feat: update lc problems

Author: yanglbme

solution/0000-0099/0015.3Sum/README_EN.md

@@ -10,20 +10,39 @@
 
 <p>&nbsp;</p>
 <p><strong>Example 1:</strong></p>
-<pre><strong>Input:</strong> nums = [-1,0,1,2,-1,-4]
+
+<pre>
+<strong>Input:</strong> nums = [-1,0,1,2,-1,-4]
 <strong>Output:</strong> [[-1,-1,2],[-1,0,1]]
-</pre><p><strong>Example 2:</strong></p>
-<pre><strong>Input:</strong> nums = []
-<strong>Output:</strong> []
-</pre><p><strong>Example 3:</strong></p>
-<pre><strong>Input:</strong> nums = [0]
+<strong>Explanation:</strong> 
+nums[0] + nums[1] + nums[1] = (-1) + 0 + 1 = 0.
+nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0.
+nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0.
+The distinct triplets are [-1,0,1] and [-1,-1,2].
+Notice that the order of the output and the order of the triplets does not matter.
+</pre>
+
+<p><strong>Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [0,1,1]
 <strong>Output:</strong> []
+<strong>Explanation:</strong> The only possible triplet does not sum up to 0.
 </pre>
+
+<p><strong>Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [0,0,0]
+<strong>Output:</strong> [[0,0,0]]
+<strong>Explanation:</strong> The only possible triplet sums up to 0.
+</pre>
+
 <p>&nbsp;</p>
 <p><strong>Constraints:</strong></p>
 
 <ul>
-	<li><code>0 &lt;= nums.length &lt;= 3000</code></li>
+	<li><code>3 &lt;= nums.length &lt;= 3000</code></li>
 	<li><code>-10<sup>5</sup> &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
 </ul>
 

solution/0100-0199/0155.Min Stack/README_EN.md

@@ -16,6 +16,8 @@
 	<li><code>int getMin()</code> retrieves the minimum element in the stack.</li>
 </ul>
 
+<p>You must implement a solution with <code>O(1)</code> time complexity for each function.</p>
+
 <p>&nbsp;</p>
 <p><strong>Example 1:</strong></p>
 

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

@@ -6,11 +6,11 @@
 
 <p>A peak element is an element that is strictly greater than its neighbors.</p>
 
-<p>Given an integer array <code>nums</code>, find a peak element, and return its index. If&nbsp;the array contains multiple peaks, return the index to <strong>any of the peaks</strong>.</p>
+<p>Given a <strong>0-indexed</strong> integer array <code>nums</code>, find a peak element, and return its index. If the array contains multiple peaks, return the index to <strong>any of the peaks</strong>.</p>
 
-<p>You may imagine that <code>nums[-1] = nums[n] = -&infin;</code>.</p>
+<p>You may imagine that <code>nums[-1] = nums[n] = -&infin;</code>. In other words, an element is always considered to be strictly greater than a neighbor that is outside the array.</p>
 
-<p>You must write an algorithm that runs in&nbsp;<code>O(log n)</code> time.</p>
+<p>You must write an algorithm that runs in <code>O(log n)</code> time.</p>
 
 <p>&nbsp;</p>
 <p><strong>Example 1:</strong></p>

solution/0200-0299/0223.Rectangle Area/README_EN.md

@@ -29,7 +29,10 @@
 <p><strong>Constraints:</strong></p>
 
 <ul>
-	<li><code>-10<sup>4</sup> &lt;= ax1, ay1, ax2, ay2, bx1, by1, bx2, by2 &lt;= 10<sup>4</sup></code></li>
+	<li><code>-10<sup>4</sup> &lt;= ax1 &lt;= ax2 &lt;= 10<sup>4</sup></code></li>
+	<li><code>-10<sup>4</sup> &lt;= ay1 &lt;= ay2 &lt;= 10<sup>4</sup></code></li>
+	<li><code>-10<sup>4</sup> &lt;= bx1 &lt;= bx2 &lt;= 10<sup>4</sup></code></li>
+	<li><code>-10<sup>4</sup> &lt;= by1 &lt;= by2 &lt;= 10<sup>4</sup></code></li>
 </ul>
 
