Minimum Number of Steps to Make Two Strings Anagram II
<p>You are given two strings <code>s</code> and <code>t</code>. In one step, you can append <strong>any character</strong> to either <code>s</code> or <code>t</code>.</p>

<p>Return <em>the minimum number of steps to make </em><code>s</code><em> and </em><code>t</code><em> <strong>anagrams</strong> of each other.</em></p>

<p>An <strong>anagram</strong> of a string is a string that contains the same characters with a different (or the same) ordering.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;<strong><u>lee</u></strong>tco<u><strong>de</strong></u>&quot;, t = &quot;co<u><strong>a</strong></u>t<u><strong>s</strong></u>&quot;
<strong>Output:</strong> 7
<strong>Explanation:</strong> 
- In 2 steps, we can append the letters in &quot;as&quot; onto s = &quot;leetcode&quot;, forming s = &quot;leetcode<strong><u>as</u></strong>&quot;.
- In 5 steps, we can append the letters in &quot;leede&quot; onto t = &quot;coats&quot;, forming t = &quot;coats<u><strong>leede</strong></u>&quot;.
&quot;leetcodeas&quot; and &quot;coatsleede&quot; 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.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;night&quot;, t = &quot;thing&quot;
<strong>Output:</strong> 0
<strong>Explanation:</strong> The given strings are already anagrams of each other. Thus, we do not need any further steps.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length, t.length &lt;= 2 * 10<sup>5</sup></code></li>
	<li><code>s</code> and <code>t</code> consist of lowercase English letters.</li>
</ul>
Hash Table; String; Counting
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;
    }
};