<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 check(vector<int> alp,vector<int> alp2)
{
    for(int i=0;i<26;i++)
    {    
            if(alp[i]!=alp2[i])
            return false;
    }
    return true;
}
    bool checkInclusion(string s1, string s2) {
      int n1=s1.size();
      int n2=s2.size();
      if(n1>n2)return false;
      vector<int> alp(26,0);
      for(char x:s1)
      {
        alp[x-'a']++;
      } 

      vector<int> alp2(26,0);
      for(int i=0;i<n1;i++)
      {  
        alp2[s2[i]-'a']++;
      }
      if(check(alp,alp2))
      return true;

      for(int i=n1;i<n2;i++)
      {
        char fir=s2[i-n1];
        char lst=s2[i];      
        alp2[fir-'a']--;
        alp2[lst-'a']++;     
        if(check(alp,alp2))
        return true;         
      }      
      return false;
    }
};
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