<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>

class Solution:
    def checkInclusion(self, s1: str, s2: str) -> bool:
        window_start, matched = 0, 0
        char_freq = {}

        for chr in s1:
            if chr not in char_freq:
                char_freq[chr] = 0
            char_freq[chr] += 1

        for window_end in range(len(s2)):
            right_char = s2[window_end]

            if right_char in char_freq:
                char_freq[right_char] -= 1
                if char_freq[right_char] == 0:
                    matched += 1
            
            if matched == len(char_freq):
                return True
            
            if window_end >= len(s1) - 1:
                left_char = s2[window_start]
                window_start += 1
                if left_char in char_freq:
                    if char_freq[left_char] == 0:
                        matched -= 1
                    char_freq[left_char] += 1
            
        return False
                    

        
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