<p>Given two strings <code>s1</code> and <code>s2</code>, return <code>true</code><em> if </em><code>s2</code><em> contains a permutation of </em><code>s1</code><em>, or </em><code>false</code><em> otherwise</em>.</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 {
public:
    bool checkInclusion(string &s1, string &s2) {
        int n1 = s1.size(), 
            n2 = s2.size();
        
        if (n1 > n2) return false;
        
        unordered_map<char,int> m1;
        
        for (int i = 0; i < n1; ++i) ++m1[s1[i]];
        
        for (int i = 0; i <= n2 - n1; ++i) {
            unordered_map<char,int> m2;
            auto str = s2.substr(i,n1);
            for (auto const &c:str) ++m2[c];
            if (m1 == m2) return true;
        }
        
        return false;
    }
};
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