<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(std::string const& s1, std::string const& s2) {
    if (s1.size() > s2.size()) return false;
    std::array<int, 26> s1_count{0}, s2_count{0};
    for (int i{0}; i < static_cast<int>(s1.size()); i++) // Construct a char count array for s1 as well as starting `s1.size()` of s2
      s1_count[s1[i] - 'a']++, s2_count[s2[i] - 'a']++;
    int matches{0}; // We don't need compare full hashmap each loop
    for (int i{0}; i < 26; i++)
      if (s1_count[i] == s2_count[i]) matches++;                                                // Window startup
    for (int l{0}, r{static_cast<int>(s1.size())}; r < static_cast<int>(s2.size()); r++, l++) { // Start r at the end of s1, window size fixed
      if (matches == 26) return true;
      // Right-side process
      int i{s2[r] - 'a'};
      if (s1_count[i] == s2_count[i]++) matches--; // If before same, now it +1, so matches -1
      if (s1_count[i] == s2_count[i]) matches++;
      // Left-side process
      i = s2[l] - 'a';
      if (s1_count[i] == s2_count[i]--) matches--;
      if (s1_count[i] == s2_count[i]) matches++; // If before bigger 1 and now same, matches +1
    }
    return matches == 26;
  }

};
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