# 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;
};


• function minSteps(s: string, t: string): number {
const cnt: number[] = Array(26).fill(0);
for (const c of s) {
++cnt[c.charCodeAt(0) - 97];
}
let ans = 0;
for (const c of t) {
if (--cnt[c.charCodeAt(0) - 97] < 0) {
++ans;
}
}
return ans;
}