Tuple with Same Product

题目

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 .

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)

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.

解法

1. Brute Force方法会TLE
2. HashMap方法

用一个HashMap记录乘积出现的次数，对于乘积出现次数n大于1的，该乘积所对应的组合数为(n-1)!

思考：像这种条件a * b = c * d的，找到这个最小粒度因素就是两个数的乘积，一个O(N2)的方法就可以找到所有乘积和它们的出现次数 --> HashMap

class Solution {
    public int tupleSameProduct(int[] nums) {
        Map<Integer, Integer> products = new HashMap<>();
        for (int i = 0; i < nums.length; i++) {
            for (int j = i+1; j < nums.length; j++) {
                int product = nums[i] * nums[j];
                products.put(product, products.getOrDefault(product, 0) + 1);
            }
        }
        
        int res = 0;
        for (Map.Entry<Integer, Integer> entry: products.entrySet()) {
            int value = entry.getValue();
            res += ((value*(value-1))*4);
        }
        return res;
    }
}