leetcode 1170. Compare Strings by Frequency of the Smallest Character 解法 python

本文介绍了一种解决LeetCode题目中关于字符串最小字符频率比较的问题，通过三种不同的方法实现，包括手动二分查找、使用库函数进行二分查找以及直接遍历比较。文章详细解释了解题思路和时间复杂度分析。

一.问题描述

Let's define a function f(s) over a non-empty string s, which calculates the frequency of the smallest character in s. For example, if s = "dcce" then f(s) = 2 because the smallest character is "c" and its frequency is 2.

Now, given string arrays queries and words, return an integer array answer, where each answer[i] is the number of words such that f(queries[i]) < f(W), where W is a word in words.

Example 1:

Input: queries = ["cbd"], words = ["zaaaz"]
Output: [1]
Explanation: On the first query we have f("cbd") = 1, f("zaaaz") = 3 so f("cbd") < f("zaaaz").

Example 2:

Input: queries = ["bbb","cc"], words = ["a","aa","aaa","aaaa"]
Output: [1,2]
Explanation: On the first query only f("bbb") < f("aaaa"). On the second query both f("aaa") and f("aaaa") are both > f("cc").

Constraints:

1 <= queries.length <= 2000
1 <= words.length <= 2000
1 <= queries[i].length, words[i].length <= 10
queries[i][j], words[i][j] are English lowercase letters.

二.解题思路

这道题就是中规中矩的走，由于leetcode限制比较松，如果你写的好的话，不用二分搜索也能过。

首先计算queries里的每个min，然后再计算words里的每个min，最后遍历每个query和words比对。

我们可以计算一下时间复杂度。

假设queries长为m，words长为n，那么时间复杂度就是O(mn)。

虽然leetcode比较松，但是如果你写的不好有很多多余的操作的话，就会TLE。

其实这道题最好使用二分查找做，把words的min数组排序。然后遍历query的时候用二分查找。

时间复杂度是O(mlogn).

如果你对二分查找不熟悉，建议自己写，如果很熟你可以直接引用库函数。

自己写的时候注意二分查找的返回的坐标含义，是小于该坐标的值全都大于target还是全都大于等于target。根据不同问题得自己弄明白。

我把三个版本都贴上仅供参考。

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欢迎大家一起套路一起刷题一起ac。

三.源码

1.自己动手写二分查找

class Solution:
    def numSmallerByFrequency(self, queries: List[str], words: List[str]) -> List[int]:
        len_w=len(words)
        def binSearch(arr, target):
            left,right=0,len(arr)-1
            while left<=right:
                mid=int((left+right)/2)
                if target>=arr[mid]:
                    left=mid+1
                elif target<arr[mid]:
                    right=mid-1
            return left
        cnts_w_min=sorted([word.count(min(word)) for word in words])
        cnts_q_min=[query.count(min(query)) for query in queries] 
        return [len_w-binSearch(cnts_w_min,cnt_q) for cnt_q in cnts_q_min]

2.库函数

库函数的bisect比自己实现的要快一些，这就很尴尬了，难道用了什么高大上的东西。

from bisect import bisect
class Solution:
    def numSmallerByFrequency(self, queries: List[str], words: List[str]) -> List[int]:
        len_w=len(words)
        cnts_w_min=sorted([word.count(min(word)) for word in words])
        cnts_q_min=[query.count(min(query)) for query in queries] 
        return [len_w-bisect(cnts_w_min,cnt_q) for cnt_q in cnts_q_min]

3.不用二分查找

class Solution:
    def numSmallerByFrequency(self, queries: List[str], words: List[str]) -> List[int]:
        cnts_w_min=[word.count(min(word)) for word in words]
        cnts_q_min=[query.count(min(query)) for query in queries] 
        return [sum([cnt_q<cnt_w for cnt_w in cnts_w_min]) for cnt_q in cnts_q_min]