<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) {
        unordered_map<char, int> charCnt;
        int num = 0;

        for(auto c : t){
            if(charCnt.find(c) == charCnt.end()) num++;
            charCnt[c]++;
        }

        int n = s.length();
        int l = 0;
        int r = 0;
        int min = n;
        string ans = "";

        while(r < n){
            while(num != 0 && r < n){
                if(charCnt.find(s[r]) != charCnt.end()){
                    charCnt[s[r]]--;
                    if(charCnt[s[r]] == 0) num -= 1;
                }
                r++;
            }

            while(num == 0 && l <= r){
                if(charCnt.find(s[l]) != charCnt.end()){
                    charCnt[s[l]]++;
                    if(charCnt[s[l]] == 1) num += 1;
                }
                l++;
            }

            if((r - l + 1) <= min){
                min = r - l + 1;
                ans = s.substr(l - 1, min);
            }
        }

        return ans;
    }
};
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