 ## Solutions

solution/0300-0399/0320.Generalized Abbreviation/README.md

@@ -10,6 +10,7 @@
 
 <ul>
 	<li>例如，<code>"abcde"</code> 可以缩写为：
+
     <ul>
     	<li><code>"a3e"</code>（<code>"bcd"</code> 变为 <code>"3"</code> ）</li>
     	<li><code>"1bcd1"</code>（<code>"a"</code> 和 <code>"e"</code> 都变为 <code>"1"</code>）<meta charset="UTF-8" /></li>
@@ -23,6 +24,7 @@
     	<li><meta charset="UTF-8" /><code>"22de"</code>&nbsp;(<code>"ab"</code> 变为&nbsp;<code>"2"</code>&nbsp;，&nbsp;<code>"bc"</code>&nbsp;变为&nbsp;<code>"2"</code>) &nbsp;是无效的，因为被选择的字符串是重叠的</li>
     </ul>
     </li>
+
 </ul>
 
 <p>给你一个字符串&nbsp;<code>word</code> ，返回&nbsp;<em>一个由</em>&nbsp;<code>word</code> 的<em>所有可能 <strong>广义缩写词</strong> 组成的列表</em>&nbsp;。按 <strong>任意顺序</strong> 返回答案。</p>

solution/0300-0399/0336.Palindrome Pairs/README.md

@@ -32,6 +32,8 @@
 <strong>输出：</strong>[[0,1],[1,0]]
 </pre>
 
+
+
 <p><strong>提示：</strong></p>
 
 <ul>

solution/0500-0599/0554.Brick Wall/README.md

@@ -28,6 +28,8 @@
 <strong>输出：</strong>3
 </pre>
 
+
+
 <p><strong>提示：</strong></p>
 
 <ul>

solution/0500-0599/0560.Subarray Sum Equals K/README.md

@@ -6,7 +6,7 @@
 
 <!-- 这里写题目描述 -->
 
-<p>给你一个整数数组 <code>nums</code> 和一个整数&nbsp;<code>k</code> ，请你统计并返回 <em>该数组中和为&nbsp;<code>k</code><strong>&nbsp;</strong>的子数组的个数&nbsp;</em>。</p>
+<p>给你一个整数数组 <code>nums</code> 和一个整数&nbsp;<code>k</code> ，请你统计并返回 <em>该数组中和为&nbsp;<code>k</code><strong>&nbsp;</strong>的连续子数组的个数&nbsp;</em>。</p>
 
 <p>&nbsp;</p>
 

solution/0500-0599/0561.Array Partition/README.md

@@ -41,7 +41,6 @@
 	<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
 </ul>
 
-
 ## 解法
 
 <!-- 这里可写通用的实现逻辑 -->

solution/0500-0599/0561.Array Partition/README_EN.md

@@ -35,7 +35,6 @@ So the maximum possible sum is 4.</pre>
 	<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
 </ul>
 
-
 ## Solutions
 
 <!-- tabs:start -->

solution/0600-0699/0655.Print Binary Tree/README.md

@@ -6,62 +6,51 @@
 
 <!-- 这里写题目描述 -->
 
-<p>在一个 m*n 的二维字符串数组中输出二叉树，并遵守以下规则：</p>
+<p>给你一棵二叉树的根节点 <code>root</code> ，请你构造一个下标从 <strong>0</strong> 开始、大小为 <code>m x n</code> 的字符串矩阵 <code>res</code> ，用以表示树的 <strong>格式化布局</strong> 。构造此格式化布局矩阵需要遵循以下规则：</p>
 
