Tuple with Same Product

Tuple with Same Product - Problem

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.

All four elements must be different from each other, and their products must be equal.

Input & Output

Example 1 — Basic Case

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

› Output: 8

💡 Note: We have 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). All satisfy a*b = c*d where 2*6 = 3*4 = 12.

Example 2 — Single Element Repeated Product

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

› Output: 16

💡 Note: Product 10 appears 3 times: 1*10, 2*5, so we get 3*(3-1) = 6 tuples. Product 20 appears 2 times: 2*10, 4*5, so we get 2*(2-1) = 2 tuples. Total is 6+2 = 8 tuples, but accounting for all permutations gives 16.

Example 3 — No Valid Tuples

$ Input: nums = [2,3,5,7]

› Output: 0

💡 Note: All numbers are prime, so no two different pairs can have the same product. Products are: 2*3=6, 2*5=10, 2*7=14, 3*5=15, 3*7=21, 5*7=35. All unique, so 0 tuples.

Constraints

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

Visualization

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Asked in

G Google 15 f Facebook 12 a Amazon 8

The optimal approach uses a hash map to count pair products. First, generate all pairs and count their product frequencies. Then, for each product appearing multiple times, the number of valid tuples is count×(count-1). Time: O(n²), Space: O(n²).

Common Approaches

✓ Brute Force

⏱️ Time: O(n⁴) Space: O(1)

Generate all combinations of 4 different elements and check if the first two multiply to the same value as the last two.

Hash Map Product Counting

⏱️ Time: O(n²) Space: O(n²)

First pass: compute all pair products and count frequencies. Second pass: for each product with count >= 2, calculate number of valid tuples.

Brute Force — Algorithm Steps

Generate all combinations of 4 elements
For each combination, check if a*b = c*d
Count valid tuples

Visualization

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Step-by-Step Walkthrough

Generate Combinations

Create all possible 4-element combinations

Check Products

For each combination, test if any two pairs have equal products

Count Valid Tuples

Add 8 for each valid combination (accounting for permutations)

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(int* nums, int numsSize) {
    int count = 0;
    
    for (int i = 0; i < numsSize; i++) {
        for (int j = i + 1; j < numsSize; j++) {
            for (int k = j + 1; k < numsSize; k++) {
                for (int l = k + 1; l < numsSize; l++) {
                    // Check all possible pairings
                    if (nums[i] * nums[j] == nums[k] * nums[l]) {
                        count += 8;
                    } else if (nums[i] * nums[k] == nums[j] * nums[l]) {
                        count += 8;
                    } else if (nums[i] * nums[l] == nums[j] * nums[k]) {
                        count += 8;
                    }
                }
            }
        }
    }
    
    return count;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    int nums[1000];
    int count = 0;
    char* token = strtok(line, "[],\n");
    while (token != NULL) {
        if (strlen(token) > 0 && token[0] != ' ') {
            nums[count++] = atoi(token);
        }
        token = strtok(NULL, "[],\n");
    }
    
    int result = solution(nums, count);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n⁴)

Four nested loops to check all combinations of 4 elements

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra space for variables

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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