<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) {
        const int m = s.size();
        const int n = t.size();

        if (m < n)
            return "";

        map<char, int> letterCnt;
        map<char, int> currentLetterCnt;

        for (const char c : t) {
            ++letterCnt[c];
        }
        // string index = ( i ... j ]
        int i = 0;
        int j = 0;
        int restCnt = n;
        int minL = m+1;
        int minI = i;

        while (j < m) {
            while (restCnt && j < m) {
                const int jc = s[j];

                if (0 != letterCnt[jc]) {
                    ++currentLetterCnt[jc];
                    if (currentLetterCnt[jc] <= letterCnt[jc]) 
                        --restCnt;
                }
                ++j;
            } 
            if (restCnt)
                break;
            while (i <= j) {
                if (minL > j-i) {
                    minL = j-i;
                    minI = i;
                }

                const int ic = s[i++];
                if (currentLetterCnt[ic]) {
                    --currentLetterCnt[ic];
                    if (currentLetterCnt[ic] < letterCnt[ic]) {
                        ++restCnt;
                        break;
                    }
                    
                }
            }
        }

        if (minL > m) {
            return "";
        }
        
        return s.substr(minI, minL);
    }
};
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