feat: add solutions to lc problems: No.2446~2449

* No.2446.Determine if Two Events Have Conflict
* No.2447.Number of Subarrays With GCD Equal to K
* No.2448.Minimum Cost to Make Array Equal
* No.2449.Minimum Number of Operations to Make Arrays Similar

Author: yanglbme

lcci/03.03.Stack of Plates/README_EN.md

@@ -86,7 +86,7 @@ class StackOfPlates {
     public StackOfPlates(int cap) {
         this.cap = cap;
     }
-    
+
     public void push(int val) {
         if (cap == 0) {
             return;
@@ -96,11 +96,11 @@ class StackOfPlates {
         }
         stk.get(stk.size() - 1).push(val);
     }
-    
+
     public int pop() {
         return popAt(stk.size() - 1);
     }
-    
+
     public int popAt(int index) {
         int ans = -1;
         if (index >= 0 && index < stk.size()) {

solution/0000-0099/0003.Longest Substring Without Repeating Characters/README_EN.md

@@ -4,26 +4,26 @@
 
 ## Description
 
-<p>Given a string <code>s</code>, find the length of the <strong>longest substring</strong> without repeating characters.</p>
+<p>Given a string <code>s</code>, find the length of the <strong>longest <span data-keyword="substring-nonempty">substring</span></strong> without repeating characters.</p>
 
 <p>&nbsp;</p>
-<p><strong class="example">Example 1:</strong></p>
+<p><strong>Example 1:</strong></p>
 
 <pre>
 <strong>Input:</strong> s = &quot;abcabcbb&quot;
 <strong>Output:</strong> 3
 <strong>Explanation:</strong> The answer is &quot;abc&quot;, with the length of 3.
 </pre>
 
-<p><strong class="example">Example 2:</strong></p>
+<p><strong>Example 2:</strong></p>
 
 <pre>
 <strong>Input:</strong> s = &quot;bbbbb&quot;
 <strong>Output:</strong> 1
 <strong>Explanation:</strong> The answer is &quot;b&quot;, with the length of 1.
 </pre>
 
-<p><strong class="example">Example 3:</strong></p>
+<p><strong>Example 3:</strong></p>
 
 <pre>
 <strong>Input:</strong> s = &quot;pwwkew&quot;

solution/0000-0099/0039.Combination Sum/README.md

@@ -6,7 +6,7 @@
 
 <!-- 这里写题目描述 -->
 
-<p>给你一个 <strong>无重复元素</strong> 的整数数组&nbsp;<code>candidates</code> 和一个目标整数&nbsp;<code>target</code>&nbsp;，找出&nbsp;<code>candidates</code>&nbsp;中可以使数字和为目标数&nbsp;<code>target</code> 的 <em>所有&nbsp;</em><strong>不同组合</strong> ，并以列表形式返回。你可以按 <strong>任意顺序</strong> 返回这些组合。</p>
+<p>给你一个 <strong>无重复元素</strong> 的整数数组&nbsp;<code>candidates</code> 和一个目标整数&nbsp;<code>target</code>&nbsp;，找出&nbsp;<code>candidates</code>&nbsp;中可以使数字和为目标数&nbsp;<code>target</code> 的 所有<em>&nbsp;</em><strong>不同组合</strong> ，并以列表形式返回。你可以按 <strong>任意顺序</strong> 返回这些组合。</p>
 
 <p><code>candidates</code> 中的 <strong>同一个</strong> 数字可以 <strong>无限制重复被选取</strong> 。如果至少一个数字的被选数量不同，则两种组合是不同的。&nbsp;</p>
 
@@ -43,9 +43,9 @@
 
 <ul>
 	<li><code>1 &lt;= candidates.length &lt;= 30</code></li>
-	<li><code>1 &lt;= candidates[i] &lt;= 200</code></li>
-	<li><code>candidate</code> 中的每个元素都 <strong>互不相同</strong></li>
-	<li><code>1 &lt;= target &lt;= 500</code></li>
+	<li><code>2 &lt;= candidates[i] &lt;= 40</code></li>
+	<li><code>candidates</code> 的所有元素 <strong>互不相同</strong></li>
+	<li><code>1 &lt;= target &lt;= 40</code></li>
 </ul>
 
