p] * nums[r]. We must ensure that the indices respect the spacing condition.

We can use dictionaries to efficiently store and retrieve the counts of valid product pairs as we iterate through the array.

Code Solutions

def count_special_subsequences(nums):
    count = 0
    n = len(nums)
    
    # Dictionary to hold counts of products
    product_map = {}
    
    # Iterate over possible p and r indices
    for p in range(n - 3):
        for r in range(p + 2, n - 1):
            product = nums[p] * nums[r]
            if product not in product_map:
                product_map[product] = 0
            
            # Iterate over q for valid pairs
            for q in range(p + 1, r):
                if nums[q] != 0 and product % nums[q] == 0:  # Check division for s
                    s_value = product // nums[q]
                    # Check for valid s
                    for s in range(q + 1, n):
                        if nums[s] == s_value:
                            count += 1
                            
            # Update the count of this product
            product_map[product] += 1
    
    return count
← Back to All Questions