feat: add weekly contest 433 and biweekly contest 148 (doocs#3970)

Author: yanglbme

solution/0200-0299/0238.Product of Array Except Self/README_EN.md

@@ -37,7 +37,7 @@ tags:
 <ul>
 	<li><code>2 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
 	<li><code>-30 &lt;= nums[i] &lt;= 30</code></li>
-	<li>The product of any prefix or suffix of <code>nums</code> is <strong>guaranteed</strong> to fit in a <strong>32-bit</strong> integer.</li>
+	<li>The input is generated such that <code>answer[i]</code> is <strong>guaranteed</strong> to fit in a <strong>32-bit</strong> integer.</li>
 </ul>
 
 <p>&nbsp;</p>

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

@@ -25,7 +25,7 @@ tags:
 
 <p>&nbsp;</p>
 <p><strong class="example">Example 1:</strong></p>
-<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0500-0599/0554.Brick%20Wall/images/cutwall-grid.jpg" style="width: 493px; height: 577px;" />
+<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0500-0599/0554.Brick%20Wall/images/a.png" style="width: 400px; height: 384px;" />
 <pre>
 <strong>Input:</strong> wall = [[1,2,2,1],[3,1,2],[1,3,2],[2,4],[3,1,2],[1,3,1,1]]
 <strong>Output:</strong> 2

solution/1400-1499/1400.Construct K Palindrome Strings/README_EN.md

@@ -21,7 +21,7 @@ tags:
 
 <!-- description:start -->
 
-<p>Given a string <code>s</code> and an integer <code>k</code>, return <code>true</code> if you can use all the characters in <code>s</code> to construct <code>k</code> <span data-keyword="palindrome-string">palindrome strings</span> or <code>false</code> otherwise.</p>
+<p>Given a string <code>s</code> and an integer <code>k</code>, return <code>true</code> if you can use all the characters in <code>s</code> to construct <strong>non-empty</strong> <code>k</code> <span data-keyword="palindrome-string">palindrome strings</span> or <code>false</code> otherwise.</p>
 
 <p>&nbsp;</p>
 <p><strong class="example">Example 1:</strong></p>

solution/2200-2299/2266.Count Number of Texts/README.md

@@ -80,7 +80,32 @@ Alice 可能发出的文字信息包括：
 
 <!-- solution:start -->
 
-### 方法一
+### 方法一：分组 + 动态规划
+
+根据题目描述，对于字符串 $\textit{pressedKeys}$ 中连续的相同字符，可以将其分为一组，然后分别计算每组的方案数，最后将所有组的方案数相乘即可。
+
+问题的关键在于如何计算每组的方案数。
+
+如果一组字符为 '7' 或 '9'，我们可以分别将该组的末尾 $1$, $2$, $3$, $4$ 个字符视为一个字母，然后将该组字符规模缩小，转化为规模更小的子问题。
+
+同样地，如果一组字符为 '2', '3', '4', '5', '6', '8'，我们可以将该组的末尾 $1$, $2$, $3$ 个字符视为一个字母，然后将该组字符规模缩小，转化为规模更小的子问题。
+
+因此，我们定义 $f[i]$ 表示长度为 $i$ 的连续相同字符，且字符不为 '7' 或 '9' 的方案数，定义 $g[i]$ 表示长度为 $i$ 的连续相同字符，且字符为 '7' 或 '9' 的方案数。
+
+初始时 $f[0] = f[1] = 1$, $f[2] = 2$, $f[3] = 4$, $g[0] = g[1] = 1$, $g[2] = 2$, $g[3] = 4$。
+
+对于 $i \ge 4$，有：
+
+$$
+\begin{aligned}
+f[i] & = f[i-1] + f[i-2] + f[i-3] \\
+g[i] & = g[i-1] + g[i-2] + g[i-3] + g[i-4]
+\end{aligned}
+$$
+
+最后，我们遍历 $\textit{pressedKeys}$，将连续相同字符分组，然后计算每组的方案数，最后将所有组的方案数相乘即可。
+
+时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为字符串 $\textit{pressedKeys}$ 的长度。
 
