<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.size()>s2.size())return false;
        unordered_map<char,int>mp;
        for(int i =0;i<s1.size();i++){
            mp[s1[i]]++;
        }
        int i =0;
        int size =s1.size();
        int k =s2.size()-s1.size();
        while(i<=k){
            unordered_map<char,int>y =mp;
            string s = s2.substr(i, size);
            int j =i;
            int v =size+i;
            while(j<v){
                if(y.find(s2[j])!=y.end()){
                    y[s2[j]]--;
                    if(y[s2[j]]==0)y.erase(s2[j]);
                }
                else break;
                j++;
            }
            if(j ==v)return true;
            i++;
        }
    return false;}
};
125065