<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 compare(vector<int> a, vector<int> b)
    {
        for(int i = 0; i<26; i++)
        {
            if(a[i]!=b[i])
            return false;
        }
        return true;
    }
    bool checkInclusion(string s1, string s2) {
        
        if(s2.size()<s1.size())
        return false;

        vector<int> freqs1(26,0);
        vector<int> freqs2(26,0);

        int start = 0;
        for(int i = 0; i<s1.size(); i++)
        {
            freqs1[s1[i]-'a']++;
        }
        for(int i = 0; i<s2.size(); i++)
        {
            freqs2[s2[i]-'a']++;

            if(i-start+1 > s1.size())
            {
                freqs2[s2[start]-'a']--;
                start++;
            }

            if(compare(freqs1, freqs2)==true)
            return true;
        }
        return false;
    }
    
};
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