<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 isallzero( unordered_map<char,int>mp){
       for(auto i:mp){
        if(i.second!=0) return false;
       }
       return true;
    }
    bool checkInclusion(string s1, string s2) {
        
        int n=s1.size();
        int m=s2.size();
        if(n>m) return false;
        unordered_map<char,int>mp;
        for(int i=0;i<n;i++){
            mp[s1[i]]++;
        }
       int i=0,j=0;
       while(j<m){
         if(j-i>=n){
            if(mp.find(s2[i])!=mp.end()){
                 mp[s2[i]]++;
            }
            i++;
         }
         if(mp.find(s2[j])!=mp.end()){
                 mp[s2[j]]--;
                  if(isallzero(mp)) return true;
            }
         j++;
       }
       return false;
    }
};
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