<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(s2.size() < s1.size())
          return 0;
        int size1 = s1.size();  
        for(int i = 0; i <(s2.size()-s1.size()+1); i++)
        {
            vector<int> v(26, 0);
            for(int j = i; j < i+size1; j++)
            {
                v[s2[j]- 'a'] +=1;
            }
            int k;
            for( k=0; k<size1;k++)
            {
                if(v[s1[k]-'a'] == 0)
                   break;
                v[s1[k]-'a']--;
            }
            if(k==size1)
              return 1;
        }
        return 0;
    }
};
20721