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
.
Example 1:
Input: nums = [2,3,4,6] Output: 8 Explanation: There are 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)
Example 2:
Input: nums = [1,2,4,5,10] Output: 16 Explanation: There are 16 valid tuples: (1,10,2,5) , (1,10,5,2) , (10,1,2,5) , (10,1,5,2) (2,5,1,10) , (2,5,10,1) , (5,2,1,10) , (5,2,10,1) (2,10,4,5) , (2,10,5,4) , (10,2,4,5) , (10,2,5,4) (4,5,2,10) , (4,5,10,2) , (5,4,2,10) , (5,4,10,2)
Constraints:
1 <= nums.length <= 1000
1 <= nums[i] <= 10<sup>4</sup>
- All elements in
nums
are distinct.
class Solution {
public:
int tupleSameProduct(vector<int>& nums) {
uint size = nums.size();
unordered_map<int, int> mp;
for( int i = 0; i < size; i++ )
for( int j = i +1; j< size; j++){
int prod = nums[i] * nums[j];
mp[prod]++;
}
int res = 0;
for( auto& [prod, cnt] : mp ){
int pairs = (cnt *(cnt-1)) / 2;
res += 8 * pairs;
}
return res;
}
};