<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 checkInclusion(string s1, string s2) {
        vector<int> charCount(26);
        for (int i = 0; i < s1.size(); i++){
            char c = s1[i];
            int ind = c - 'a';
            charCount[ind]++;
        }
        
        vector<vector<int>> pChar(s2.size() + 1, vector<int>(26, 0));
        for (int i = 0; i < s2.size(); i++){
            for (int j = 0; j < 26; j++){
                pChar[i + 1][j] = pChar[i][j];
            }
            int cPos = s2[i] - 'a';
            pChar[i + 1][cPos]++;
        }
        int start = s1.size();
        for (int i = start; i < s2.size() + 1; i++){
            bool countEqual = true;
            for (int j = 0; j < 26; j++){
                int s2C = pChar[i][j] - pChar[i - start][j];
                if (s2C != charCount[j]){
                    countEqual = false;
                    break;
                }
            }
            if (countEqual == true){
                return true;
            }
        }
        return false;
    }
};
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