[leetcode] 1711. Count Good Meals

[leetcode] 1711. Count Good Meals

Description

A good meal is a meal that contains exactly two different food items with a sum of deliciousness equal to a power of two.

You can pick any two different foods to make a good meal.

Given an array of integers deliciousness where deliciousness[i] is the deliciousness of the ith item of food, return the number of different good meals you can make from this list modulo 109 + 7.

Note that items with different indices are considered different even if they have the same deliciousness value.

Example 1:

Input: deliciousness = [1,3,5,7,9] Output: 4 Explanation: The good meals are (1,3), (1,7), (3,5) and, (7,9). Their respective sums are 4, 8, 8, and 16, all of which are powers of 2. Example 2:

Input: deliciousness = [1,1,1,3,3,3,7] Output: 15 Explanation: The good meals are (1,1) with 3 ways, (1,3) with 9 ways, and (1,7) with 3 ways.

Constraints:

1 <= deliciousness.length <= 1050 <= deliciousness[i] <= 220

分析

题目的意思是：给你一个数组，求出其中能够构成2的n次方的对数，我参考了一下别人的思路，很巧妙，首先用字典d来统计遍历过得deliciousness的数字，res来统计符合条件的对数，其中2的n次方最小是0，最大是2的22次方，因此一个一个的遍历是否满足构成条件就行了。核心代码如下

res+=d[2**i-num]

直接在字典里面找出能够构成2的i次方的数进行统计就行了。如果有不懂的，可以参考代码二，在本地调试一下

代码

class Solution: def countPairs(self, deliciousness: List[int]) -> int: d=defaultdict(int) res=0 for num in deliciousness: for i in range(22): res+=d[2**i-num] d[num]+=1 return res%(10**9+7)

代码二

from collections import defaultdictclass Solution: def countPairs(self, deliciousness) -> int: d=defaultdict(int) res=0 for num in deliciousness: for i in range(22): res+=d[2**i-num] d[num]+=1 print(d) return res%(10**9+7)if __name__=="__main__": solution=Solution() deliciousness = [1,3,5,7,9] res=solution.countPairs(deliciousness) print(res)

参考文献

​​[Python3] frequency table​​

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