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Formatted question description: https://leetcode.ca/all/1347.html

1347. Minimum Number of Steps to Make Two Strings Anagram (Medium)

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.

Related Topics:
String

Solution 1.

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

• class Solution {
      public int minSteps(String s, String t) {
          int length = s.length();
          int[] counts = new int[26];
          for (int i = 0; i < length; i++) {
              char c1 = s.charAt(i), c2 = t.charAt(i);
              counts[c1 - 'a']++;
              counts[c2 - 'a']--;
          }
          int steps = 0;
          for (int i = 0; i < 26; i++) {
              if (counts[i] > 0)
                  steps += counts[i];
          }
          return steps;
      }
  }
  
  ############
  
  class Solution {
      public int minSteps(String s, String t) {
          int[] cnt = new int[26];
          for (int i = 0; i < s.length(); ++i) {
              ++cnt[s.charAt(i) - 'a'];
          }
          int ans = 0;
          for (int i = 0; i < t.length(); ++i) {
              if (--cnt[t.charAt(i) - 'a'] < 0) {
                  ++ans;
              }
          }
          return ans;
      }
  }
  

• // OJ: https://leetcode.com/problems/minimum-number-of-steps-to-make-two-strings-anagram/
  // Time: O(N)
  // Space: O(1)
  class Solution {
  public:
      int minSteps(string s, string t) {
          int cnt[26] = {0}, ans = 0;
          for (char c : s) cnt[c - 'a']++;
          for (char c : t) cnt[c - 'a']--;
          for (int n : cnt) {
              if (n > 0) ans += n;
          }
          return ans;
      }
  };
  

• class Solution:
      def minSteps(self, s: str, t: str) -> int:
          cnt = Counter(s)
          ans = 0
          for c in t:
              if cnt[c] > 0:
                  cnt[c] -= 1
              else:
                  ans += 1
          return ans
  
  
  

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

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

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