<p>Given two strings <code>s</code> and <code>t</code> of lengths <code>m</code> and <code>n</code> respectively, return <em>the <strong>minimum window</strong></em> <span data-keyword="substring-nonempty"><strong><em>substring</em></strong></span><em> of </em><code>s</code><em> such that every character in </em><code>t</code><em> (<strong>including duplicates</strong>) is included in the window</em>. If there is no such substring, return <em>the empty string </em><code>&quot;&quot;</code>.</p>

<p>The testcases will be generated such that the answer is <strong>unique</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;ADOBECODEBANC&quot;, t = &quot;ABC&quot;
<strong>Output:</strong> &quot;BANC&quot;
<strong>Explanation:</strong> The minimum window substring &quot;BANC&quot; includes &#39;A&#39;, &#39;B&#39;, and &#39;C&#39; from string t.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;a&quot;, t = &quot;a&quot;
<strong>Output:</strong> &quot;a&quot;
<strong>Explanation:</strong> The entire string s is the minimum window.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;a&quot;, t = &quot;aa&quot;
<strong>Output:</strong> &quot;&quot;
<strong>Explanation:</strong> Both &#39;a&#39;s from t must be included in the window.
Since the largest window of s only has one &#39;a&#39;, return empty string.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == s.length</code></li>
	<li><code>n == t.length</code></li>
	<li><code>1 &lt;= m, n &lt;= 10<sup>5</sup></code></li>
	<li><code>s</code> and <code>t</code> consist of uppercase and lowercase English letters.</li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong> Could you find an algorithm that runs in <code>O(m + n)</code> time?</p>

class Solution {
public:
  static string minWindow(const string &str, const string &pattern) {
    array<int, 58> required{};
    for (const char letter : pattern) {
      ++required[letter - 'A'];
    }

    const auto not_enough = [&]() -> bool {
      for (array<int, 58>::size_type i = 0; i < 58; ++i) {
        if (required[i] > 0) {
          return true;
        }
      }
      return false;
    };

    string::size_type min_dist = numeric_limits<string::size_type>::max();
    string::const_iterator min_first, min_last;
    for (auto first = str.cbegin(), last = str.cbegin(); ; ++first) {
      while (not_enough() && last != str.cend()) {
        --required[*last - 'A'];
        ++last;
      }
      if (not_enough()) {
        return {min_first, min_last};
      }

      const string::size_type dist = last - first;
      if (dist < min_dist) {
        min_dist = dist;
        min_first = first;
        min_last = last;
      }
      ++required[*first - 'A'];
    }
    return {min_first, min_last};
  }
};
9200