 ## 解法

solution/0100-0199/0140.Word Break II/README_EN.md

@@ -9,22 +9,22 @@
 <p><strong>Note</strong> that the same word in the dictionary may be reused multiple times in the segmentation.</p>
 
 <p>&nbsp;</p>
-<p><strong class="example">Example 1:</strong></p>
+<p><strong>Example 1:</strong></p>
 
 <pre>
 <strong>Input:</strong> s = &quot;catsanddog&quot;, wordDict = [&quot;cat&quot;,&quot;cats&quot;,&quot;and&quot;,&quot;sand&quot;,&quot;dog&quot;]
 <strong>Output:</strong> [&quot;cats and dog&quot;,&quot;cat sand dog&quot;]
 </pre>
 
-<p><strong class="example">Example 2:</strong></p>
+<p><strong>Example 2:</strong></p>
 
 <pre>
 <strong>Input:</strong> s = &quot;pineapplepenapple&quot;, wordDict = [&quot;apple&quot;,&quot;pen&quot;,&quot;applepen&quot;,&quot;pine&quot;,&quot;pineapple&quot;]
 <strong>Output:</strong> [&quot;pine apple pen apple&quot;,&quot;pineapple pen apple&quot;,&quot;pine applepen apple&quot;]
 <strong>Explanation:</strong> Note that you are allowed to reuse a dictionary word.
 </pre>
 
-<p><strong class="example">Example 3:</strong></p>
+<p><strong>Example 3:</strong></p>
 
 <pre>
 <strong>Input:</strong> s = &quot;catsandog&quot;, wordDict = [&quot;cats&quot;,&quot;dog&quot;,&quot;sand&quot;,&quot;and&quot;,&quot;cat&quot;]

solution/0100-0199/0141.Linked List Cycle/README_EN.md

@@ -11,23 +11,23 @@
 <p>Return&nbsp;<code>true</code><em> if there is a cycle in the linked list</em>. Otherwise, return <code>false</code>.</p>
 
 <p>&nbsp;</p>
-<p><strong class="example">Example 1:</strong></p>
+<p><strong>Example 1:</strong></p>
 <img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0100-0199/0141.Linked%20List%20Cycle/images/circularlinkedlist.png" style="width: 300px; height: 97px; margin-top: 8px; margin-bottom: 8px;" />
 <pre>
 <strong>Input:</strong> head = [3,2,0,-4], pos = 1
 <strong>Output:</strong> true
 <strong>Explanation:</strong> There is a cycle in the linked list, where the tail connects to the 1st node (0-indexed).
 </pre>
 
-<p><strong class="example">Example 2:</strong></p>
+<p><strong>Example 2:</strong></p>
 <img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0100-0199/0141.Linked%20List%20Cycle/images/circularlinkedlist_test2.png" style="width: 141px; height: 74px;" />
 <pre>
 <strong>Input:</strong> head = [1,2], pos = 0
 <strong>Output:</strong> true
 <strong>Explanation:</strong> There is a cycle in the linked list, where the tail connects to the 0th node.
 </pre>
 
-<p><strong class="example">Example 3:</strong></p>
+<p><strong>Example 3:</strong></p>
 <img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0100-0199/0141.Linked%20List%20Cycle/images/circularlinkedlist_test3.png" style="width: 45px; height: 45px;" />
 <pre>
 <strong>Input:</strong> head = [1], pos = -1

solution/0100-0199/0162.Find Peak Element/README_EN.md

@@ -13,14 +13,14 @@
 <p>You must write an algorithm that runs in <code>O(log n)</code> time.</p>
 
 <p>&nbsp;</p>
-<p><strong class="example">Example 1:</strong></p>
+<p><strong>Example 1:</strong></p>
 
 <pre>
 <strong>Input:</strong> nums = [1,2,3,1]
 <strong>Output:</strong> 2
 <strong>Explanation:</strong> 3 is a peak element and your function should return the index number 2.</pre>
 
-<p><strong class="example">Example 2:</strong></p>
+<p><strong>Example 2:</strong></p>
 
