<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)
    {
        if (s1.size() > s2.size())
            return false;
        
        unordered_map<char, int> dict;

        for (int i = 0; i < s1.size(); i++)
        {
            dict[s1[i]]++;
        }

        for (int j = 0; j <= s2.size() - s1.size(); j++)
        {
            bool permutation_found = true;
            unordered_map<char, int> temp(dict);
            
            for (int k = j; k < j + s1.size(); k++)
            {
                if (temp[s2[k]] <= 0)
                {
                    permutation_found = false;
                    break;
                }

                temp[s2[k]]--;
            }

            if (permutation_found)
                return true;
        }

        return false;
    }
};
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