<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:
    vector<int> countstring(string &str){
        vector<int> arr(26,0);
        for(int i =0 ; i < str.length() ; i++){
            int c = str[i] - 'a';
            arr[c]++;
        }
        return arr;
    }
    bool checkequal(vector<int> a , vector<int> b){
        for(int i = 0 ; i < 26; i++){
            if (a[i] != b[i]){
                return 0;
            }
        }
        return 1;
    }
    bool checkInclusion(string s1, string s2) {
        vector<int> s1c = countstring(s1);
        if (s2.length() < s1.length()){
            return false;
        }
        for (int i = 0 ; i < s2.length() - s1.length()+1; i++){
            string temp = s2.substr(i,s1.length());
            vector<int> s2c = countstring(temp);
            if (checkequal(s1c , s2c )){
                return true;
            }
        }
        return false;
    }
};
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