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Leetcode 1577. Number of Ways Wh

Leetcode 1577. Number of Ways Wh

作者: SnailTyan | 来源:发表于2021-09-01 14:03 被阅读0次

    文章作者:Tyan
    博客:noahsnail.com  |  CSDN  |  简书

    1. Description

    Number of Ways Where Square of Number Is Equal to Product of Two Numbers

    2. Solution

    解析: Version 1,分别计算两个数组的平方和以及所有组合乘积并统计对应值的个数,遍历每个数组平方和的个数,找到另一个数组对应的积的个数,二者相乘,加到三元组总个数中。Version 2进行进一步优化。

    • Version 1
    class Solution:
        def numTriplets(self, nums1: List[int], nums2: List[int]) -> int:
            count = 0
            square1 = collections.defaultdict(int)
            square2 = collections.defaultdict(int)
            product1 = collections.defaultdict(int)
            product2 = collections.defaultdict(int)
            for x in nums1:
                temp = x ** 2
                square1[temp] = square1[temp] + 1
            for x in nums2:
                temp = x ** 2
                square2[temp] = square2[temp] + 1
            for i in range(len(nums1)):
                for j in range(i+1, len(nums1)):
                    temp = nums1[i] * nums1[j]
                    product1[temp] = product1[temp] + 1
            for i in range(len(nums2)):
                for j in range(i+1, len(nums2)):
                    temp = nums2[i] * nums2[j]
                    product2[temp] = product2[temp] + 1
            for k, v in square1.items():
                count += v * product2[k]
            for k, v in square2.items():
                count += v * product1[k]
            return count
    
    • Version 2
    class Solution:
        def numTriplets(self, nums1: List[int], nums2: List[int]) -> int:
            count = 0
            product1 = collections.defaultdict(int)
            product2 = collections.defaultdict(int)
            for i in range(len(nums1)):
                for j in range(i+1, len(nums1)):
                    temp = nums1[i] * nums1[j]
                    product1[temp] = product1[temp] + 1
            for i in range(len(nums2)):
                for j in range(i+1, len(nums2)):
                    temp = nums2[i] * nums2[j]
                    product2[temp] = product2[temp] + 1
            for x in nums1:
                temp = x ** 2
                count += product2[temp]
            for x in nums2:
                temp = x ** 2
                square2[temp] = square2[temp] + 1
                count += product1[temp]
            return count
    

    Reference

    1. https://leetcode.com/problems/number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers/

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