How do you solve Tuple with Same Product on LeetCode?

Tuple with Same Product (LeetCode #1726) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n) space complexity. Key insight: Distinct integers mean that each product will be unique to a pair of numbers.

What is the brute force solution for Tuple with Same Product?

The brute force approach for Tuple with Same Product is Brute Force, with O(n²) time complexity and O(1) space complexity. We can find all possible pairs of numbers in the array and calculate their products. Then, we can check how many pairs of products are equal, which will give us the number of valid tuples. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Tuple with Same Product?

Tuple with Same Product uses the following concepts: Array, Hash Table, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Tuple with Same Product?

Clearly explain your thought process while coding. Consider edge cases, such as the smallest and largest possible inputs. Practice writing clean, efficient code to avoid unnecessary complexity.

What are common mistakes when solving Tuple with Same Product?

Forgetting to check that a, b, c, and d are distinct. Not accounting for the order of tuples when calculating the total count.

#1726

Tuple with Same Product

Medium

ArrayHash TableCountingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n²)
Space	O(1)	O(n)

💡

Intuition

Time O(n²)Space O(n)

Instead of checking all pairs, we can use a hash map to count the occurrences of each product formed by pairs of numbers. This allows us to efficiently calculate the number of valid tuples without redundant checks.

⚙️

Algorithm

3 steps

1Step 1: Initialize a hash map to count the frequency of each product formed by pairs of numbers.
2Step 2: Iterate through each unique pair of numbers, calculate their product, and update the hash map.
3Step 3: For each product that has a frequency greater than 1, calculate the number of valid tuples using the formula: count * (count - 1) * 2.

solution.py15 lines

1# Full working Python code
2from collections import defaultdict
3
4def tupleSameProduct(nums):
5    product_count = defaultdict(int)
6    n = len(nums)
7    for i in range(n):
8        for j in range(i + 1, n):
9            product = nums[i] * nums[j]
10            product_count[product] += 1
11    count = 0
12    for value in product_count.values():
13        if value > 1:
14            count += value * (value - 1) * 2
15    return count

ℹ

Complexity note: The time complexity remains O(n²) due to the nested loops, but we use a hash map to store product counts, which increases space complexity to O(n).

1Distinct integers mean that each product will be unique to a pair of numbers.
2The number of valid tuples can be derived from the frequency of each product.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.