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

Description

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.

Solutions

• Java
• C++
• Python
• Go
• TypeScript
• Javascript

• class Solution {
      public int minSteps(String s, String t) {
          int[] cnt = new int[26];
          for (char c : s.toCharArray()) {
              ++cnt[c - 'a'];
          }
          for (char c : t.toCharArray()) {
              --cnt[c - 'a'];
          }
          int ans = 0;
          for (int v : cnt) {
              ans += Math.abs(v);
          }
          return ans;
      }
  }
  

• class Solution {
  public:
      int minSteps(string s, string t) {
          vector<int> cnt(26);
          for (char& c : s) ++cnt[c - 'a'];
          for (char& c : t) --cnt[c - 'a'];
          int ans = 0;
          for (int& v : cnt) ans += abs(v);
          return ans;
      }
  };
  

• class Solution:
      def minSteps(self, s: str, t: str) -> int:
          cnt = Counter(s)
          for c in t:
              cnt[c] -= 1
          return sum(abs(v) for v in cnt.values())
  
  

• func minSteps(s string, t string) int {
  	cnt := make([]int, 26)
  	for _, c := range s {
  		cnt[c-'a']++
  	}
  	for _, c := range t {
  		cnt[c-'a']--
  	}
  	ans := 0
  	for _, v := range cnt {
  		ans += abs(v)
  	}
  	return ans
  }
  
  func abs(x int) int {
  	if x < 0 {
  		return -x
  	}
  	return x
  }
  

• function minSteps(s: string, t: string): number {
      let cnt = new Array(128).fill(0);
      for (const c of s) {
          ++cnt[c.charCodeAt(0)];
      }
      for (const c of t) {
          --cnt[c.charCodeAt(0)];
      }
      let ans = 0;
      for (const v of cnt) {
          ans += Math.abs(v);
      }
      return ans;
  }
  
  

• /**
   * @param {string} s
   * @param {string} t
   * @return {number}
   */
  var minSteps = function (s, t) {
      let cnt = new Array(26).fill(0);
      for (const c of s) {
          ++cnt[c.charCodeAt() - 'a'.charCodeAt()];
      }
      for (const c of t) {
          --cnt[c.charCodeAt() - 'a'.charCodeAt()];
      }
      let ans = 0;
      for (const v of cnt) {
          ans += Math.abs(v);
      }
      return ans;
  };
  
  

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