<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:
    bool isContained(unordered_map<char, int>& smap, unordered_map<char, int>& tmap) {
        for (auto& e: tmap) {
            if (smap.find(e.first) != smap.end()) {
                //cout << "Char: " << e.first << ", " << smap[e.first] << ", " << tmap[e.first] << endl; 
                if (smap[e.first] < tmap[e.first])
                    return false;
            }
            else return false;
        }
        return true;
    }
    string minWindow(string s, string t) {
        unordered_map<char, int> tmap;
        unordered_map<char, int> smap;
        for (int i = 0; i < t.size(); i++) {
            tmap[t[i]]++;
        }
        int tsz = t.size();
        int i = 0, j = 0, result = s.size();
        string str = "";
        bool contained = false;
        while (i < s.size()) {
            smap[s[i]]++;
            while (j < s.size() && smap[s[j]] > tmap[s[j]]) {
                smap[s[j++]]--;
            }
            contained = isContained(smap, tmap);
            if (contained) {
                //cout << "CurrSize:" << i-j+1 << ", " << result << endl; 
                if (i-j+1 <= result) {
                    result = min(result, i-j+1);
                    str = s.substr(j, i-j+1);
                }
            }
            i++;
        }
        contained = isContained(smap, tmap);
        if (contained) {
            //cout << "CurrSize:" << i-j << ", " << result << endl; 
            if (i-j <= result) {
                result = min(result, i-j);
                str = s.substr(j, i-j);
            }
        }
        return str;
    }
};
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