<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) {
      int n=s.length();
      if(s.length()<t.length())return "";
      int i=0, j=0;
      string ans="";
      int len=n;
      unordered_map<char,int>mp;
      for(auto it : t){
        mp[it]++;
      }
      int k=t.length();
      int count=mp.size();
      while(j<n){
        if(mp.find(s[j])!=mp.end()){
          mp[s[j]]--;
          if(mp[s[j]]==0)count--;
        }
        if(count>0){
          j++;
        }
        else if(count==0){
          while(i<j){
            if(mp.find(s[i])==mp.end())i++;
            else if(mp.find(s[i])!=mp.end() and mp[s[i]]<0){
              mp[s[i]]++;
              i++;
            }
            else if(mp.find(s[i])!=mp.end() and mp[s[i]]==0){
              break;
            }
          }
          if(len>=j-i+1){
            ans=s.substr(i,j-i+1);
            len=j-i+1;
          }
          j++;
        }
      }
      return ans;
    }
};
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