1347. Minimum Number of Steps to Make Two Strings Anagram

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

Example 2:

Input: s = "leetcode", t = "practice"
Output: 5
Explanation: 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: 0
Explanation: "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 <= 50000
• s.length == t.length
• s and t contain lower-case English letters only.

class Solution {
    public int minSteps(String s, String t) {
        int le = s.length();
        Map<Character, Integer> map1 = new HashMap();
        Map<Character, Integer> map2 = new HashMap();
        int cont = 0;
            
        for(int i = 0; i < le; i++){
            map1.put(s.charAt(i), map1.getOrDefault(s.charAt(i), 0) + 1);
            map2.put(t.charAt(i), map2.getOrDefault(t.charAt(i), 0) + 1);
        }
        for(Character c: map2.keySet()){
            if(map1.containsKey(c)) cont += Math.min(map1.get(c), map2.get(c));
        }
        return le - cont;
    }
}

用两个hashmap存放character的出现次数，返回总长度 - 重合部分

class Solution {
    public int minSteps(String s, String t) {
        int n = s.length(), ans = 0;
        int[] arr = new int[26];
        for(int i = 0; i < n; i++) {
            arr[s.charAt(i) - 'a']++;
            arr[t.charAt(i) - 'a']--;
        }
        for(int i = 0; i < 26; i++) if(arr[i] > 0) ans += arr[i];
        return ans;
    }
}

但这种比较字符类型的题目用一个长26的array更好处理