Today I have done leetcode's February LeetCoding Challenge with cpp.

February LeetCoding Challenge 6

Description

Tuple with Same Product

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] <= 104
• All elements in nums are distinct.

Solution

class Solution {
public:
  int tupleSameProduct(vector<int>& nums) {
    int len = nums.size();
    if(len < 4) return 0;
    map<int, int> productCount;
    for(int i = 0; i < nums.size(); ++i) {
      for(int j = i + 1; j < nums.size(); ++j) {
        productCount[nums[i] * nums[j]] += 1;
      }
    }
    int answer = 0;
    for(auto [_, c] : productCount) {
      answer += 4 * c * (c - 1);
    }
    return answer;
  }
};

// Accepted
// 37/37 cases passed (681 ms)
// Your runtime beats 14.73 % of cpp submissions
// Your memory usage beats 15.67 % of cpp submissions (99.2 MB)