**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 :

` ````
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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