Question

1347. Minimum Number of Steps to Make Two Strings Anagram

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Description

You are given two strings of the same length 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. 

 

Constraints:

• 1 <= s.length <= 5 * 104
• s.length == t.length
• s and t consist of lowercase English letters only.

Solutions

Solution 1

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

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;
  }
}

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