<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_multiset<char> setik;

        for (int i = 0; i< s1.size();i++) {
            setik.insert(s1[i]);
        }

        unordered_multiset<char> buf;

        int p1 = 1;
        int p2 = s1.size();

        if (s1.size() > s2.size()) {
            return false;
        }

        for (int i = 0; i< s1.size(); i++) {
            buf.insert(s2[i]);
        }

        if (buf == setik)
            return true;

        int n = 0;
        while (p2<s2.size()){
            n = buf.count(s2[p1-1]);
            n--;
            buf.erase(s2[p1-1]);
            for (int i = 0; i< n; i++) {
                buf.insert(s2[p1-1]);
            }
            buf.insert(s2[p2]);
            if (buf == setik)
                return true;
            p2++;
            p1++;
        }

        return false;
    }
};
82003