<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 check(string &str, string &s) {
        vector<int> count(26, 0);
        for (char c : s) {
            count[c - 'a']++;
        }
        for (char c : str) {
            count[c - 'a']--;
        }
        for (int i : count) {
            if (i != 0) {
                return false;
            }
        }
        return true;
    }

    bool checkInclusion(string s1, string s2) {
        int k = s1.size();
        int n = s2.size();
        if (k > n) return false;

        for (int i = 0; i <= n - k; i++) {
            string str = s2.substr(i, k);
            if (check(str, s1)) {
                return true;
            }
        }

        return false;
    }
};

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