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
func checkInclusion(s1 string, s2 string) bool {
	need := 0
	cnt := [26]int{}

	for _, c := range s1 {
		if cnt[c-'a']++; cnt[c-'a'] == 1 {
			need++
		}
	}

	m, n := len(s1), len(s2)
	for i := 0; i < n; i++ {
		c := s2[i] - 'a'
		if cnt[c]--; cnt[c] == 0 {
			need--
		}
		if i >= m {
			c = s2[i-m] - 'a'
			if cnt[c]++; cnt[c] == 1 {
				need++
			}
		}
		if need == 0 {
			return true
		}
	}
	return false
}
