<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(s1.length()>s2.length())return false;


        int len1 = s1.length();
        int len2 = s2.length();

        int i=0;
     

        unordered_map<char,int>frequencyMap1;
        unordered_map<char, int>frequencyMap2;

        for(char c:s1){
            frequencyMap1[c]++;
        }

        while(i<=len2 - len1){

           string temp = s2.substr(i, len1);

        for(char c:temp){
            frequencyMap2[c]++;
        }

        if(frequencyMap1==frequencyMap2){
            return true;
        }
        else{
            frequencyMap2.clear();
        }
        i++;
        }
        
return false;
        }

   
        
    };
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