<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.length() > s2.length()) {
            return false;
        }

        unordered_map<char, int> map1;
        for(char c : s1)  {
            ++map1[c];
        }

        int left = 0;
        int right = s1.length() - 1;
        bool skip = false;

        unordered_map<char, int> map2;
        for(int i = left; i <= right; ++i) {
            char c = s2[i];
            ++map2[c];
        }
        
        bool notEqual = false;
        while(right < s2.length()) {
            for(auto p : map1) {
                if(map2[p.first] != p.second) {
                    notEqual = true;
                    break;
                }
            }

            if(notEqual) {
                --map2[s2[left]];
                ++left;
                ++right;
                ++map2[s2[right]];
                notEqual = false;
            } else {
                return true;
            }
        }

        return false;
    }
};
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