minimum-window-substring

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Algorithms

Given two strings <code>s</code> and <code>t</code> of lengths <code>m</code> and <code>n</code> respectively, return the minimum window substring of <code>s</code> such that every character in <code>t</code> (including duplicates) is included in the window. If there is no such substring, return the empty string <code>&quot;&quot;</code>.

The testcases will be generated such that the answer is unique.

&nbsp;
Example 1:

<pre>
Input: s = &quot;ADOBECODEBANC&quot;, t = &quot;ABC&quot;
Output: &quot;BANC&quot;
Explanation: The minimum window substring &quot;BANC&quot; includes &#39;A&#39;, &#39;B&#39;, and &#39;C&#39; from string t.
</pre>

Example 2:

<pre>
Input: s = &quot;a&quot;, t = &quot;a&quot;
Output: &quot;a&quot;
Explanation: The entire string s is the minimum window.
</pre>

Example 3:

<pre>
Input: s = &quot;a&quot;, t = &quot;aa&quot;
Output: &quot;&quot;
Explanation: 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>

&nbsp;
Constraints:

<ul>
	<li><code>m == s.length</code></li>
	<li><code>n == t.length</code></li>
	<li><code>1 &lt;= m, n &lt;= 105</code></li>
	<li><code>s</code> and <code>t</code> consist of uppercase and lowercase English letters.</li>
</ul>

&nbsp;
Follow up: Could you find an algorithm that runs in <code>O(m + n)</code> time?

minWindow

Minimum Window Substring

Substring with Concatenation of All Words

substring-with-concatenation-of-all-words

串联所有单词的子串

Minimum Size Subarray Sum

minimum-size-subarray-sum

长度最小的子数组

Sliding Window Maximum

sliding-window-maximum

滑动窗口最大值

Permutation in String

permutation-in-string

字符串的排列

Smallest Range Covering Elements from K Lists

smallest-range-covering-elements-from-k-lists

最小区间

Minimum Window Subsequence

minimum-window-subsequence

最小窗口子序列

Count Substrings That Can Be Rearranged to Contain a String II

count-substrings-that-can-be-rearranged-to-contain-a-string-ii

统计重新排列后包含另一个字符串的子字符串数目 II

Count Substrings That Can Be Rearranged to Contain a String I

count-substrings-that-can-be-rearranged-to-contain-a-string-i

统计重新排列后包含另一个字符串的子字符串数目 I

Hash Table

String

Sliding Window

&#32473;&#23450;&#20004;&#20010;&#23383;&#31526;&#20018;&nbsp;<code>s</code> &#21644;&nbsp;<code>t</code>&#65292;&#38271;&#24230;&#20998;&#21035;&#26159;&nbsp;<code>m</code> &#21644;&nbsp;<code>n</code>&#65292;&#36820;&#22238; s &#20013;&#30340;&nbsp;&#26368;&#30701;&#31383;&#21475; &#23376;&#20018;&#65292;&#20351;&#24471;&#35813;&#23376;&#20018;&#21253;&#21547; <code>t</code> &#20013;&#30340;&#27599;&#19968;&#20010;&#23383;&#31526;&#65288;&#21253;&#25324;&#37325;&#22797;&#23383;&#31526;&#65289;&#12290;&#22914;&#26524;&#27809;&#26377;&#36825;&#26679;&#30340;&#23376;&#20018;&#65292;&#36820;&#22238;&#31354;&#23383;&#31526;&#20018;&nbsp;<code>""</code>&#12290;

&#27979;&#35797;&#29992;&#20363;&#20445;&#35777;&#31572;&#26696;&#21807;&#19968;&#12290;

&nbsp;

&#31034;&#20363; 1&#65306;

<pre>
&#36755;&#20837;&#65306;s = "ADOBECODEBANC", t = "ABC"
&#36755;&#20986;&#65306;"BANC"
&#35299;&#37322;&#65306;&#26368;&#23567;&#35206;&#30422;&#23376;&#20018; "BANC" &#21253;&#21547;&#26469;&#33258;&#23383;&#31526;&#20018; t &#30340; 'A'&#12289;'B' &#21644; 'C'&#12290;
</pre>

&#31034;&#20363; 2&#65306;

<pre>
&#36755;&#20837;&#65306;s = "a", t = "a"
&#36755;&#20986;&#65306;"a"
&#35299;&#37322;&#65306;&#25972;&#20010;&#23383;&#31526;&#20018; s &#26159;&#26368;&#23567;&#35206;&#30422;&#23376;&#20018;&#12290;
</pre>

&#31034;&#20363; 3:

<pre>
&#36755;&#20837;: s = "a", t = "aa"
&#36755;&#20986;: ""
&#35299;&#37322;: t &#20013;&#20004;&#20010;&#23383;&#31526; 'a' &#22343;&#24212;&#21253;&#21547;&#22312; s &#30340;&#23376;&#20018;&#20013;&#65292;
&#22240;&#27492;&#27809;&#26377;&#31526;&#21512;&#26465;&#20214;&#30340;&#23376;&#23383;&#31526;&#20018;&#65292;&#36820;&#22238;&#31354;&#23383;&#31526;&#20018;&#12290;</pre>

&nbsp;

&#25552;&#31034;&#65306;

<ul>
	<li><code>m == s.length</code></li>
	<li><code>n == t.length</code></li>
	<li><code>1 &lt;= m, n &lt;= 105</code></li>
	<li><code>s</code> &#21644; <code>t</code> &#30001;&#33521;&#25991;&#23383;&#27597;&#32452;&#25104;</li>
</ul>

&nbsp;
&#36827;&#38454;&#65306;&#20320;&#33021;&#35774;&#35745;&#19968;&#20010;&#22312; <code>O(m + n)</code> &#26102;&#38388;&#20869;&#35299;&#20915;&#27492;&#38382;&#39064;&#30340;&#31639;&#27861;&#21527;&#65311;

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