<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(unordered_map<char, int >mp1 , string s){
        //   if(mp1.size()  != mp2.size())return false;

          for(auto it  : s){

              if(mp1.find(it) != mp1.end()){
              if(--mp1[it] == -1)return false;
              }
              else{
                     return false;
              }
          }
          return true;
    }
    bool checkInclusion(string s1, string s2) {
        unordered_map<char , int >mp1;
        // unordered_map<char ,int>mp2;

        for(auto it  : s1){
             mp1[it]++;
        }

        // int  i =0; 
        int  j =0; 

        while(j + s1.length() <= s2.length()){
               if(check(mp1 , s2.substr(j, s1.length())))return true;
               j++;
        }

        return false;
    }
};
126721