<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(s1.size() > s2.size())
        return false;
       unordered_map<char,int> mp;
       for(auto ch:s1)
        mp[ch]++;
       int w = s1.size();
    //    sort(s1.begin(),s1.end());
    //    for(int i =0;i<s2.size()-w+1;i++){
    //         string s = s2.substr(i,w);
    //         sort(s.begin(),s.end());
    //         if(s1 == s)
    //         return true;
    //    }

    //    return false;
      for(int i=0;i<s2.size()-w+1;i++){
        unordered_map<char,int> temp;
        for(int j = i;j<i+w;j++){
            temp[s2[j]]++;
        }
        bool found = true;
        if(temp.size() == mp.size()){
            for(auto ch:s1){
                if(temp[ch] != mp[ch])
                    found = false;
            }
            if(found)
            return true;
        }
      }
      return false;
    }
};
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