<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) {
        auto&& checkIfPermutation = [] (std::string& str1, std::string& str2) -> bool {
            if (str1.size() != str2.size()) {
                return false;
            }

            std::vector<int> countVec(26);
            for (auto ch : str1) {
                countVec[ch - 'a']++;
            }
            
            for (auto ch : str2) {
                countVec[ch - 'a']--;
            }

            for (int i = 0; i < 26; ++i) {
                if (countVec[i] != 0) {
                    return false;
                }
            }
            return true;
        };

        int l = 0, r = s1.size() - 1;
        while (r < s2.size()) {
            std::string tempStr(s2.begin() + l, s2.begin() + r + 1);
            if (checkIfPermutation(tempStr, s1)) {
                return true;
            }
            l++;
            r++;
        }
        return false;
    }
};
72065