feat: update lc problems (#3653)

Author: yanglbme

solution/0100-0199/0191.Number of 1 Bits/README_EN.md

@@ -17,7 +17,7 @@ tags:
 
 <!-- description:start -->
 
-<p>Write a function that takes the binary representation of a positive integer and returns the number of <span data-keyword="set-bit">set bits</span> it has (also known as the <a href="http://en.wikipedia.org/wiki/Hamming_weight" target="_blank">Hamming weight</a>).</p>
+<p>Given a positive integer <code>n</code>, write a function that returns the number of <span data-keyword="set-bit">set bits</span> in its binary representation (also known as the <a href="http://en.wikipedia.org/wiki/Hamming_weight" target="_blank">Hamming weight</a>).</p>
 
 <p>&nbsp;</p>
 <p><strong class="example">Example 1:</strong></p>

solution/3000-3099/3038.Maximum Number of Operations With the Same Score I/README.md

@@ -47,7 +47,7 @@ tags:
 
 <pre>
 <b>输入：</b>nums = [3,2,6,1,4]
-<b>输出：</b>2
+<b>输出：1</b>
 <b>解释：</b>我们执行以下操作：
 - 删除前两个元素，分数为 3 + 2 = 5 ，nums = [6,1,4] 。
 由于下一次操作的分数与前一次不相等，我们无法继续进行任何操作。

solution/3300-3399/3323.Minimize Connected Groups by Inserting Interval/README.md

@@ -16,7 +16,7 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3300-3399/3323.Mi
 
 <p>给定一个 2 维数组&nbsp;<code>intervals</code>，其中&nbsp;<code>intervals[i] = [start<sub>i</sub>, end<sub>i</sub>]</code>&nbsp;表示区间&nbsp;<code>i</code>&nbsp;的开头和结尾。另外还给定一个整数&nbsp;<code>k</code>。</p>
 
-<p>你必须向数组添加 <strong>恰好一个</strong>&nbsp;新的区间&nbsp;<code>[start<sub>new</sub>, end<sub>new</sub>]</code>&nbsp;使得：</p>
+<p>你必须向数组 <strong>恰好添加一个</strong>&nbsp;新的区间&nbsp;<code>[start<sub>new</sub>, end<sub>new</sub>]</code>&nbsp;使得：</p>
 
 <ul>
 	<li>新区间的长度，<code>end<sub>new</sub> - start<sub>new</sub></code>&nbsp;最多为&nbsp;<code>k</code>。</li>
@@ -30,7 +30,7 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3300-3399/3323.Mi
 	<li>然而，区间组&nbsp;<code>[[1, 2], [3, 4]]</code>&nbsp;不是连通的，因为&nbsp;<code>(2, 3)</code>&nbsp;段没有被覆盖。</li>
 </ul>
 
-<p>返回在数组添加 <strong>恰好一个</strong> 新区间后，连通组的 <strong>最小</strong> 数量。</p>
+<p>返回在数组&nbsp;<strong>恰好添加一个</strong> 新区间后，连通组的 <strong>最小</strong> 数量。</p>
 
 <p>&nbsp;</p>
 

solution/3300-3399/3324.Find the Sequence of Strings Appeared on the Screen/README.md

