Tuple with Same Product - Practice Coding Problems

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

Tap to expand

Understanding the Visualization

Input

Array of distinct positive integers

Process

Find all pairs with matching products

Output

Count of valid 4-element tuples

Key Takeaway

🎯 Key Insight: Count pairs by their product - if n pairs share the same product, they form n×(n-1) valid tuples

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

Approach	Time	Space	Notes
✓ Hash Map Product Counting	O(n²)	O(n²)	Count pairs by product, then calculate combinations
Brute Force	O(n⁴)	O(1)	Check all possible 4-element combinations

Hash Map Product Counting — Algorithm Steps

Generate all pairs and their products
Count frequency of each product in hash map
For each product with count >= 2, add count*(count-1) to result

Visualization

Tap to expand

Step-by-Step Walkthrough

Count Pair Products

Generate all pairs and count product frequencies

Find Duplicates

Identify products that appear multiple times

Calculate Tuples

For each duplicate product, add count*(count-1) tuples

Code -

solution.c — C

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

struct HashNode {
    int key;
    int count;
    struct HashNode* next;
};

#define HASH_SIZE 10007

struct HashNode* hashTable[HASH_SIZE];

int hash(int key) {
    return abs(key) % HASH_SIZE;
}

void insert(int key) {
    int index = hash(key);
    struct HashNode* node = hashTable[index];
    
    while (node != NULL) {
        if (node->key == key) {
            node->count++;
            return;
        }
        node = node->next;
    }
    
    struct HashNode* newNode = (struct HashNode*)malloc(sizeof(struct HashNode));
    newNode->key = key;
    newNode->count = 1;
    newNode->next = hashTable[index];
    hashTable[index] = newNode;
}

int solution(int* nums, int numsSize) {
    memset(hashTable, 0, sizeof(hashTable));
    
    // Count all pair products
    for (int i = 0; i < numsSize; i++) {
        for (int j = i + 1; j < numsSize; j++) {
            int product = nums[i] * nums[j];
            insert(product);
        }
    }
    
    int result = 0;
    // For each product with count >= 2, we can form count*(count-1) tuples
    for (int i = 0; i < HASH_SIZE; i++) {
        struct HashNode* node = hashTable[i];
        while (node != NULL) {
            if (node->count >= 2) {
                result += node->count * (node->count - 1);
            }
            node = node->next;
        }
    }
    
    return result;
}

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(n²)

Two nested loops to generate all pairs, then iterate through hash map

⚠ Quadratic Growth

Space Complexity

O(n²)

Hash map can store up to n²/2 different products

⚠ Quadratic Space

31.2K Views

Medium Frequency

~25 min Avg. Time

892 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Hash Map Product Counting — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler