<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 ls = s.size(), lt = t.size();
        map<char, int> my;
        for(char ch : t) my[ch]++;
        int i = 0;
        string res = "";
        for(int j=0; j<=ls; j++) {
            char ch = s[j];
            if(my.find(ch) == my.end()) continue;
            my[ch]--;
            if(my[ch] >= 0) lt--;
            if(lt == 0) {
                while(i < j) {
                    if(my.find(s[i]) == my.end()) i++;
                    else if(my[s[i]] < 0) {
                        my[s[i]]++;
                        i++;
                    }
                    else break;
                }
                if(res == "" || res.size() > (j-i+1))
                    res = s.substr(i, j-i+1);
            }
        }
        return res;
    }

    string minWindow2(string s, string t) {
        if(s=="" || t=="") return "";
        string res = "";
        map<char, int> my;
        for(char c : t) my[c]++;
        int cnt = t.size();
        for(int i=0, start=0; i<=s.size(); i++) {
            if(my.find(s[i]) == my.end() ) continue;
            my[s[i]]--;
            if(my[s[i]] >= 0) {
                cnt--;
            }
            if(cnt==0) {
                while(start<i) {
                    if(my.find(s[start])==my.end() ) start++;
                    else if(my[s[start]] < 0) {
                        my[s[start]]++;
                        start++;
                    }
                    else break;
                }
                if(res=="" || res.size() > (i-start+1))
                    res = s.substr(start, i-start+1);
            }
        }
        return res;
    }
};
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