Leetcode 1577. Number of Ways Where Square of Number Is Equal to Product of Two Numbers

文章作者:Tyan
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1. Description

Maximum Number of Coins You Can Get

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
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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
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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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