<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 n=s2.size();
        if(n< s1.size()) return false;
        for(int i=0; i<=n-s1.size(); i++){
            int cnt=0;
            unordered_map<char,int> mpp;
            for(int j=0; j<s1.size(); j++){
                mpp[s1[j]]++;
            }
            for(int k=i; k<i+s1.size(); k++){
                if(mpp.find(s2[k])!=mpp.end()){
                    cnt++;
                    mpp[s2[k]]--;
                    if(mpp[s2[k]]==0) mpp.erase(s2[k]);
                }
                else break;
            }
            if(cnt==s1.size()) return true;
        }
        return false;
    }
};
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