<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>

//Mehmet Fuat Karkaya

class Solution {
public:
    bool checkInclusion(string s1, string s2) {
        if(s1.size() > s2.size())
            return false;
        unordered_map<char, int> s1mapInit;
        set<char> s1set;
        for(char c : s1){
            s1set.insert(c);
            s1mapInit[c]++;
        }
        for(int i=0; i<s2.size()-s1.size()+1; i++){
            unordered_map<char, int> s1map = s1mapInit;
            for(int j=0; j<s1.size(); j++){
                char ch = s2[i+j];
                if(s1set.find(ch) == s1set.end()){//if does not consists
                    i = i+j;
                    break;
                }
                s1map[ch]--;
                if(s1map[ch]<0)
                    break;
                if(j==s1.size()-1)
                    return true; 
            }
        }
        return false;            
    }
};

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