<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> hashSet, tempSet;

        for (char c : s1) {
            hashSet[c]++;
        }

        int L = 0, R = 0;
        int s1_length = s1.size();
        int s2_length = s2.size();

        while (R < s2_length) {
            tempSet[s2[R]]++;

           
            while (tempSet[s2[R]] > hashSet[s2[R]]) {
                tempSet[s2[L]]--;
                if (tempSet[s2[L]] == 0) {
                    tempSet.erase(s2[L]);
                }
                L++;
            }

            if (R - L + 1 == s1_length) {
                if (tempSet == hashSet) {
                    return true;
                }
            }

            R++;
        }

        return false;
    }
};
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