 <pre>
 <strong>Input:</strong> nums = [1,2,1,3,5,6,4]

solution/0200-0299/0219.Contains Duplicate II/README_EN.md

@@ -7,21 +7,21 @@
 <p>Given an integer array <code>nums</code> and an integer <code>k</code>, return <code>true</code> if there are two <strong>distinct indices</strong> <code>i</code> and <code>j</code> in the array such that <code>nums[i] == nums[j]</code> and <code>abs(i - j) &lt;= k</code>.</p>
 
 <p>&nbsp;</p>
-<p><strong class="example">Example 1:</strong></p>
+<p><strong>Example 1:</strong></p>
 
 <pre>
 <strong>Input:</strong> nums = [1,2,3,1], k = 3
 <strong>Output:</strong> true
 </pre>
 
-<p><strong class="example">Example 2:</strong></p>
+<p><strong>Example 2:</strong></p>
 
 <pre>
 <strong>Input:</strong> nums = [1,0,1,1], k = 1
 <strong>Output:</strong> true
 </pre>
 
-<p><strong class="example">Example 3:</strong></p>
+<p><strong>Example 3:</strong></p>
 
 <pre>
 <strong>Input:</strong> nums = [1,2,3,1,2,3], k = 2

solution/0400-0499/0411.Minimum Unique Word Abbreviation/README_EN.md

@@ -21,7 +21,7 @@
 <p>Given a target string <code>target</code> and an array of strings <code>dictionary</code>, return <em>an <strong>abbreviation</strong> of </em><code>target</code><em> with the <strong>shortest possible length</strong> such that it is <strong>not an abbreviation</strong> of <strong>any</strong> string in </em><code>dictionary</code><em>. If there are multiple shortest abbreviations, return any of them</em>.</p>
 
 <p>&nbsp;</p>
-<p><strong class="example">Example 1:</strong></p>
+<p><strong>Example 1:</strong></p>
 
 <pre>
 <strong>Input:</strong> target = &quot;apple&quot;, dictionary = [&quot;blade&quot;]
@@ -31,7 +31,7 @@ The next shortest abbreviations are &quot;a4&quot; and &quot;4e&quot;. &quot;4e&
 Hence, return &quot;a4&quot;.
 </pre>
 
-<p><strong class="example">Example 2:</strong></p>
+<p><strong>Example 2:</strong></p>
 
 <pre>
 <strong>Input:</strong> target = &quot;apple&quot;, dictionary = [&quot;blade&quot;,&quot;plain&quot;,&quot;amber&quot;]
@@ -54,6 +54,7 @@ Since none of them are abbreviations of words in the dictionary, returning any o
 	<li><code>1 &lt;= dictionary[i].length &lt;= 100</code></li>
 	<li><code>log<sub>2</sub>(n) + m &lt;= 21</code> if <code>n &gt; 0</code></li>
 	<li><code>target</code> and <code>dictionary[i]</code> consist of lowercase English letters.</li>
+	<li><code>dictionary</code> does not contain <code>target</code>.</li>
 </ul>
 
 ## Solutions

solution/0700-0799/0744.Find Smallest Letter Greater Than Target/README_EN.md

@@ -14,15 +14,15 @@
 <pre>
 <strong>Input:</strong> letters = [&quot;c&quot;,&quot;f&quot;,&quot;j&quot;], target = &quot;a&quot;
 <strong>Output:</strong> &quot;c&quot;
-<strong>Explanation:</strong> The smallest character that is lexicogrpahically greater than &#39;a&#39; in letters is &#39;c&#39;.
+<strong>Explanation:</strong> The smallest character that is lexicographically greater than &#39;a&#39; in letters is &#39;c&#39;.
 </pre>
 
 <p><strong class="example">Example 2:</strong></p>
 
 <pre>
 <strong>Input:</strong> letters = [&quot;c&quot;,&quot;f&quot;,&quot;j&quot;], target = &quot;c&quot;
 <strong>Output:</strong> &quot;f&quot;
-<strong>Explanation:</strong> The smallest character that is lexicogrpahically greater than &#39;c&#39; in letters is &#39;f&#39;.
+<strong>Explanation:</strong> The smallest character that is lexicographically greater than &#39;c&#39; in letters is &#39;f&#39;.
 </pre>
 