 <!-- tabs:start -->
 
@@ -98,9 +123,9 @@ for _ in range(100000):
 class Solution:
     def countTexts(self, pressedKeys: str) -> int:
         ans = 1
-        for ch, s in groupby(pressedKeys):
+        for c, s in groupby(pressedKeys):
             m = len(list(s))
-            ans = ans * (g[m] if ch in "79" else f[m]) % mod
+            ans = ans * (g[m] if c in "79" else f[m]) % mod
         return ans
 ```
 
@@ -113,12 +138,10 @@ class Solution {
     private static long[] f = new long[N];
     private static long[] g = new long[N];
     static {
-        f[0] = 1;
-        f[1] = 1;
+        f[0] = f[1] = 1;
         f[2] = 2;
         f[3] = 4;
-        g[0] = 1;
-        g[1] = 1;
+        g[0] = g[1] = 1;
         g[2] = 2;
         g[3] = 4;
         for (int i = 4; i < N; ++i) {
@@ -130,10 +153,11 @@ class Solution {
     public int countTexts(String pressedKeys) {
         long ans = 1;
         for (int i = 0, n = pressedKeys.length(); i < n; ++i) {
-            int j = i;
             char c = pressedKeys.charAt(i);
-            for (; j + 1 < n && pressedKeys.charAt(j + 1) == c; ++j)
-                ;
+            int j = i;
+            while (j + 1 < n && pressedKeys.charAt(j + 1) == c) {
+                ++j;
+            }
             int cnt = j - i + 1;
             ans = c == '7' || c == '9' ? ans * g[cnt] : ans * f[cnt];
             ans %= MOD;
@@ -144,6 +168,45 @@ class Solution {
 }
 ```
 
+#### C++
+
+```cpp
+const int mod = 1e9 + 7;
+const int n = 1e5 + 10;
+long long f[n], g[n];
+
+int init = []() {
+    f[0] = g[0] = 1;
+    f[1] = g[1] = 1;
+    f[2] = g[2] = 2;
+    f[3] = g[3] = 4;
+    for (int i = 4; i < n; ++i) {
+        f[i] = (f[i - 1] + f[i - 2] + f[i - 3]) % mod;
+        g[i] = (g[i - 1] + g[i - 2] + g[i - 3] + g[i - 4]) % mod;
+    }
+    return 0;
+}();
+
+class Solution {
+public:
+    int countTexts(string pressedKeys) {
+        long long ans = 1;
+        for (int i = 0, n = pressedKeys.length(); i < n; ++i) {
+            char c = pressedKeys[i];
+            int j = i;
+            while (j + 1 < n && pressedKeys[j + 1] == c) {
+                ++j;
+            }
+            int cnt = j - i + 1;
+            ans = c == '7' || c == '9' ? ans * g[cnt] : ans * f[cnt];
+            ans %= mod;
+            i = j;
+        }
+        return ans;
+    }
+};
+```
+
 #### Go
 
 ```go

solution/2200-2299/2266.Count Number of Texts/README_EN.md

@@ -76,7 +76,32 @@ Since we need to return the answer modulo 10<sup>9</sup> + 7, we return 20828761
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Grouping + Dynamic Programming
+
+According to the problem description, for consecutive identical characters in the string $\textit{pressedKeys}$, we can group them together and then calculate the number of ways for each group. Finally, we multiply the number of ways for all groups.
+
+The key problem is how to calculate the number of ways for each group.
+
+If a group of characters is '7' or '9', we can consider the last $1$, $2$, $3$, or $4$ characters of the group as one letter, then reduce the size of the group and transform it into a smaller subproblem.
+
+Similarly, if a group of characters is '2', '3', '4', '5', '6', or '8', we can consider the last $1$, $2$, or $3$ characters of the group as one letter, then reduce the size of the group and transform it into a smaller subproblem.
+
+Therefore, we define $f[i]$ to represent the number of ways for a group of length $i$ with identical characters that are not '7' or '9', and $g[i]$ to represent the number of ways for a group of length $i$ with identical characters that are '7' or '9'.
+
+Initially, $f[0] = f[1] = 1$, $f[2] = 2$, $f[3] = 4$, $g[0] = g[1] = 1$, $g[2] = 2$, $g[3] = 4$.
+
+For $i \ge 4$, we have:
+
+$$
+\begin{aligned}
+f[i] & = f[i-1] + f[i-2] + f[i-3] \\
+g[i] & = g[i-1] + g[i-2] + g[i-3] + g[i-4]
+\end{aligned}
+$$
+
+Finally, we traverse $\textit{pressedKeys}$, group consecutive identical characters, calculate the number of ways for each group, and multiply the number of ways for all groups.
+
+The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the string $\textit{pressedKeys}$.
 
 <!-- tabs:start -->
 
@@ -94,9 +119,9 @@ for _ in range(100000):
 class Solution:
     def countTexts(self, pressedKeys: str) -> int:
         ans = 1
-        for ch, s in groupby(pressedKeys):
+        for c, s in groupby(pressedKeys):
             m = len(list(s))
-            ans = ans * (g[m] if ch in "79" else f[m]) % mod
+            ans = ans * (g[m] if c in "79" else f[m]) % mod
         return ans
 ```
 
