<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.length() > s2.length())
          return false;
        unordered_map<char,int> mp;
        for(int i =0; i< s1.length(); i++)
          mp[s1[i]]++;
        int i =0;
        int j =0;
        int k = s1.length();
        unordered_map<char,int> mp1;
        while(i < s2.length() && j < s2.length())
        {
             while(j < s2.length() && (j-i+1) <= k)
             {
                 mp1[s2[j]]++;
                 j++;  
             }
             if(mp1 == mp)
               return true;
             mp1.clear();  
             i++;
             j = i;
        }    
        return false;
    }
};
//if s1's length is greater than s2 return false
//now use a map to count the frequency of chars in s1
//now run a fixed sized sliding window on s2: window size is s1's length
//if char s2[j] is in map, we increase window length by j++
//ab ba
//a_ b_ c_  a c b b a c b c a c a b c b a
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