#1726

Medium

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Tuple with Same Product

Given an array of distinct integers, return the number of tuples where the product of two pairs is the same.

Problem Statement

Given an array of distinct positive integers, your task is to return the number of valid tuples (a, b, c, d) where the product of a and b equals the product of c and d. The four integers must be distinct, and they should all be from the given array. For example, given nums = [2,3,4,6], the output is 8 because there are 8 valid tuples such as (2,6,3,4) and (6,2,3,4).

Constraints ensure that the array will contain distinct integers. The size of the array can be up to 1000, and each element is a positive integer not exceeding 10,000. As the problem involves counting tuples, it is important to avoid rechecking the same pairs by using hash lookups to optimize performance.

Examples

Example 1

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

Output: 8

There are 8 valid tuples: (2,6,3,4) , (2,6,4,3) , (6,2,3,4) , (6,2,4,3) (3,4,2,6) , (4,3,2,6) , (3,4,6,2) , (4,3,6,2)

Example 2

Input: nums = [1,2,4,5,10]

Output: 16

There are 16 valid tuples: (1,10,2,5) , (1,10,5,2) , (10,1,2,5) , (10,1,5,2) (2,5,1,10) , (2,5,10,1) , (5,2,1,10) , (5,2,10,1) (2,10,4,5) , (2,10,5,4) , (10,2,4,5) , (10,2,5,4) (4,5,2,10) , (4,5,10,2) , (5,4,2,10) , (5,4,10,2)

Constraints

1 <= nums.length <= 1000
1 <= nums[i] <= 104
All elements in nums are distinct.

Solution Approach

Pair Products Hashing

The key approach for this problem is to use a hash map to store products of pairs. For each pair of distinct integers (a, b), calculate their product and store it in a hash map, where the key is the product and the value is the number of times that product has been seen. This allows us to quickly identify if the same product appears again for another pair, reducing the problem to a simple lookup and count.

Counting Tuples

Once pairs and their products are stored in the hash map, we can count how many valid tuples can be formed. For each product found in the hash map, if it appears more than once, the count of tuples is updated by considering all unique combinations of these pairs, ensuring no repetition of indices.

Efficiently Handle Large Input Sizes

Since the array can have up to 1000 elements, the brute-force approach of checking all quadruples would take O(n^4) time, which is inefficient. By leveraging hash maps and considering only distinct pairs, we reduce the time complexity to O(n^2), making it feasible to handle large inputs.

Complexity Analysis

Metric	Value
Time	O(n^2)
Space	O(n^2)

The time complexity of this solution is O(n^2) because we are iterating over all pairs of distinct integers in the array. For each pair, we perform a constant-time hash lookup and insertion. The space complexity is also O(n^2) because, in the worst case, we may store up to n^2 distinct products in the hash map.

What Interviewers Usually Probe

The candidate demonstrates understanding of hashing techniques for counting distinct pairs.
The candidate proposes an efficient solution with O(n^2) time complexity, recognizing the inefficiency of brute force.
The candidate properly identifies the trade-off between time complexity and space complexity in the solution.

Common Pitfalls or Variants

Common pitfalls

Forgetting to handle distinctness of the elements in the pairs, which could lead to incorrect tuple counts.
Overlooking the need to use hash maps efficiently, potentially causing unnecessary recomputation.
Not considering space complexity when dealing with large input sizes, which may lead to inefficient solutions.

Follow-up variants

Consider implementing the same problem but allowing repeated elements in the array.
Optimize the solution by using a more advanced data structure, like a balanced binary search tree, instead of a hash map.
Expand the problem to include tuples of more than four elements with the same product.

FAQ

What is the key pattern in the Tuple with Same Product problem?

The key pattern is array scanning combined with hash lookups to count distinct pairs of products efficiently.

How do I handle large inputs in the Tuple with Same Product problem?

To handle large inputs, use hash maps to store pair products and avoid brute-force solutions that would take too long.

Can I optimize the space complexity of this problem?

The space complexity is O(n^2) due to the hash map storing pair products. While it's hard to reduce, you can consider pruning unnecessary entries to optimize.

What is the time complexity of the Tuple with Same Product problem?

The time complexity is O(n^2), as we compute products for each pair of distinct integers and perform hash map operations.

How does hashing help in solving the Tuple with Same Product problem?

Hashing allows efficient storage and lookup of pair products, enabling the counting of valid tuples without redundant calculations.

terminal

Solution

Solution 1: Combination + Hash Table

Assuming there are $n$ pairs of numbers, for any two pairs of numbers $a, b$ and $c, d$ that satisfy the condition $a \times b = c \times d$, there are a total of $\mathrm{C}_n^2 = \frac{n \times (n-1)}{2}$ such combinations.

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class Solution:
    def tupleSameProduct(self, nums: List[int]) -> int:
        cnt = defaultdict(int)
        for i in range(1, len(nums)):
            for j in range(i):
                x = nums[i] * nums[j]
                cnt[x] += 1
        return sum(v * (v - 1) // 2 for v in cnt.values()) << 3

Continue Practicing

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