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Description
tab: English
<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>""</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>
> [!tip]- Hint 1
>
> Use two pointers to create a window of letters in s, which would have all the characters from t.
> [!tip]- Hint 2
>
> Expand the right pointer until all the characters of t are covered.
> [!tip]- Hint 3
>
> Once all the characters are covered, move the left pointer and ensure that all the characters are still covered to minimize the subarray size.
> [!tip]- Hint 4
>
> Continue expanding the right and left pointers until you reach the end of s.
---
[submissions](https://leetcode.com/problems/minimum-window-substring/submissions/) | [solutions](https://leetcode.com/problems/minimum-window-substring/solutions/)
tab: 中文
<p>给你一个字符串 <code>s</code> 、一个字符串 <code>t</code> 。返回 <code>s</code> 中涵盖 <code>t</code> 所有字符的最小子串。如果 <code>s</code> 中不存在涵盖 <code>t</code> 所有字符的子串，则返回空字符串 <code>""</code> 。</p>
<p> </p>
<p><strong>注意：</strong></p>
<ul>
<li>对于 <code>t</code> 中重复字符，我们寻找的子字符串中该字符数量必须不少于 <code>t</code> 中该字符数量。</li>
<li>如果 <code>s</code> 中存在这样的子串，我们保证它是唯一的答案。</li>
</ul>
<p> </p>
<p><strong>示例 1：</strong></p>
<pre>
<strong>输入：</strong>s = "ADOBECODEBANC", t = "ABC"
<strong>输出：</strong>"BANC"
<strong>解释：</strong>最小覆盖子串 "BANC" 包含来自字符串 t 的 'A'、'B' 和 'C'。
</pre>
<p><strong>示例 2：</strong></p>
<pre>
<strong>输入：</strong>s = "a", t = "a"
<strong>输出：</strong>"a"
<strong>解释：</strong>整个字符串 s 是最小覆盖子串。
</pre>
<p><strong>示例 3:</strong></p>
<pre>
<strong>输入:</strong> s = "a", t = "aa"
<strong>输出:</strong> ""
<strong>解释:</strong> t 中两个字符 'a' 均应包含在 s 的子串中，
因此没有符合条件的子字符串，返回空字符串。</pre>
<p> </p>
<p><strong>提示：</strong></p>
<ul>
<li><code><sup>m == s.length</sup></code></li>
<li><code><sup>n == t.length</sup></code></li>
<li><code>1 <= m, n <= 10<sup>5</sup></code></li>
<li><code>s</code> 和 <code>t</code> 由英文字母组成</li>
</ul>
<p> </p>
<strong>进阶：</strong>你能设计一个在 <code>o(m+n)</code> 时间内解决此问题的算法吗？
> [!tip]- 提示 1
>
> Use two pointers to create a window of letters in s, which would have all the characters from t.
> [!tip]- 提示 2
>
> Expand the right pointer until all the characters of t are covered.
> [!tip]- 提示 3
>
> Once all the characters are covered, move the left pointer and ensure that all the characters are still covered to minimize the subarray size.
> [!tip]- 提示 4
>
> Continue expanding the right and left pointers until you reach the end of s.
---
[提交记录](https://leetcode.cn/problems/minimum-window-substring/submissions/) | [题解](https://leetcode.cn/problems/minimum-window-substring/solution/)
Solutions & Notes
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