<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){
        map<char,int> m1;
    for(int i=0;i<s1.length();i++){
        m1[s1[i]]++;
    }
        int n=s1.length();
        int m=s2.length();
        if(n>m){
            return  false;
        }
        int j=0;
        for(int i=0;i<m-n+1;i++){
            map<char,int> m2;
            for(j=0;j<n;j++){
                if(m1.find(s2[i+j])==m1.end()){
                    break;
                }
                m2[s2[i+j]]++;
                
            }
            if(j==n && m1==m2){
                return true;
            }
        }
        return false;
    }
};
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