<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 (t.length() == 0) return "";
  map < char, int > mpt;
  map < char, int > mps;
  for (int i = 0; i < t.length(); i++) {
    mpt[t[i]]++;
  }
  int cur = 0, total = mpt.size();
  int i = 0, j = 0;
  int res = INT_MAX;
  pair < int, int > indices;
  while (j < s.length()) {
    if (mpt.find(s[j]) != mpt.end()) mps[s[j]]++;
    if (mpt.find(s[j]) != mpt.end() && mps[s[j]] == mpt[s[j]]) cur++;

    while (cur == total) {
      if (j - i + 1 < res) {
        res = j - i + 1;

        indices = make_pair(i, j);
      }
      if (mps.find(s[i]) != mps.end()) {
        mps[s[i]]--;
        if (mps.find(s[i]) != mps.end() && mps[s[i]] < mpt[s[i]]) cur--;
      }
      i++;
    }
    j++;
  }
  if (res == INT_MAX) return "";
  string str = "";
  for (int i = indices.first; i <= indices.second; i++) {
    str += s[i];
  }
  return str;
    }
};
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