<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) {
        string ans="";
    if (t.size() > s.size()) return "";
    vector<int> hash(256,0);
    int l =0;
    int r=0;
    int m = s.size();
    int n = t.size();
    int minl = 1e5;
    int cnt  = 0;
    int si=-1;

    for(int i = 0 ;i<n; i++ ){
        hash[t[i]]++;
    }
    while(r<m){
        if(hash[s[r]]>0){
            cnt++;
            }
            
            hash[s[r]]--;
        
        while(cnt==n){
            if(r-l+1 < minl){
                minl = r-l+1;
            si = l;
            }
            hash[s[l]]++;
            if(hash[s[l]]>0){
                cnt--;
            }
            l++;
        }
        r++;
    }
    if(si==-1){
        return ans;
    }
    return s.substr(si,minl);

    }
};
10200