<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) {
        unordered_map<char, int> s1Count;
        unordered_map<char, int> s2Count;

        int left = 0;
        int right = left + (s1.size() - 1);

        for (char c : s1) {
            s1Count[c] += 1;
        }

        while (right < s2.size()) {
            for (int i = left; i <= right; i++) {
                s2Count[s2[i]] += 1;
            }

            if (!isPermutation(s1, s2, s1Count, s2Count)) {
                s2Count.clear();
                left++;
                right++;
                continue;
            }

            return true;
        }

        return false;
    }

    bool isPermutation(
        string& s1,
        string& s2,
        unordered_map<char, int> &s1Count, 
        unordered_map<char, int> &s2Count
    ) {
        for (char c : s1) {
            if (s1Count[c] != s2Count[c]) {
                return false;
            }
        }

        return true;
    }
};
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