<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:
    string minWindow(string& s, string& t) {
        if (s.size() < t.size()) return "";
        short counts[58] = {0};
        for (char ch : t) counts[ch-'A']++;
        short mismatches = 0;
        for (int c : counts) if (c != 0) mismatches++;
        int i = 0, j = 0, ansi = 0;
        for (; j < s.size(); j++) {
            if (--counts[s[j]-'A'] == 0) {
                mismatches--;
                if (mismatches == 0) break;
            }
        }
        if (mismatches != 0) return "";
        for (; i < s.size(); i++) {
            if (counts[s[i]-'A']++ == 0) {
                mismatches++;
                for (j++, i++; j < s.size(); j++, i++) {
                    if (--counts[s[j]-'A'] == 0) {
                        mismatches--;
                    }
                    if (counts[s[i]-'A']++ == 0) {
                        mismatches++;
                    }
                    if (mismatches == 0) break;
                }
            }
            if (mismatches == 0) ansi = i + 1;
            else break;
        }

        return s.substr(ansi, j - i + 1);
    }
};
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