LeetCode 2186. Minimum Number of Steps to Make Two Strings Anagram II

原题链接在这里：https://leetcode.com/problems/minimum-number-of-steps-to-make-two-strings-anagram-ii/

题目：

You are given two strings s and t. In one step, you can append any character to either s or t.

Return the minimum number of steps to make s and t anagrams of each other.

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

Example 1:

Input: s = "leetcode", t = "coats"
Output: 7
Explanation: 
- In 2 steps, we can append the letters in "as" onto s = "leetcode", forming s = "leetcodeas".
- In 5 steps, we can append the letters in "leede" onto t = "coats", forming t = "coatsleede".
"leetcodeas" and "coatsleede" are now anagrams of each other.
We used a total of 2 + 5 = 7 steps.
It can be shown that there is no way to make them anagrams of each other with less than 7 steps.

Example 2:

Input: s = "night", t = "thing"
Output: 0
Explanation: The given strings are already anagrams of each other. Thus, we do not need any further steps.

Constraints:

• 1 <= s.length, t.length <= 2 * 105
• s and t consist of lowercase English letters.

题解：

Map the count for char in s and t.

Accumlate the difference to res.

Time Complexity: O(n). n = s.length().

Space: O(1).

AC Java:

class Solution {
    public int minSteps(String s, String t) {
        int [] map = new int[26];
        for(int i = 0; i < s.length(); i++){
            map[s.charAt(i) - 'a']++;
        }
        
        for(int i = 0; i < t.length(); i++){
            map[t.charAt(i) - 'a']--;
        }
        
        int res = 0;
        for(int num : map){
            res += Math.abs(num);
        }
        
        return res;
    }
}

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