<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) {
        unordered_map<char, int> umap;
        unordered_map<char, int> umap2;
        char x;
        
        for(int i=0; i < s1.length(); i++ ){
            x = s1[i];
            umap[x] += 1;
        }    
        int l = 0;
        int r = 0;

        int window = 0;

        for(r;r < s2.length();r++){
            
            if((umap.find(s2[r]) != umap.end() && window < s1.length())){
                window++; 
                if(window == s1.length()){
                    
                    for(int i = l; i <= r; i++){
                        x = s2[i]; 
                        umap2[x] += 1;
                    }
        
                    if( umap == umap2){
                        return true;
                    }
                    else{
                        umap2.erase(umap2.begin(),umap2.end());
                        while (window == s1.length()){
                            window--;
                            l++;
                        } 
                        
                    }
            
                }
            }
            else{
                window = 0;
                l = r + 1;
            }
        }
        return false;
    }
    
};
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