Is Tuple with Same Product easy or hard?

Tuple with Same Product is generally classified as a medium difficulty problem. The challenge lies in recognizing that equal products between pairs can be grouped and counted instead of checking every possible quadruple.

Tuple with Same Product Python/Java solution

Most implementations compute all pair products and store counts in a dictionary or hash map. Python uses a dictionary, Java uses HashMap, and C++ typically uses unordered_map. After counting frequencies, the result is calculated using the combination formula and multiplied by 8 for tuple permutations.

How to solve Tuple with Same Product in O(n)?

An O(n) solution is not achievable because the problem fundamentally depends on examining pair relationships between elements. At minimum, all O(n^2) pairs must be considered to detect equal products. The best practical approach uses a hash map to process these pairs efficiently in O(n^2) time.

What is the best approach for Tuple with Same Product?

The hash map pair-product approach is the most efficient. Compute the product for every pair of elements and store its frequency in a hash table. If a product appears k times, it forms k choose 2 pair combinations, and each combination produces 8 valid tuples. This results in O(n^2) time and O(n^2) space complexity.

Is Tuple with Same Product asked at Google/Amazon/Meta?

Tuple with Same Product represents a common interview pattern involving pair enumeration and hash maps. Variations of this problem appear in interviews at companies like Amazon, Google, and Meta where candidates are expected to recognize product collisions and optimize pair counting.

What data structure is used in Tuple with Same Product?

A hash table (hash map) is the primary data structure. It stores the frequency of each product generated by element pairs. This enables constant-time lookups and allows counting how many pair combinations share the same product.

What is the time complexity of Tuple with Same Product?

The optimal solution runs in O(n^2) time because it iterates through all pairs in the array and records their products in a hash map. Space complexity is also O(n^2) in the worst case due to storing pair product frequencies.

Description Hints 2 Solution Video 19:27

1726. Tuple with Same Product

Medium70.1% acceptance5 companies

Given an array nums of distinct positive integers, return the number of tuples (a, b, c, d) such that a * b = c * d where a, b, c, and d are elements of nums, and a != b != c != d.

Example 1:

Input: nums = [2,3,4,6]
Output: 8
Explanation: 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
Explanation: 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.

Asked at 5 companies

Google Bloomberg Amazon Microsoft Meta

Topics

Hints open one at a time

Hint 1

Note that all of the integers are distinct. This means that each time a product is formed it must be formed by two unique integers.

Hint 2

Count the frequency of each product of 2 distinct numbers. Then calculate the permutations formed.

19:27

Still stuck?Read 3 approaches, or watch a walkthrough.

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1class Solution:
2    def tupleSameProduct(self, nums: List[int]) -> int:
3        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2

Input

[2,3,4,6]

Input

[1,2,4,5,10]

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