<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.empty()||t.empty()){
            return "";
        }
        unordered_map<char,int>required;
        for(char c : t){
            required[c]++;
        }
        int requiredCount = required.size();

        int left = 0;
        int right = 0;
        int formed = 0;
        unordered_map<char,int>window;


        int minLength = INT_MAX;
        int minLeft = 0;
        int minRight = 0;
        while(right < s.size()){
            char c = s[right];
            window[c]++;
            if(required.count(c)&&window[c]==required[c]){
                formed++;
            }
            while(left <= right && formed == requiredCount){
                if(right - left + 1 < minLength){
                    minLength = right - left + 1;
                    minLeft = left;
                    minRight = right;
                }

                char leftChar = s[left];
                window[leftChar]--;
                if(required.count(leftChar)&&window[leftChar]<required[leftChar]){
                    formed--;
                }
                left++;
            }
            right++;
        }
        return (minLength==INT_MAX)? "": s.substr(minLeft,minLength);
    }
};
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