<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:
         #brute force: O(n*m) comparing exact characters
        #O(26*n) and O(n) solutions
        #basically looking for anagrams using sliding window
        #problem limited to lowercase letters: hashmap for a-z
        #O(26*n) solution compares both s1 and s2 hashmaps to each other

        if len(s1) > len(s2):
            return False

        def isAnagram(str1, str2):
            strMap1 = {}
            strMap2 = {}
            for c1 in str1:
                strMap1[c1] = 1 + strMap1.get(c1, 0)
                
            for c2 in str2:
                strMap2[c2] = 1 + strMap2.get(c2, 0)
            
            return strMap1 == strMap2
        
        left = 0
        right = len(s1)
        while right < (len(s2) + 1):
            # print((left, right))
            # print((s1, s2[left:right]))
            if isAnagram(s1, s2[left:right]):
                return True
            left += 1
            right += 1
        
        return False


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