# 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

• 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;
};