[leetcode] 1347. Minimum Number of Steps to Make Two Strings Anagram

[leetcode] 1347. Minimum Number of Steps to Make Two Strings Anagram

Description

Given two equal-size strings s and t. In one step you can choose any character of t and replace it with another character.

Return the minimum number of steps to make t an anagram of s.

An Anagram of a string is a string that contains the same characters with a different (or the same) ordering.

Example 1:

Input: s = "bab", t = "aba"Output: 1Explanation: Replace the first 'a' in t with b, t = "bba" which is anagram of s.

Example 2:

Input: s = "leetcode", t = "practice"Output: 5Explanation: Replace 'p', 'r', 'a', 'i' and 'c' from t with proper characters to make t anagram of s.

Example 3:

Input: s = "anagram", t = "mangaar"Output: 0Explanation: "anagram" and "mangaar" are anagrams.

Example 4:

Input: s = "xxyyzz", t = "xxyyzz"Output: 0

Example 5:

Input: s = "friend", t = "family"Output: 4

Constraints:

1 <= s.length <= 50000s.length == t.lengths and t contain lower-case English letters only.

分析

题目的意思是：给定两个字符串，可以替换其中一个字符串的字符，问最小的替换数使得两字符串相等，允许字符串的位置不一样。我的思路很直接，直接统计第一个字符串s的字符频率，然后再去比较第二个字符串t，然后找出字典中不存在的字符的个数就是结果了。这道题有点简单，我有点紧张。

代码

class Solution: def minSteps(self, s: str, t: str) -> int: d=collections.defaultdict(int) for ch in s: d[ch]+=1 res=0 for ch in t: if(ch in d and d[ch]>0): d[ch]-=1 else: res+=1 return res

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