-<ol>
-	<li>行数&nbsp;<code>m</code>&nbsp;应当等于给定二叉树的高度。</li>
-	<li>列数&nbsp;<code>n</code>&nbsp;应当总是奇数。</li>
-	<li>根节点的值（以字符串格式给出）应当放在可放置的第一行正中间。根节点所在的行与列会将剩余空间划分为两部分（<strong>左下部分和右下部分</strong>）。你应该将左子树输出在左下部分，右子树输出在右下部分。左下和右下部分应当有相同的大小。即使一个子树为空而另一个非空，你不需要为空的子树输出任何东西，但仍需要为另一个子树留出足够的空间。然而，如果两个子树都为空则不需要为它们留出任何空间。</li>
-	<li>每个未使用的空间应包含一个空的字符串<code>&quot;&quot;</code>。</li>
-	<li>使用相同的规则输出子树。</li>
-</ol>
+<ul>
+	<li>树的 <strong>高度</strong> 为 <code>height</code> ，矩阵的行数 <code>m</code> 应该等于 <code>height + 1</code> 。</li>
+	<li>矩阵的列数 <code>n</code> 应该等于 <code>2<sup>height+1</sup> - 1</code> 。</li>
+	<li><strong>根节点</strong> 需要放置在 <strong>顶行</strong> 的 <strong>正中间</strong> ，对应位置为 <code>res[0][(n-1)/2]</code> 。</li>
+	<li>对于放置在矩阵中的每个节点，设对应位置为 <code>res[r][c]</code> ，将其左子节点放置在 <code>res[r+1][c-2<sup>height-r-1</sup>]</code> ，右子节点放置在 <code>res[r+1][c+2<sup>height-r-1</sup>]</code> 。</li>
+	<li>继续这一过程，直到树中的所有节点都妥善放置。</li>
+	<li>任意空单元格都应该包含空字符串 <code>""</code> 。</li>
+</ul>
 
-<p><strong>示例 1:</strong></p>
+<p>返回构造得到的矩阵<em> </em><code>res</code> 。</p>
 
-<pre>
-<strong>输入:</strong>
-     1
-    /
-   2
-<strong>输出:</strong>
-[[&quot;&quot;, &quot;1&quot;, &quot;&quot;],
- [&quot;2&quot;, &quot;&quot;, &quot;&quot;]]
-</pre>
+<p>&nbsp;</p>
 
-<p><strong>示例 2:</strong></p>
+<p>&nbsp;</p>
 
+<p><strong>示例 1：</strong></p>
+<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0600-0699/0655.Print%20Binary%20Tree/images/print1-tree.jpg" style="width: 141px; height: 181px;" />
 <pre>
-<strong>输入:</strong>
-     1
-    / \
-   2   3
-    \
-     4
-<strong>输出:</strong>
-[[&quot;&quot;, &quot;&quot;, &quot;&quot;, &quot;1&quot;, &quot;&quot;, &quot;&quot;, &quot;&quot;],
- [&quot;&quot;, &quot;2&quot;, &quot;&quot;, &quot;&quot;, &quot;&quot;, &quot;3&quot;, &quot;&quot;],
- [&quot;&quot;, &quot;&quot;, &quot;4&quot;, &quot;&quot;, &quot;&quot;, &quot;&quot;, &quot;&quot;]]
+<strong>输入：</strong>root = [1,2]
+<strong>输出：</strong>
+[["","1",""],
+&nbsp;["2","",""]]
 </pre>
 
