<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(vector<int> a1, vector<int> a2) {
            for (int i = 0; i < 26; i++) {
                if (a1[i] != a2[i]) {
                    return false;
                }
            }
            return true;
        }
    public:
        bool checkInclusion(string s1, string s2) {
            vector<int> countS1(26, 0);
            for (int i = 0; i < s1.size(); i++) {
                countS1[s1[i] - 'a']++;
            }
            int windowSize = s1.size();
            vector<int> countS2(26, 0);
            for (int i = 0; i < s2.size(); i++) {
                countS2[s2[i] - 'a']++;
                if (i >= windowSize) {
                    countS2[s2[i - windowSize] - 'a']--;
                }
                if (checkEqual(countS1, countS2)) {
                    return true;
                }
            }
            return false;
        }
};

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