LeetCode 3404 - Count Special Subsequences

This problem asks us to count the number of special subsequences of length four in a given array nums of positive integers. A special subsequence (p, q, r, s) must satisfy two requirements.

leetcodemediumarrayhash-tablemathenumeration

LeetCode Problem 3404

Difficulty: 🟡 Medium
Topics: Array, Hash Table, Math, Enumeration

Solution

Problem Understanding

This problem asks us to count the number of special subsequences of length four in a given array nums of positive integers. A special subsequence (p, q, r, s) must satisfy two requirements. First, the product of the first and third elements must equal the product of the second and fourth elements: nums[p] * nums[r] == nums[q] * nums[s]. Second, each selected index must be separated by at least one element from the previous index: q - p > 1, r - q > 1, and s - r > 1.

The input is a list of integers with length between 7 and 1000, and each element is between 1 and 1000. The output is a single integer: the number of valid special subsequences. The constraints indicate that a brute-force O(n^4) solution will be too slow, and the problem size is small enough that an O(n^2) or O(n^3) approach may be feasible with some optimization.

Important edge cases include arrays with repeated numbers, arrays where no four-element subsequence satisfies the product condition, and arrays with the minimum length of 7.

Approaches

A naive brute-force solution would iterate over all possible quadruples (p, q, r, s) that satisfy the separation constraints. For each quadruple, we would check if the product condition holds. This is correct because it explicitly checks every candidate subsequence. However, this is extremely inefficient because the number of quadruples grows roughly as O(n^4), which is impractical for n up to 1000.

The key insight for a more efficient solution is to enumerate pairs (p, r) and (q, s) separately and use a hash map to count potential products. By fixing (p, r) pairs and recording their products, we can later iterate over (q, s) pairs and quickly find matches in the hash map. This reduces the number of operations dramatically because we are effectively reducing the four-nested-loop problem into a combination of two double loops and a hash map lookup. This approach leverages the mathematical property nums[p] * nums[r] == nums[q] * nums[s] and uses additional space to store intermediate counts.

Approach Time Complexity Space Complexity Notes
Brute Force O(n^4) O(1) Check all quadruples explicitly; correct but slow
Hash Map Pair Counting O(n^2) O(n^2) Count products of (p, r) and match with (q, s) efficiently

Algorithm Walkthrough

  1. Initialize an empty hash map pair_product_counts. This map will store the frequency of products formed by (p, r) pairs with r - p > 1.
  2. Iterate through all valid (p, r) pairs such that r - p > 1. For each pair, calculate nums[p] * nums[r] and increment its count in pair_product_counts. This precomputes all possible "left-side" products.
  3. Initialize a counter total_count to zero. This will accumulate the number of special subsequences.
  4. Iterate through all valid (q, s) pairs such that s - q > 1. For each pair, compute the product nums[q] * nums[s] and check if it exists in pair_product_counts. If it does, increment total_count by the count stored in the map. This counts all subsequences where (p, r) product matches (q, s) product while respecting the separation constraint.
  5. Return total_count.

Why it works: By storing all (p, r) products first, we ensure that any matching (q, s) pair counted later will have indices that satisfy the gap constraints automatically because we only consider indices separated by at least one element. The hash map guarantees that we count all valid combinations efficiently without missing any subsequences or double-counting.

Python Solution

from typing import List
from collections import defaultdict

class Solution:
    def numberOfSubsequences(self, nums: List[int]) -> int:
        n = len(nums)
        pair_product_counts = defaultdict(int)

        # Count products for (p, r) pairs
        for p in range(n):
            for r in range(p + 2, n):
                product = nums[p] * nums[r]
                pair_product_counts[product] += 1

        total_count = 0
        # Count matching products for (q, s) pairs
        for q in range(1, n - 2):
            for s in range(q + 2, n):
                product = nums[q] * nums[s]
                total_count += pair_product_counts.get(product, 0)

        return total_count

Implementation explanation: The first nested loop enumerates all (p, r) pairs with r - p > 1 and counts products in pair_product_counts. The second nested loop enumerates (q, s) pairs with s - q > 1 and checks if the product exists in pair_product_counts. The result accumulates in total_count and is returned. Using defaultdict(int) simplifies incrementing counts and handling missing keys.

Go Solution

package main

func numberOfSubsequences(nums []int) int64 {
    n := len(nums)
    pairProductCounts := make(map[int]int64)

    // Count products for (p, r) pairs
    for p := 0; p < n; p++ {
        for r := p + 2; r < n; r++ {
            product := nums[p] * nums[r]
            pairProductCounts[product]++
        }
    }

    var totalCount int64 = 0
    // Count matching products for (q, s) pairs
    for q := 1; q < n-2; q++ {
        for s := q + 2; s < n; s++ {
            product := nums[q] * nums[s]
            totalCount += pairProductCounts[int(product)]
        }
    }

    return totalCount
}

Go-specific notes: We use a map[int]int64 to handle potentially large counts without overflow. The iteration indices and separation constraints are handled identically to Python. Go requires explicit type conversions when indexing maps, but otherwise the logic mirrors the Python version.

Worked Examples

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

First, count (p, r) products:

p r nums[p]*nums[r]
0 2 1*3 = 3
0 3 1*4 = 4
0 4 1*3 = 3
0 5 1*6 = 6
0 6 1*1 = 1
1 3 2*4 = 8
1 4 2*3 = 6
... ... ...

Count products for (q, s):

q s nums[q]*nums[s] Matches?
2 4 3*3 = 9 No
2 6 3*1 = 3 Yes, increment total_count

Finally, total_count = 1.

Example 2: nums = [3,4,3,4,3,4,3,4]

Following the same procedure, we find three matches:

  • (0,2,4,6) → 9 = 9
  • (1,3,5,7) → 16 = 16
  • (0,2,5,7) → 12 = 12

Complexity Analysis

Measure Complexity Explanation
Time O(n^2) Two nested loops of size n for (p,r) and (q,s) pairs
Space O(n^2) Hash map stores up to O(n^2) different products

The algorithm reduces the naive O(n^4) brute-force check to O(n^2) by using a hash map for fast lookups, which is feasible for n up to 1000.

Test Cases

# Example cases
assert Solution().numberOfSubsequences([1,2,3,4,3,6,1]) == 1
assert Solution().numberOfSubsequences([3,4,3,4,3,4,3,4]) == 3

# Edge cases
assert Solution().numberOfSubsequences([1,1,1,1,1,1,1]) == 1 # all identical numbers
assert Solution().numberOfSubsequences([1,2,3,4,5,6,7]) == 0 # strictly increasing, no products match
assert Solution().numberOfSubsequences([1000]*10) == 35 # repeated large numbers, multiple combinations

# Minimum length
assert Solution().numberOfSubsequences([1,2,3,4,5,6,7]) == 0

# Mixed values
assert Solution().numberOfSubsequences([2,3,2,3,6,6