<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 {
    bool areVectorsEqual(vector<int> a, vector<int> b){
        for(int i=0; i<26; i++){
            if(a[i]!=b[i]) return false;
        }
        return true;
    }
public:
    bool checkInclusion(string s1, string s2) {
        if(s2.size()<s1.size()) return false;
        vector<int> freqS1(26, 0);
        for(char c: s1) freqS1[c-'a']++;
        
        vector<int> freqS2(26, 0);
        int i=0, j=0;
        
        while(j<s2.size()){
            freqS2[s2[j]-'a']++;
            
            if(j-i+1==s1.size()){
                if(areVectorsEqual(freqS1, freqS2)) return true;
            }
            
            if(j-i+1<s1.size()) j++;
            else{
                freqS2[s2[i]-'a']--;
                i++;
                j++;
            }
        }
        return false;
    }
};
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