Equivalent Product Pairs - Google Top Interview Questions


Problem Statement :


You are given a list of unique positive integers nums. 
Return the number of quadruples (a, b, c, d) in nums such that a * b = c * d and a, b, c, and d are all pairwise distinct.

Constraints

0 ≤ n ≤ 1,000 where n is the length of nums

Example 1

Input

nums = [2, 3, 4, 6]

Output

8

Explanation

We have the following (a, b, c, d)



[3,4,2,6]

[4,3,2,6]

[3,4,6,2]

[4,3,6,2]

[2,6,3,4]

[2,6,4,3]

[6,2,3,4]

[6,2,4,3]


Solution :



title-img 




                        Solution in C++ :

int solve(vector<int>& A) {
    int res = 0;
    unordered_map<int, int> f;
    for (int i = 0; i < A.size(); i++) {
        for (int j = i + 1; j < A.size(); j++) {
            f[A[i] * A[j]]++;
        }
    }

    for (auto it = f.begin(); it != f.end(); it++) {
        int cnt = it->second;
        int pairs = (cnt) * (cnt - 1) / 2;
        res += pairs * 8;
    }
    return res;
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    public int solve(int[] nums) {
        if (nums == null || nums.length == 0)
            return 0;

        Map<Integer, Integer> map = new HashMap();
        int ans = 0;
        for (int i = 0; i < nums.length; i++) {
            for (int j = i + 1; j < nums.length; j++) {
                int product = nums[i] * nums[j];
                if (map.containsKey(product)) {
                    ans += (8 * map.get(product));
                }
                map.put(product, map.getOrDefault(product, 0) + 1);
            }
        }
        return ans;
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, nums):
        c = {}
        n = len(nums)
        for i in range(n):
            for j in range(i + 1, n):
                x = nums[i] * nums[j]
                c[x] = c.get(x, 0) + 1
        ret = 0
        for x in c.values():
            ret += x * (x - 1)
        return ret * 4


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