feat: add biweekly contest 143 and weeky contest 423

solution/0700-0799/0777.Swap Adjacent in LR String/README_EN.md

@@ -17,15 +17,15 @@ tags:
 
 <!-- description:start -->
 
-<p>In a string composed of <code>&#39;L&#39;</code>, <code>&#39;R&#39;</code>, and <code>&#39;X&#39;</code> characters, like <code>&quot;RXXLRXRXL&quot;</code>, a move consists of either replacing one occurrence of <code>&quot;XL&quot;</code> with <code>&quot;LX&quot;</code>, or replacing one occurrence of <code>&quot;RX&quot;</code> with <code>&quot;XR&quot;</code>. Given the starting string <code>start</code> and the ending string <code>end</code>, return <code>True</code> if and only if there exists a sequence of moves to transform <code>start</code> to <code>end</code>.</p>
+<p>In a string composed of <code>&#39;L&#39;</code>, <code>&#39;R&#39;</code>, and <code>&#39;X&#39;</code> characters, like <code>&quot;RXXLRXRXL&quot;</code>, a move consists of either replacing one occurrence of <code>&quot;XL&quot;</code> with <code>&quot;LX&quot;</code>, or replacing one occurrence of <code>&quot;RX&quot;</code> with <code>&quot;XR&quot;</code>. Given the starting string <code>start</code> and the ending string <code>result</code>, return <code>True</code> if and only if there exists a sequence of moves to transform <code>start</code> to <code>result</code>.</p>
 
 <p>&nbsp;</p>
 <p><strong class="example">Example 1:</strong></p>
 
 <pre>
-<strong>Input:</strong> start = &quot;RXXLRXRXL&quot;, end = &quot;XRLXXRRLX&quot;
+<strong>Input:</strong> start = &quot;RXXLRXRXL&quot;, result = &quot;XRLXXRRLX&quot;
 <strong>Output:</strong> true
-<strong>Explanation:</strong> We can transform start to end following these steps:
+<strong>Explanation:</strong> We can transform start to result following these steps:
 RXXLRXRXL -&gt;
 XRXLRXRXL -&gt;
 XRLXRXRXL -&gt;
@@ -36,7 +36,7 @@ XRLXXRRLX
 <p><strong class="example">Example 2:</strong></p>
 
 <pre>
-<strong>Input:</strong> start = &quot;X&quot;, end = &quot;L&quot;
+<strong>Input:</strong> start = &quot;X&quot;, result = &quot;L&quot;
 <strong>Output:</strong> false
 </pre>
 
@@ -45,8 +45,8 @@ XRLXXRRLX
 
 <ul>
 	<li><code>1 &lt;= start.length&nbsp;&lt;= 10<sup>4</sup></code></li>
-	<li><code>start.length == end.length</code></li>
-	<li>Both <code>start</code> and <code>end</code> will only consist of characters in <code>&#39;L&#39;</code>, <code>&#39;R&#39;</code>, and&nbsp;<code>&#39;X&#39;</code>.</li>
+	<li><code>start.length == result.length</code></li>
+	<li>Both <code>start</code> and <code>result</code> will only consist of characters in <code>&#39;L&#39;</code>, <code>&#39;R&#39;</code>, and&nbsp;<code>&#39;X&#39;</code>.</li>
 </ul>
 
 <!-- description:end -->

solution/0900-0999/0909.Snakes and Ladders/README_EN.md

@@ -36,13 +36,13 @@ tags:
 
 <p>A board square on row <code>r</code> and column <code>c</code> has a snake or ladder if <code>board[r][c] != -1</code>. The destination of that snake or ladder is <code>board[r][c]</code>. Squares <code>1</code> and <code>n<sup>2</sup></code> are not the starting points of any snake or ladder.</p>
 
