<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) {
        int left = 0;
        int right = s1.length() - 1;

        unordered_map<char,int> mp1;
        unordered_map<char, int> mp2;

        for(auto a: s1){
            mp1[a]++;
        }

        while(right < s2.length()){
            mp2.clear();
            string cur = s2.substr(left, right - left + 1);
            int count = 0;

            for(auto a: cur){
                mp2[a]++;
            }

            for(auto a: mp1){
                if(mp2.count(a.first) && mp2[a.first] == a.second){
                    count++;
                }
            }

            if(count == mp1.size()){
                return true;
            }

            left++;
            right++;
        }

        return false;
    }
};
103534