<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) {
        vector<int> buckets = vector<int>(256);
        int numUniques = 0;
        
        for(char c : t)
        {
            if(buckets[c] == 0)
                numUniques++;
            
            buckets[c]++;
        }
        
        int start = 0;
        int end = 0;
        
        int minWindowLen=-1;
        int minWindowIdx=-1;
        
        while(end < s.size())
        {
            buckets[s[end]]--;
            if(buckets[s[end]] == 0)
                numUniques--;
            
            while(numUniques == 0) // Seen all needed characters
            {
                buckets[s[start]] ++;
                if(buckets[s[start]] > 0)
                    numUniques++;
                
                if(numUniques != 0)
                {
                    if(minWindowLen == -1 || end - start + 1 < minWindowLen)
                    {
                        minWindowLen = end - start + 1;
                        minWindowIdx = start;
                    }
                }
                start++;
            }

            end++;
        }
        
        if(minWindowLen != -1)
            return s.substr(minWindowIdx, minWindowLen);

        return "";
    }
};
10000