-<p>Note that you only take a snake or ladder at most once per move. If the destination to a snake or ladder is the start of another snake or ladder, you do <strong>not</strong> follow the subsequent&nbsp;snake or ladder.</p>
+<p>Note that you only take a snake or ladder at most once per dice roll. If the destination to a snake or ladder is the start of another snake or ladder, you do <strong>not</strong> follow the subsequent&nbsp;snake or ladder.</p>
 
 <ul>
 	<li>For example, suppose the board is <code>[[-1,4],[-1,3]]</code>, and on the first move, your destination square is <code>2</code>. You follow the ladder to square <code>3</code>, but do <strong>not</strong> follow the subsequent ladder to <code>4</code>.</li>
 </ul>
 
-<p>Return <em>the least number of moves required to reach the square </em><code>n<sup>2</sup></code><em>. If it is not possible to reach the square, return </em><code>-1</code>.</p>
+<p>Return <em>the least number of dice rolls required to reach the square </em><code>n<sup>2</sup></code><em>. If it is not possible to reach the square, return </em><code>-1</code>.</p>
 
 <p>&nbsp;</p>
 <p><strong class="example">Example 1:</strong></p>

solution/1200-1299/1257.Smallest Common Region/README_EN.md

@@ -64,6 +64,7 @@ region2 = "New York"
 	<li><code>1 &lt;= regions[i][j].length, region1.length, region2.length &lt;= 20</code></li>
 	<li><code>region1 != region2</code></li>
 	<li><code>regions[i][j]</code>, <code>region1</code>, and <code>region2</code> consist of English letters.</li>
+	<li>The input is generated such that there exists a region which contains all the other regions, either directly or indirectly.</li>
 </ul>
 
 <!-- description:end -->

solution/2600-2699/2623.Memoize/README_EN.md

@@ -85,7 +85,7 @@ values = [[5],[]]
 <ul>
 	<li><code>0 &lt;= a, b &lt;= 10<sup>5</sup></code></li>
 	<li><code>1 &lt;= n &lt;= 10</code></li>
-	<li><code>0 &lt;= actions.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= actions.length &lt;= 10<sup>5</sup></code></li>
 	<li><code>actions.length === values.length</code></li>
 	<li><code>actions[i]</code> is one of &quot;call&quot; and &quot;getCallCount&quot;</li>
 	<li><code>fnName</code> is one of &quot;sum&quot;, &quot;factorial&quot; and&nbsp;&quot;fib&quot;</li>

solution/3000-3099/3011.Find if Array Can Be Sorted/README_EN.md

@@ -24,7 +24,7 @@ tags:
 
 <p>In one <strong>operation</strong>, you can swap any two <strong>adjacent</strong> elements if they have the <strong>same</strong> number of <span data-keyword="set-bit">set bits</span>. You are allowed to do this operation <strong>any</strong> number of times (<strong>including zero</strong>).</p>
 
-<p>Return <code>true</code> <em>if you can sort the array, else return </em><code>false</code>.</p>
+<p>Return <code>true</code> <em>if you can sort the array in ascending order, else return </em><code>false</code>.</p>
 
 <p>&nbsp;</p>
 <p><strong class="example">Example 1:</strong></p>

solution/3200-3299/3255.Find the Power of K-Size Subarrays II/README.md

@@ -24,7 +24,7 @@ tags:
 <p>一个数组的 <strong>能量值</strong> 定义为：</p>
 
 <ul>
-	<li>如果 <strong>所有</strong>&nbsp;元素都是依次&nbsp;<strong>连续</strong> 且 <strong>上升</strong> 的，那么能量值为 <strong>最大</strong>&nbsp;的元素。</li>
+	<li>如果 <strong>所有</strong>&nbsp;元素都是依次&nbsp;<strong>连续</strong>（即 <code>nums[i] + 1 = nums[i + 1]</code>，<code>i &lt; n</code>）且 <strong>上升</strong> 的，那么能量值为 <strong>最大</strong>&nbsp;的元素。</li>
 	<li>否则为 -1 。</li>
 </ul>
 