-<p><strong>示例 3:</strong></p>
-
+<p><strong>示例 2：</strong></p>
+<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0600-0699/0655.Print%20Binary%20Tree/images/print2-tree.jpg" style="width: 207px; height: 302px;" />
 <pre>
-<strong>输入:</strong>
-      1
-     / \
-    2   5
-   / 
-  3 
- / 
-4 
-<strong>输出:</strong>
-[[&quot;&quot;,  &quot;&quot;,  &quot;&quot;, &quot;&quot;,  &quot;&quot;, &quot;&quot;, &quot;&quot;, &quot;1&quot;, &quot;&quot;,  &quot;&quot;,  &quot;&quot;,  &quot;&quot;,  &quot;&quot;, &quot;&quot;, &quot;&quot;]
- [&quot;&quot;,  &quot;&quot;,  &quot;&quot;, &quot;2&quot;, &quot;&quot;, &quot;&quot;, &quot;&quot;, &quot;&quot;,  &quot;&quot;,  &quot;&quot;,  &quot;&quot;,  &quot;5&quot;, &quot;&quot;, &quot;&quot;, &quot;&quot;]
- [&quot;&quot;,  &quot;3&quot;, &quot;&quot;, &quot;&quot;,  &quot;&quot;, &quot;&quot;, &quot;&quot;, &quot;&quot;,  &quot;&quot;,  &quot;&quot;,  &quot;&quot;,  &quot;&quot;,  &quot;&quot;, &quot;&quot;, &quot;&quot;]
- [&quot;4&quot;, &quot;&quot;,  &quot;&quot;, &quot;&quot;,  &quot;&quot;, &quot;&quot;, &quot;&quot;, &quot;&quot;,  &quot;&quot;,  &quot;&quot;,  &quot;&quot;,  &quot;&quot;,  &quot;&quot;, &quot;&quot;, &quot;&quot;]]
+<strong>输入：</strong>root = [1,2,3,null,4]
+<strong>输出：</strong>
+[["","","","1","","",""],
+&nbsp;["","2","","","","3",""],
+&nbsp;["","","4","","","",""]]
 </pre>
 
-<p><strong>注意:</strong> 二叉树的高度在范围 [1, 10] 中。</p>
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li>树中节点数在范围 <code>[1, 2<sup>10</sup>]</code> 内</li>
+	<li><code>-99 &lt;= Node.val &lt;= 99</code></li>
+	<li>树的深度在范围 <code>[1, 10]</code> 内</li>
+</ul>
 
 ## 解法
 

solution/0600-0699/0692.Top K Frequent Words/README_EN.md

@@ -31,7 +31,7 @@ Note that "i" comes before "love" due to a lower alphabetica
 
 <ul>
 	<li><code>1 &lt;= words.length &lt;= 500</code></li>
-	<li><code>1 &lt;= words[i] &lt;= 10</code></li>
+	<li><code>1 &lt;= words[i].length &lt;= 10</code></li>
 	<li><code>words[i]</code> consists of lowercase English letters.</li>
 	<li><code>k</code> is in the range <code>[1, The number of <strong>unique</strong> words[i]]</code></li>
 </ul>

solution/0700-0799/0716.Max Stack/README_EN.md

@@ -17,6 +17,8 @@
 	<li><code>int popMax()</code> Retrieves the maximum element in the stack and removes it. If there is more than one maximum element, only remove the <strong>top-most</strong> one.</li>
 </ul>
 
+<p>You must come up with a solution that supports <code>O(1)</code> for each <code>top</code> call and <code>O(logn)</code> for each other call?&nbsp;</p>
+
 <p>&nbsp;</p>
 <p><strong>Example 1:</strong></p>
 
@@ -49,9 +51,6 @@ stk.top();     // return 5, [<strong><u>5</u></strong>] the stack did not change
 	<li>There will be <strong>at least one element</strong> in the stack when <code>pop</code>, <code>top</code>, <code>peekMax</code>, or <code>popMax</code> is called.</li>
 </ul>
 
-<p>&nbsp;</p>
-<strong>Follow up:</strong> Could you come up with a solution that supports <code>O(1)</code> for each <code>top</code> call and <code>O(logn)</code> for each other call?&nbsp;
-
 ## Solutions
 
