# [76. Minimum Window Substring (Hard)](https://leetcode.com/problems/minimum-window-substring)
<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" class=" cursor-pointer relative text-dark-blue-s text-sm"><div class="popover-wrapper inline-block" data-headlessui-state=""><div><div id="headlessui-popover-button-:r7s:" aria-expanded="false" data-headlessui-state=""><strong><em>substring</em></strong></div></div></div></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>""</code>.</p>
<p>The testcases will be generated such that the answer is <strong>unique</strong>.</p>
<p> </p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> s = "ADOBECODEBANC", t = "ABC"
<strong>Output:</strong> "BANC"
<strong>Explanation:</strong> The minimum window substring "BANC" includes 'A', 'B', and 'C' from string t.
</pre>
<p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> s = "a", t = "a"
<strong>Output:</strong> "a"
<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 = "a", t = "aa"
<strong>Output:</strong> ""
<strong>Explanation:</strong> Both 'a's from t must be included in the window.
Since the largest window of s only has one 'a', return empty string.
</pre>
<p> </p>
<p><strong>Constraints:</strong></p>
<ul>
<li><code>m == s.length</code></li>
<li><code>n == t.length</code></li>
<li><code>1 <= m, n <= 10<sup>5</sup></code></li>
<li><code>s</code> and <code>t</code> consist of uppercase and lowercase English letters.</li>
</ul>
<p> </p>
<p><strong>Follow up:</strong> Could you find an algorithm that runs in <code>O(m + n)</code> time?</p>
**Related Topics**:
[Hash Table](https://leetcode.com/tag/hash-table/), [String](https://leetcode.com/tag/string/), [Sliding Window](https://leetcode.com/tag/sliding-window/)
**Similar Questions**:
* [Substring with Concatenation of All Words (Hard)](https://leetcode.com/problems/substring-with-concatenation-of-all-words/)
* [Minimum Size Subarray Sum (Hard)](https://leetcode.com/problems/minimum-size-subarray-sum/)
* [Sliding Window Maximum (Hard)](https://leetcode.com/problems/sliding-window-maximum/)
* [Permutation in String (Hard)](https://leetcode.com/problems/permutation-in-string/)
* [Smallest Range Covering Elements from K Lists (Hard)](https://leetcode.com/problems/smallest-range-covering-elements-from-k-lists/)
* [Minimum Window Subsequence (Hard)](https://leetcode.com/problems/minimum-window-subsequence/)
## Solution 1. Minimum Sliding Window
```cpp
// OJ: https://leetcode.com/problems/minimum-window-substring/
// Author: github.com/lzl124631x
// Time: O(M+N)
// Space: O(C) where C is the range of characters
class Solution {
public:
string minWindow(string s, string t) {
int M = s.size(), N = t.size(), cnt[128] = {}, matched = 0, minLen =
INT_MAX, start = -1;
for (char c : t) cnt[c]++;
for (int i = 0, j = 0; j < M; ++j) {
matched += cnt[s[j]]-- > 0;
while (matched >= N) {
if (j - i + 1 < minLen) {
minLen = j - i + 1;
start = i;
}
matched -= ++cnt[s[i++]] > 0;
}
}
return start == -1 ? "" : s.substr(start, minLen);
}
};
```
