1726. Tuple with Same Product

package LeetCode_1726

/**
 * 1726. Tuple with Same Product
 * https://leetcode.com/problems/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 valids 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) , (10,2,4,5)
(4,5,2,10) , (4,5,10,2) , (5,4,2,10) , (5,4,10,2)

Example 3:
Input: nums = [2,3,4,6,8,12]
Output: 40

Example 4:
Input: nums = [2,3,5,7]
Output: 0

Constraints:
1. 1 <= nums.length <= 1000
2. 1 <= nums[i] <= 10^4
3. All elements in nums are distinct.
 * */
class Solution {
    /*
    * solution: HashMap, accumulate the number of that every two number have a same product,
    * Time:O(n^2), Space:O(n)
    * */
    fun tupleSameProduct(nums: IntArray): Int {
        val map = HashMap<Int, Int>()
        var count = 0
        for (i in nums.indices) {
            for (j in i + 1 until nums.size) {
                val product = nums[i] * nums[j]
                val countOfProduct = map.getOrDefault(product, 0)
                //accumulate the count if have same product, for example [1,10,2,5]: 1*10 and 2*5
                count += countOfProduct
                map.put(product, countOfProduct + 1)
            }
        }
        //each tuple [a,b,c,d] can be arranged in 8 different ways ot matching
        return 8 * count
    }
}
原文地址:https://www.cnblogs.com/johnnyzhao/p/14292106.html