<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 {
private:
    string ft_strncpy(string dest,string src,int n){
        for(int i=0;i<n;i++){
            dest[i] = src[i];
            i++;
        }
       return dest; 
    }
public:
    bool checkInclusion(string s1, string s2) {
        map<char,int> set;
        auto str = s2.begin();
        sort(s1.begin(),s1.end());
        int n = s1.size();
        for(auto c : s1)
            set[c]++;
        for(int i =0;i < s2.size();i++){
            while (str < s2.end() && !set.count(*str)) str++;
            string temp = s2.substr(i,n);
            sort(temp.begin(),temp.end());
            if(temp==s1) return true;
            str++;
        }
        return false;
    }
};
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