@@ -0,0 +1,175 @@
+---
+comments: true
+difficulty: 中等
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3300-3399/3324.Find%20the%20Sequence%20of%20Strings%20Appeared%20on%20the%20Screen/README.md
+---
+
+<!-- problem:start -->
+
+# [3324. 出现在屏幕上的字符串序列](https://leetcode.cn/problems/find-the-sequence-of-strings-appeared-on-the-screen)
+
+[English Version](/solution/3300-3399/3324.Find%20the%20Sequence%20of%20Strings%20Appeared%20on%20the%20Screen/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>给你一个字符串 <code>target</code>。</p>
+
+<p>Alice 将会使用一种特殊的键盘在她的电脑上输入 <code>target</code>，这个键盘<strong> 只有两个 </strong>按键：</p>
+
+<ul>
+	<li>按键 1：在屏幕上的字符串后追加字符 <code>'a'</code>。</li>
+	<li>按键 2：将屏幕上字符串的 <strong>最后一个 </strong>字符更改为英文字母表中的 <strong>下一个</strong> 字符。例如，<code>'c'</code> 变为 <code>'d'</code>，<code>'z'</code> 变为 <code>'a'</code>。</li>
+</ul>
+
+<p><strong>注意</strong>，最初屏幕上是一个<em>空</em>字符串 <code>""</code>，所以她<strong> 只能</strong> 按按键 1。</p>
+
+<p>请你考虑按键次数 <strong>最少</strong> 的情况，按字符串出现顺序，返回 Alice 输入 <code>target</code> 时屏幕上出现的所有字符串列表。</p>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1：</strong></p>
+
+<div class="example-block">
+<p><strong>输入：</strong> <span class="example-io">target = "abc"</span></p>
+
+<p><strong>输出：</strong> <span class="example-io">["a","aa","ab","aba","abb","abc"]</span></p>
+
+<p><strong>解释：</strong></p>
+
+<p>Alice 按键的顺序如下：</p>
+
+<ul>
+	<li>按下按键 1，屏幕上的字符串变为 <code>"a"</code>。</li>
+	<li>按下按键 1，屏幕上的字符串变为 <code>"aa"</code>。</li>
+	<li>按下按键 2，屏幕上的字符串变为 <code>"ab"</code>。</li>
+	<li>按下按键 1，屏幕上的字符串变为 <code>"aba"</code>。</li>
+	<li>按下按键 2，屏幕上的字符串变为 <code>"abb"</code>。</li>
+	<li>按下按键 2，屏幕上的字符串变为 <code>"abc"</code>。</li>
+</ul>
+</div>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<div class="example-block">
+<p><strong>输入：</strong> <span class="example-io">target = "he"</span></p>
+
+<p><strong>输出：</strong> <span class="example-io">["a","b","c","d","e","f","g","h","ha","hb","hc","hd","he"]</span></p>
+</div>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= target.length &lt;= 400</code></li>
+	<li><code>target</code> 仅由小写英文字母组成。</li>
+</ul>
+
+<!-- description:end -->
+
+## 解法
+
+<!-- solution:start -->
+
+### 方法一：模拟
+
+我们可以模拟 Alice 按键的过程，从空字符串开始，每次按键后更新字符串，直到得到目标字符串。
+
+时间复杂度 $O(n^2 \times |\Sigma|)$，其中 $n$ 是目标字符串的长度，而 $\Sigma$ 是字符集，这里是小写字母集合，因此 $|\Sigma| = 26$。
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+class Solution:
+    def stringSequence(self, target: str) -> List[str]:
+        ans = []
+        for c in target:
+            s = ans[-1] if ans else ""
+            for a in ascii_lowercase:
+                t = s + a
+                ans.append(t)
+                if a == c:
+                    break
+        return ans
+```
+
+#### Java
+
+```java
+class Solution {
+    public List<String> stringSequence(String target) {
+        List<String> ans = new ArrayList<>();
+        for (char c : target.toCharArray()) {
+            String s = ans.isEmpty() ? "" : ans.get(ans.size() - 1);
+            for (char a = 'a'; a <= c; ++a) {
+                String t = s + a;
+                ans.add(t);
+            }
+        }
+        return ans;
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    vector<string> stringSequence(string target) {
+        vector<string> ans;
+        for (char c : target) {
+            string s = ans.empty() ? "" : ans.back();
+            for (char a = 'a'; a <= c; ++a) {
+                string t = s + a;
+                ans.push_back(t);
+            }
+        }
+        return ans;
+    }
+};
+```
+
+#### Go
+
+```go
+func stringSequence(target string) (ans []string) {
+	for _, c := range target {
+		s := ""
+		if len(ans) > 0 {
+			s = ans[len(ans)-1]
+		}
+		for a := 'a'; a <= c; a++ {
+			t := s + string(a)
+			ans = append(ans, t)
+		}
+	}
+	return
+}
+```
+
+#### TypeScript
+
+```ts
+function stringSequence(target: string): string[] {
+    const ans: string[] = [];
+    for (const c of target) {
+        let s = ans.length > 0 ? ans[ans.length - 1] : '';
+        for (let a = 'a'.charCodeAt(0); a <= c.charCodeAt(0); a++) {
+            const t = s + String.fromCharCode(a);
+            ans.push(t);
+        }
+    }
+    return ans;
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->

solution/3300-3399/3324.Find the Sequence of Strings Appeared on the Screen/README_EN.md

