<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 f(unordered_map<char, int>mp , string s , int n1){
        for(auto it :s){
             if(mp.find(it)  != mp.end()){
                 if(--mp[it] == -1)return false;
             }
             else return false;
        }
        return true;
    }
    bool checkInclusion(string s1, string s2) {
        int n1 = s1.length();
        int n2 = s2.length();

        unordered_map<char  ,int>mp;

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

        int i=0; 
        while( i + n1 <= s2.length()){
              if(f(mp , s2.substr(i , n1) , n1)){
                cout<<s2.substr(i, n1)<<endl;
                return true;
              }
              i++;
        }
        return false;;
    }
};
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