<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> c;
        for(auto x: s1){
            c[x]++;
        }
        int i,j;
        int z=s1.size();
        for(int i=0;i<(s2.size()-z+1);i++){
            if(c.count(s2[i]) && c.count(s2[i+z-1])){
                unordered_map<char,int> d;
                d=c;
                bool check=true;
                for(int j=i;j<(i+z);j++){
                    if(d.count(s2[j])){
                        //cout<<" ";
                    d[s2[j]]--;
                    if(d[s2[j]]<0){
                        check =false;
                        break;
                    }
                    }
                    else{
                        check = false;
                        break;
                    }
                }
                if(check){
                    bool c2=true;
                    for(auto x:d){
                        if(x.second!=0){
                            c2=false;
                            break;
                        }
                    }
                    if(c2) return true;
                }
            }
        }
        return false;
    }
};
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