@@ -109,12 +134,10 @@ class Solution {
     private static long[] f = new long[N];
     private static long[] g = new long[N];
     static {
-        f[0] = 1;
-        f[1] = 1;
+        f[0] = f[1] = 1;
         f[2] = 2;
         f[3] = 4;
-        g[0] = 1;
-        g[1] = 1;
+        g[0] = g[1] = 1;
         g[2] = 2;
         g[3] = 4;
         for (int i = 4; i < N; ++i) {
@@ -126,10 +149,11 @@ class Solution {
     public int countTexts(String pressedKeys) {
         long ans = 1;
         for (int i = 0, n = pressedKeys.length(); i < n; ++i) {
-            int j = i;
             char c = pressedKeys.charAt(i);
-            for (; j + 1 < n && pressedKeys.charAt(j + 1) == c; ++j)
-                ;
+            int j = i;
+            while (j + 1 < n && pressedKeys.charAt(j + 1) == c) {
+                ++j;
+            }
             int cnt = j - i + 1;
             ans = c == '7' || c == '9' ? ans * g[cnt] : ans * f[cnt];
             ans %= MOD;
@@ -140,6 +164,45 @@ class Solution {
 }
 ```
 
+#### C++
+
+```cpp
+const int mod = 1e9 + 7;
+const int n = 1e5 + 10;
+long long f[n], g[n];
+
+int init = []() {
+    f[0] = g[0] = 1;
+    f[1] = g[1] = 1;
+    f[2] = g[2] = 2;
+    f[3] = g[3] = 4;
+    for (int i = 4; i < n; ++i) {
+        f[i] = (f[i - 1] + f[i - 2] + f[i - 3]) % mod;
+        g[i] = (g[i - 1] + g[i - 2] + g[i - 3] + g[i - 4]) % mod;
+    }
+    return 0;
+}();
+
+class Solution {
+public:
+    int countTexts(string pressedKeys) {
+        long long ans = 1;
+        for (int i = 0, n = pressedKeys.length(); i < n; ++i) {
+            char c = pressedKeys[i];
+            int j = i;
+            while (j + 1 < n && pressedKeys[j + 1] == c) {
+                ++j;
+            }
+            int cnt = j - i + 1;
+            ans = c == '7' || c == '9' ? ans * g[cnt] : ans * f[cnt];
+            ans %= mod;
+            i = j;
+        }
+        return ans;
+    }
+};
+```
+
 #### Go
 
 ```go

solution/2200-2299/2266.Count Number of Texts/Solution.cpp

@@ -0,0 +1,34 @@
+const int mod = 1e9 + 7;
+const int n = 1e5 + 10;
+long long f[n], g[n];
+
+int init = []() {
+    f[0] = g[0] = 1;
+    f[1] = g[1] = 1;
+    f[2] = g[2] = 2;
+    f[3] = g[3] = 4;
+    for (int i = 4; i < n; ++i) {
+        f[i] = (f[i - 1] + f[i - 2] + f[i - 3]) % mod;
+        g[i] = (g[i - 1] + g[i - 2] + g[i - 3] + g[i - 4]) % mod;
+    }
+    return 0;
+}();
+
+class Solution {
+public:
+    int countTexts(string pressedKeys) {
+        long long ans = 1;
+        for (int i = 0, n = pressedKeys.length(); i < n; ++i) {
+            char c = pressedKeys[i];
+            int j = i;
+            while (j + 1 < n && pressedKeys[j + 1] == c) {
+                ++j;
+            }
+            int cnt = j - i + 1;
+            ans = c == '7' || c == '9' ? ans * g[cnt] : ans * f[cnt];
+            ans %= mod;
+            i = j;
+        }
+        return ans;
+    }
+};

