<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:
    int check(string s1,string s2)
    {
        vector<int> mp1(27,0);
        for(char c:s1)
        {
            mp1[c-'a']++;
        }
        vector<int> mp2(27,0);
        for(char c:s2)
        {
            mp2[c-'a']++;
        }
        
        for(char c:s1)
        {
            if(mp1[c-'a']!=mp2[c-'a'])
                return 0;
        }
        return 1;

    }
    bool checkInclusion(string s1, string s2) {
        int n=s2.length();
        int m=s1.length();
        for(int i=0;i<=n-m;i++)
        {
            //cout<<s2.substr(i,m)<<endl;
            //cout<<check(s1,s2.substr(i,m))<<endl;
            if(check(s1,s2.substr(i,m)))
                return true;
        }
        return false;
    }
};
135003