solution/3300-3399/3345.Smallest Divisible Digit Product I/README.md

@@ -0,0 +1,156 @@
+---
+comments: true
+difficulty: 简单
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3300-3399/3345.Smallest%20Divisible%20Digit%20Product%20I/README.md
+---
+
+<!-- problem:start -->
+
+# [3345. 最小可整除数位乘积 I](https://leetcode.cn/problems/smallest-divisible-digit-product-i)
+
+[English Version](/solution/3300-3399/3345.Smallest%20Divisible%20Digit%20Product%20I/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>给你两个整数&nbsp;<code>n</code> 和&nbsp;<code>t</code>&nbsp;。请你返回大于等于&nbsp;<code>n</code>&nbsp;的&nbsp;<strong>最小</strong>&nbsp;整数，且该整数的&nbsp;<strong>各数位之积</strong>&nbsp;能被&nbsp;<code>t</code>&nbsp;整除。</p>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1：</strong></p>
+
+<div class="example-block">
+<p><span class="example-io"><b>输入：</b>n = 10, t = 2</span></p>
+
+<p><span class="example-io"><b>输出：</b>10</span></p>
+
+<p><strong>解释：</strong></p>
+
+<p>10 的数位乘积为 0 ，可以被 2 整除，所以它是大于等于 10 且满足题目要求的最小整数。</p>
+</div>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<div class="example-block">
+<p><span class="example-io"><b>输入：</b>n = 15, t = 3</span></p>
+
+<p><span class="example-io"><b>输出：</b>16</span></p>
+
+<p><strong>解释：</strong></p>
+
+<p>16 的数位乘积为 6 ，可以被 3 整除，所以它是大于等于 15 且满足题目要求的最小整数。</p>
+</div>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= n &lt;= 100</code></li>
+	<li><code>1 &lt;= t &lt;= 10</code></li>
+</ul>
+
+<!-- description:end -->
+
+## 解法
+
+<!-- solution:start -->
+
+### 方法一：枚举
+
+我们注意到，每 $10$ 个数里一定会出现数位乘积为 $0$ 的整数，因此我们可以直接枚举大于等于 $n$ 的整数，直到找到一个数位乘积能被 $t$ 整除的整数。
+
+时间复杂度 $O(\log n)$，空间复杂度 $O(1)$。
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+class Solution:
+    def smallestNumber(self, n: int, t: int) -> int:
+        for i in count(n):
+            p = 1
+            x = i
+            while x:
+                p *= x % 10
+                x //= 10
+            if p % t == 0:
+                return i
+```
+
+#### Java
+
+```java
+class Solution {
+    public int smallestNumber(int n, int t) {
+        for (int i = n;; ++i) {
+            int p = 1;
+            for (int x = i; x > 0; x /= 10) {
+                p *= (x % 10);
+            }
+            if (p % t == 0) {
+                return i;
+            }
+        }
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    int smallestNumber(int n, int t) {
+        for (int i = n;; ++i) {
+            int p = 1;
+            for (int x = i; x > 0; x /= 10) {
+                p *= (x % 10);
+            }
+            if (p % t == 0) {
+                return i;
+            }
+        }
+    }
+};
+```
+
+#### Go
+
+```go
+func smallestNumber(n int, t int) int {
+	for i := n; ; i++ {
+		p := 1
+		for x := i; x > 0; x /= 10 {
+			p *= x % 10
+		}
+		if p%t == 0 {
+			return i
+		}
+	}
+}
+```
+
+#### TypeScript
+
+```ts
+function smallestNumber(n: number, t: number): number {
+    for (let i = n; ; ++i) {
+        let p = 1;
+        for (let x = i; x; x = Math.floor(x / 10)) {
+            p *= x % 10;
+        }
+        if (p % t === 0) {
+            return i;
+        }
+    }
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->