<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> count2;
        for (char i : s1)
            count2[i]++;

        int l = 0, flag = 1;
        while (l + s1.size() <= s2.size()) {
            unordered_map<char, int> count = count2;
            if (count.find(s2[l]) != count.end()) {
                for (int i = l; i < s1.size() + l; i++) {
                    if (count.find(s2[i]) == count.end()) {
                        l = i;
                        goto point;
                    } else
                        count[s2[i]]--;
                }
                flag = 1;
                for (auto i : count) 
                    if (i.second != 0)
                        flag = 0;
                if (flag)
                    return true;
            }
            point:
            l++;
        }
        return false;
    }
};

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