<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 fn(string &a, string &b){
  
    unordered_map<char,int>mpp;
    for(auto i:b)
    mpp[i]++;
    for(auto i:a){
        if( mpp.find(i)==mpp.end())
        return false;
        mpp[i]--;
        if(mpp[i]==0)
        mpp.erase(i);
    }
    return true;

}
    bool checkInclusion(string s1, string s2) {
        int n=s1.size();
        int m=s2.size();
        if(m<n)
        return false;
        int i=0;
        while(i<=m-n){
            string p=s2.substr(i,n);
            // cout<<p<<endl;
            if(fn(p,s1))
            return true;
            i++;
        }
        return false;
        
    }
};
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