 <!-- tabs:start -->

solution/0700-0799/0745.Prefix and Suffix Search/README.md

@@ -12,33 +12,34 @@
 
 <ul>
 	<li><code>WordFilter(string[] words)</code> 使用词典中的单词 <code>words</code> 初始化对象。</li>
-	<li><code>f(string prefix, string suffix)</code> 返回词典中具有前缀 <code>prefix</code> 和后缀<code>suffix</code> 的单词的下标。如果存在不止一个满足要求的下标，返回其中 <strong>最大的下标</strong> 。如果不存在这样的单词，返回 <code>-1</code> 。</li>
+	<li><code>f(string pref, string suff)</code> 返回词典中具有前缀&nbsp;<code>prefix</code>&nbsp;和后缀 <code>suff</code>&nbsp;的单词的下标。如果存在不止一个满足要求的下标，返回其中 <strong>最大的下标</strong> 。如果不存在这样的单词，返回 <code>-1</code> 。</li>
 </ul>
 
-<p> </p>
+<p>&nbsp;</p>
 
-<p><strong>示例</strong></p>
+<p><strong>示例：</strong></p>
 
 <pre>
-<strong>输入：</strong>
+<strong>输入</strong>
 ["WordFilter", "f"]
 [[["apple"]], ["a", "e"]]
-<strong>输出：</strong>
+<strong>输出</strong>
 [null, 0]
-
-<strong>解释：</strong>
+<strong>解释</strong>
 WordFilter wordFilter = new WordFilter(["apple"]);
-wordFilter.f("a", "e"); // 返回 0 ，因为下标为 0 的单词的 prefix = "a" 且 suffix = 'e" 。
+wordFilter.f("a", "e"); // 返回 0 ，因为下标为 0 的单词：前缀 prefix = "a" 且 后缀 suff = "e" 。
 </pre>
 
+&nbsp;
+
 <p><strong>提示：</strong></p>
 
 <ul>
-	<li><code>1 <= words.length <= 15000</code></li>
-	<li><code>1 <= words[i].length <= 10</code></li>
-	<li><code>1 <= prefix.length, suffix.length <= 10</code></li>
-	<li><code>words[i]</code>、<code>prefix</code> 和 <code>suffix</code> 仅由小写英文字母组成</li>
-	<li>最多对函数 <code>f</code> 进行 <code>15000</code> 次调用</li>
+	<li><code>1 &lt;= words.length &lt;= 10<sup>4</sup></code></li>
+	<li><code>1 &lt;= words[i].length &lt;= 7</code></li>
+	<li><code>1 &lt;= pref.length, suff.length &lt;= 7</code></li>
+	<li><code>words[i]</code>、<code>pref</code> 和 <code>suff</code> 仅由小写英文字母组成</li>
+	<li>最多对函数 <code>f</code> 执行 <code>10<sup>4</sup></code> 次调用</li>
 </ul>
 
 ## 解法

solution/0700-0799/0747.Largest Number At Least Twice of Others/README_EN.md

@@ -24,14 +24,7 @@ The index of value 6 is 1, so we return 1.
 <pre>
 <strong>Input:</strong> nums = [1,2,3,4]
 <strong>Output:</strong> -1
-<strong>Explanation:</strong> 4 is less than twice the value of 3, so we return -1.</pre>
-
-<p><strong>Example 3:</strong></p>
-
-<pre>
-<strong>Input:</strong> nums = [1]
-<strong>Output:</strong> 0
-<strong>Explanation:</strong> 1 is trivially at least twice the value as any other number because there are no other numbers.
+<strong>Explanation:</strong> 4 is less than twice the value of 3, so we return -1.
 </pre>
 
 <p>&nbsp;</p>

solution/0700-0799/0775.Global and Local Inversions/README.md

@@ -42,6 +42,8 @@
 <strong>解释：</strong>有 2 个全局倒置，和 1 个局部倒置。
 </pre>
 
+
+
 <p><strong>提示：</strong></p>
 
 <ul>

solution/0800-0899/0802.Find Eventual Safe States/README_EN.md

@@ -6,7 +6,7 @@
 
 <p>There is a directed graph of <code>n</code> nodes with each node labeled from <code>0</code> to <code>n - 1</code>. The graph is represented by a <strong>0-indexed</strong> 2D integer array <code>graph</code> where <code>graph[i]</code> is an integer array of nodes adjacent to node <code>i</code>, meaning there is an edge from node <code>i</code> to each node in <code>graph[i]</code>.</p>
 
