<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 {
private:
    bool checkEqual(int a[26], int b[26])
    {
        for(int i = 0; i < 26; i++)
        {
            if(a[i] != b[i])
            {
                return 0;
            }
        }
        return 1;
    }
public:
    bool checkInclusion(string s1, string s2) {
        //character count array
        int count1[26] = {0};
        for(int i = 0; i < s1.length(); i++)
        {
            int index = s1[i] - 'a';
            count1[index]++;
        }
        //traverse s2 string in window of size s1 length and compare
        int i = 0;
        int windowSize = s1.length();
        int count2[26] = {0};
        //running for first window
        while(i < windowSize && i < s2.length())
        {
            int index = s2[i] - 'a';
            count2[index]++;
            i++;
        }

        if(checkEqual(count1,count2))
        {
            return 1;
        }
        //next window process to be done
        while(i < s2.length())
        {
            char newChar = s2[i];
            int index = newChar - 'a';
            count2[index]++;
            
            char oldChar = s2[i-windowSize];
            index = oldChar - 'a';
            count2[index]--;

            i++;

            if(checkEqual(count1,count2))
            {
                return 1;
            }
        }
        return 0;
    }
};
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