 <p><strong class="example">Example 3:</strong></p>

solution/0700-0799/0786.K-th Smallest Prime Fraction/README.md

@@ -43,6 +43,10 @@
 	<li><code>1 &lt;= k &lt;= arr.length * (arr.length - 1) / 2</code></li>
 </ul>
 
+<p>&nbsp;</p>
+
+<p><strong>进阶：</strong>你可以设计并实现时间复杂度小于 <code>O(n<sup>2</sup>)</code> 的算法解决此问题吗？</p>
+
 ## 解法
 
 <!-- 这里可写通用的实现逻辑 -->

solution/0900-0999/0901.Online Stock Span/README.md

@@ -122,7 +122,7 @@ public:
     StockSpanner() {
 
     }
-    
+
     int next(int price) {
         int cnt = 1;
         while (!stk.empty() && stk.top().first <= price) {

solution/0900-0999/0901.Online Stock Span/README_EN.md

@@ -107,7 +107,7 @@ public:
     StockSpanner() {
 
     }
-    
+
     int next(int price) {
         int cnt = 1;
         while (!stk.empty() && stk.top().first <= price) {

solution/1300-1399/1312.Minimum Insertion Steps to Make a String Palindrome/README_EN.md

@@ -11,23 +11,23 @@
 <p>A&nbsp;<b>Palindrome String</b>&nbsp;is one that reads the same backward as well as forward.</p>
 
 <p>&nbsp;</p>
-<p><strong class="example">Example 1:</strong></p>
+<p><strong>Example 1:</strong></p>
 
 <pre>
 <strong>Input:</strong> s = &quot;zzazz&quot;
 <strong>Output:</strong> 0
-<strong>Explanation:</strong> The string &quot;zzazz&quot; is already palindrome we don&#39;t need any insertions.
+<strong>Explanation:</strong> The string &quot;zzazz&quot; is already palindrome we do not need any insertions.
 </pre>
 
-<p><strong class="example">Example 2:</strong></p>
+<p><strong>Example 2:</strong></p>
 
 <pre>
 <strong>Input:</strong> s = &quot;mbadm&quot;
 <strong>Output:</strong> 2
 <strong>Explanation:</strong> String can be &quot;mbdadbm&quot; or &quot;mdbabdm&quot;.
 </pre>
 
-<p><strong class="example">Example 3:</strong></p>
+<p><strong>Example 3:</strong></p>
 
 <pre>
 <strong>Input:</strong> s = &quot;leetcode&quot;

solution/1700-1799/1755.Closest Subsequence Sum/README.md

@@ -110,7 +110,7 @@ class Solution:
                 return
             dfs(arr, res, i + 1, s)
             dfs(arr, res, i + 1, s + arr[i])
-        
+
         n = len(nums)
         left, right = set(), set()
         dfs(nums[: n >> 1], left, 0, 0)

solution/2400-2499/2434.Using a Robot to Print the Lexicographically Smallest String/README_EN.md

@@ -14,7 +14,7 @@
 <p>Return <em>the lexicographically smallest string that can be written on the paper.</em></p>
 
 <p>&nbsp;</p>
-<p><strong class="example">Example 1:</strong></p>
+<p><strong>Example 1:</strong></p>
 
 <pre>
 <strong>Input:</strong> s = &quot;zza&quot;
@@ -25,7 +25,7 @@ Perform first operation three times p=&quot;&quot;, s=&quot;&quot;, t=&quot;zza&
 Perform second operation three times p=&quot;azz&quot;, s=&quot;&quot;, t=&quot;&quot;.
 </pre>
 
-<p><strong class="example">Example 2:</strong></p>
+<p><strong>Example 2:</strong></p>
 
 <pre>
 <strong>Input:</strong> s = &quot;bac&quot;
@@ -37,7 +37,7 @@ Perform first operation p=&quot;ab&quot;, s=&quot;&quot;, t=&quot;c&quot;.
 Perform second operation p=&quot;abc&quot;, s=&quot;&quot;, t=&quot;&quot;.
 </pre>
 
-<p><strong class="example">Example 3:</strong></p>
+<p><strong>Example 3:</strong></p>
 
 <pre>
 <strong>Input:</strong> s = &quot;bdda&quot;