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1347. Minimum Number of Steps to Make Two Strings Anagram

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

  • 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;
    }
};

  • 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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