@@ -0,0 +1,173 @@
+---
+comments: true
+difficulty: Medium
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3300-3399/3324.Find%20the%20Sequence%20of%20Strings%20Appeared%20on%20the%20Screen/README_EN.md
+---
+
+<!-- problem:start -->
+
+# [3324. Find the Sequence of Strings Appeared on the Screen](https://leetcode.com/problems/find-the-sequence-of-strings-appeared-on-the-screen)
+
+[中文文档](/solution/3300-3399/3324.Find%20the%20Sequence%20of%20Strings%20Appeared%20on%20the%20Screen/README.md)
+
+## Description
+
+<!-- description:start -->
+
+<p>You are given a string <code>target</code>.</p>
+
+<p>Alice is going to type <code>target</code> on her computer using a special keyboard that has <strong>only two</strong> keys:</p>
+
+<ul>
+	<li>Key 1 appends the character <code>&quot;a&quot;</code> to the string on the screen.</li>
+	<li>Key 2 changes the <strong>last</strong> character of the string on the screen to its <strong>next</strong> character in the English alphabet. For example, <code>&quot;c&quot;</code> changes to <code>&quot;d&quot;</code> and <code>&quot;z&quot;</code> changes to <code>&quot;a&quot;</code>.</li>
+</ul>
+
+<p><strong>Note</strong> that initially there is an <em>empty</em> string <code>&quot;&quot;</code> on the screen, so she can <strong>only</strong> press key 1.</p>
+
+<p>Return a list of <em>all</em> strings that appear on the screen as Alice types <code>target</code>, in the order they appear, using the <strong>minimum</strong> key presses.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">target = &quot;abc&quot;</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">[&quot;a&quot;,&quot;aa&quot;,&quot;ab&quot;,&quot;aba&quot;,&quot;abb&quot;,&quot;abc&quot;]</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>The sequence of key presses done by Alice are:</p>
+
+<ul>
+	<li>Press key 1, and the string on the screen becomes <code>&quot;a&quot;</code>.</li>
+	<li>Press key 1, and the string on the screen becomes <code>&quot;aa&quot;</code>.</li>
+	<li>Press key 2, and the string on the screen becomes <code>&quot;ab&quot;</code>.</li>
+	<li>Press key 1, and the string on the screen becomes <code>&quot;aba&quot;</code>.</li>
+	<li>Press key 2, and the string on the screen becomes <code>&quot;abb&quot;</code>.</li>
+	<li>Press key 2, and the string on the screen becomes <code>&quot;abc&quot;</code>.</li>
+</ul>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">target = &quot;he&quot;</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">[&quot;a&quot;,&quot;b&quot;,&quot;c&quot;,&quot;d&quot;,&quot;e&quot;,&quot;f&quot;,&quot;g&quot;,&quot;h&quot;,&quot;ha&quot;,&quot;hb&quot;,&quot;hc&quot;,&quot;hd&quot;,&quot;he&quot;]</span></p>
+</div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= target.length &lt;= 400</code></li>
+	<li><code>target</code> consists only of lowercase English letters.</li>
+</ul>
+
+<!-- description:end -->
+
+## Solutions
+
+<!-- solution:start -->
+
+### Solution 1: Simulation
+
+We can simulate Alice's typing process, starting from an empty string and updating the string after each keystroke until the target string is obtained.
+
+The time complexity is $O(n^2 \times |\Sigma|)$, where $n$ is the length of the target string and $\Sigma$ is the character set, which in this case is the set of lowercase letters, so $|\Sigma| = 26$.
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+class Solution:
+    def stringSequence(self, target: str) -> List[str]:
+        ans = []
+        for c in target:
+            s = ans[-1] if ans else ""
+            for a in ascii_lowercase:
+                t = s + a
+                ans.append(t)
+                if a == c:
+                    break
+        return ans
+```
+
+#### Java
+
+```java
+class Solution {
+    public List<String> stringSequence(String target) {
+        List<String> ans = new ArrayList<>();
+        for (char c : target.toCharArray()) {
+            String s = ans.isEmpty() ? "" : ans.get(ans.size() - 1);
+            for (char a = 'a'; a <= c; ++a) {
+                String t = s + a;
+                ans.add(t);
+            }
+        }
+        return ans;
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    vector<string> stringSequence(string target) {
+        vector<string> ans;
+        for (char c : target) {
+            string s = ans.empty() ? "" : ans.back();
+            for (char a = 'a'; a <= c; ++a) {
+                string t = s + a;
+                ans.push_back(t);
+            }
+        }
+        return ans;
+    }
+};
+```
+
+#### Go
+
+```go
+func stringSequence(target string) (ans []string) {
+	for _, c := range target {
+		s := ""
+		if len(ans) > 0 {
+			s = ans[len(ans)-1]
+		}
+		for a := 'a'; a <= c; a++ {
+			t := s + string(a)
+			ans = append(ans, t)
+		}
+	}
+	return
+}
+```
+
+#### TypeScript
+
+```ts
+function stringSequence(target: string): string[] {
+    const ans: string[] = [];
+    for (const c of target) {
+        let s = ans.length > 0 ? ans[ans.length - 1] : '';
+        for (let a = 'a'.charCodeAt(0); a <= c.charCodeAt(0); a++) {
+            const t = s + String.fromCharCode(a);
+            ans.push(t);
+        }
+    }
+    return ans;
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->