<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 solve(string a, string b) {
    // cout<<a<<" "<<b<<"\n";
    vector<char>v(26, 0);
    for(int i=0; i<a.size(); i++) {
        v[a[i]-'a']++;
        v[b[i]-'a']--;
    }

    for(int i=0; i<26; i++) {
        if(v[i]>0) return false;
    }
    return true;
}
    bool checkInclusion(string s1, string s2) {
        int n = s1.size();
        int m = s2.size();
        string p = "";
        for(int i=0; i<m-n+1; i++) {
            for(int j=i; j<n+i; j++) {
                p+=s2[j];
            }
            if(solve(p, s1)){
                return true;
            }
            p = "";
        }
        return false;
    }
};
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