-<p>A node is a <strong>terminal node</strong> if there are no outgoing edges. A node is a <strong>safe node</strong> if every possible path starting from that node leads to a <strong>terminal node</strong>.</p>
+<p>A node is a <strong>terminal node</strong> if there are no outgoing edges. A node is a <strong>safe node</strong> if every possible path starting from that node leads to a <strong>terminal node</strong> (or another safe node).</p>
 
 <p>Return <em>an array containing all the <strong>safe nodes</strong> of the graph</em>. The answer should be sorted in <strong>ascending</strong> order.</p>
 

solution/0800-0899/0828.Count Unique Characters of All Substrings of a Given String/README.md

@@ -10,9 +10,9 @@
 
 <p>例如：<code>s = "LEETCODE"</code> ，则其中 <code>"L"</code>, <code>"T"</code>,<code>"C"</code>,<code>"O"</code>,<code>"D"</code> 都是唯一字符，因为它们只出现一次，所以 <code>countUniqueChars(s) = 5</code> 。</p>
 
-<p>本题将会给你一个字符串 <code>s</code> ，我们需要返回 <code>countUniqueChars(t)</code> 的总和，其中 <code>t</code> 是 <code>s</code> 的子字符串。注意，某些子字符串可能是重复的，但你统计时也必须算上这些重复的子字符串（也就是说，你必须统计 <code>s</code> 的所有子字符串中的唯一字符）。</p>
+<p>本题将会给你一个字符串 <code>s</code> ，我们需要返回 <code>countUniqueChars(t)</code> 的总和，其中 <code>t</code> 是 <code>s</code> 的子字符串。输入用例保证返回值为&nbsp;32 位整数。</p>
 
-<p>由于答案可能非常大，请将结果 <strong>mod 10 ^ 9 + 7</strong> 后再返回。</p>
+<p>注意，某些子字符串可能是重复的，但你统计时也必须算上这些重复的子字符串（也就是说，你必须统计 <code>s</code> 的所有子字符串中的唯一字符）。</p>
 
 <p>&nbsp;</p>
 
@@ -46,7 +46,7 @@
 <p><strong>提示：</strong></p>
 
 <ul>
-	<li><code>0 &lt;= s.length &lt;= 10^5</code></li>
+	<li><code>1 &lt;= s.length &lt;= 10^5</code></li>
 	<li><code>s</code> 只包含大写英文字符</li>
 </ul>
 

solution/0800-0899/0828.Count Unique Characters of All Substrings of a Given String/README_EN.md

@@ -10,7 +10,7 @@
 	<li>For example, calling <code>countUniqueChars(s)</code> if <code>s = &quot;LEETCODE&quot;</code> then <code>&quot;L&quot;</code>, <code>&quot;T&quot;</code>, <code>&quot;C&quot;</code>, <code>&quot;O&quot;</code>, <code>&quot;D&quot;</code> are the unique characters since they appear only once in <code>s</code>, therefore <code>countUniqueChars(s) = 5</code>.</li>
 </ul>
 
-<p>Given a string <code>s</code>, return the sum of <code>countUniqueChars(t)</code> where <code>t</code> is a substring of <code>s</code>.</p>
+<p>Given a string <code>s</code>, return the sum of <code>countUniqueChars(t)</code> where <code>t</code> is a substring of <code>s</code>. The test cases are generated such that the answer fits in a 32-bit integer.</p>
 
 <p>Notice that some substrings can be repeated so in this case you have to count the repeated ones too.</p>
 

solution/0800-0899/0830.Positions of Large Groups/README.md

@@ -48,6 +48,8 @@
 <strong>输出：</strong>[]
 </pre>
 
+
+
 <p><strong>提示：</strong></p>
 
 <ul>