<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) {
        if(s2.length() < s1.length()){
            return false;
        }

        vector<int> alpha(26, 0);

        for(char c: s1){
            alpha[c - 'a']++;
        }

        for(int i = 0; i < s2.size(); ++i){
            char c = s2[i];
            if(alpha[c - 'a'] != 0){
                string str = s2.substr(i, s1.length());
                vector<int> strChars(26, 0);
                for(char testChar: str){
                    strChars[testChar - 'a']++;
                }

                bool isPerm = true;
                for(int k = 0; k < 26; ++k){
                    if(alpha[k] != strChars[k]){
                        isPerm = false;
                        break;
                    }
                }

                if(isPerm){
                    return true;
                }
            }
        }

        return false;
    }
};
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