<p>Given two strings <code>s1</code> and <code>s2</code>, return <code>true</code> if <code>s2</code> contains a <span data-keyword="permutation-string">permutation</span> of <code>s1</code>, or <code>false</code> otherwise.</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:
    def checkInclusion(self, s1: str, s2: str) -> bool:
        dic1 = {}
        for i in s1:
            if i in dic1:
                dic1[i] += 1
            else:
                dic1[i] = 1

        cur = {}
        left = 0
        for right in range(len(s2)):
            if s2[right] in cur:
                cur[s2[right]] += 1
            else:
                cur[s2[right]] = 1

            if right - left + 1 == len(s1):
                if cur == dic1: return True

                cur[s2[left]] -= 1
                if cur[s2[left]] == 0: del cur[s2[left]]
                left += 1

        return False
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