Permutation in String
<p>Given two strings <code>s1</code> and <code>s2</code>, return <code>true</code> if <code>s2</code> contains a <span data-keyword="permutation-string">permutation</span> of <code>s1</code>, or <code>false</code> otherwise.</p>

<p>In other words, return <code>true</code> if one of <code>s1</code>&#39;s permutations is the substring of <code>s2</code>.</p>

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

<pre>
<strong>Input:</strong> s1 = &quot;ab&quot;, s2 = &quot;eidbaooo&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong> s2 contains one permutation of s1 (&quot;ba&quot;).
</pre>

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

<pre>
<strong>Input:</strong> s1 = &quot;ab&quot;, s2 = &quot;eidboaoo&quot;
<strong>Output:</strong> false
</pre>

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

<ul>
	<li><code>1 &lt;= s1.length, s2.length &lt;= 10<sup>4</sup></code></li>
	<li><code>s1</code> and <code>s2</code> consist of lowercase English letters.</li>
</ul>
Hash Table; Two Pointers; String; Sliding Window
class Solution {
public:
    bool checkInclusion(string s1, string s2) {
        int need = 0;
        int cnt[26]{};
        for (char c : s1) {
            if (++cnt[c - 'a'] == 1) {
                ++need;
            }
        }
        int m = s1.size(), n = s2.size();
        for (int i = 0; i < n; ++i) {
            int c = s2[i] - 'a';
            if (--cnt[c] == 0) {
                --need;
            }
            if (i >= m) {
                c = s2[i - m] - 'a';
                if (++cnt[c] == 1) {
                    ++need;
                }
            }
            if (need == 0) {
                return true;
            }
        }
        return false;
    }
};