solution/2200-2299/2266.Count Number of Texts/Solution.java

@@ -4,12 +4,10 @@ class Solution {
     private static long[] f = new long[N];
     private static long[] g = new long[N];
     static {
-        f[0] = 1;
-        f[1] = 1;
+        f[0] = f[1] = 1;
         f[2] = 2;
         f[3] = 4;
-        g[0] = 1;
-        g[1] = 1;
+        g[0] = g[1] = 1;
         g[2] = 2;
         g[3] = 4;
         for (int i = 4; i < N; ++i) {
@@ -21,15 +19,16 @@ class Solution {
     public int countTexts(String pressedKeys) {
         long ans = 1;
         for (int i = 0, n = pressedKeys.length(); i < n; ++i) {
-            int j = i;
             char c = pressedKeys.charAt(i);
-            for (; j + 1 < n && pressedKeys.charAt(j + 1) == c; ++j)
-                ;
+            int j = i;
+            while (j + 1 < n && pressedKeys.charAt(j + 1) == c) {
+                ++j;
+            }
             int cnt = j - i + 1;
             ans = c == '7' || c == '9' ? ans * g[cnt] : ans * f[cnt];
             ans %= MOD;
             i = j;
         }
         return (int) ans;
     }
-}
+}

solution/2200-2299/2266.Count Number of Texts/Solution.py

@@ -9,7 +9,7 @@
 class Solution:
     def countTexts(self, pressedKeys: str) -> int:
         ans = 1
-        for ch, s in groupby(pressedKeys):
+        for c, s in groupby(pressedKeys):
             m = len(list(s))
-            ans = ans * (g[m] if ch in "79" else f[m]) % mod
+            ans = ans * (g[m] if c in "79" else f[m]) % mod
         return ans

solution/3400-3499/3421.Find Students Who Improved/README.md

@@ -34,8 +34,8 @@ tags:
 <p>编写一个解决方案来查找 <strong>进步的学生</strong>。如果 <strong>同时</strong> 满足以下两个条件，则该学生被认为是进步的：</p>
 
 <ul>
-	<li>在 <strong>同一科目</strong>&nbsp;至少有两个不同日期的考试。</li>
-	<li>他们在该学科&nbsp;<strong>最近的分数</strong>&nbsp;比他们 <strong>第一次的分数更高。</strong></li>
+	<li>在 <strong>同一科目</strong> 至少参加过两个不同日期的考试。</li>
+	<li>他们在该学科<strong> 最近的分数 </strong>比他们 第一次该学科考试的分数更高。</li>
 </ul>
 
 <p>返回结果表以&nbsp;<code>student_id</code>，<code>subject</code> <strong>升序</strong>&nbsp;排序。</p>

solution/3400-3499/3421.Find Students Who Improved/README_EN.md

@@ -54,15 +54,15 @@ Each row contains information about a student's score in a specific subject
 +------------+----------+-------+------------+
 | student_id | subject  | score | exam_date  |
 +------------+----------+-------+------------+
-| 101        | Math     | 70    | 15-01-2023 |
-| 101        | Math     | 85    | 15-02-2023 |
-| 101        | Physics  | 65    | 15-01-2023 |
-| 101        | Physics  | 60    | 15-02-2023 |
-| 102        | Math     | 80    | 15-01-2023 |
-| 102        | Math     | 85    | 15-02-2023 |
-| 103        | Math     | 90    | 15-01-2023 |
-| 104        | Physics  | 75    | 15-01-2023 |
-| 104        | Physics  | 85    | 15-02-2023 |
+| 101        | Math     | 70    | 2023-01-15 |
+| 101        | Math     | 85    | 2023-02-15 |
+| 101        | Physics  | 65    | 2023-01-15 |
+| 101        | Physics  | 60    | 2023-02-15 |
+| 102        | Math     | 80    | 2023-01-15 |
+| 102        | Math     | 85    | 2023-02-15 |
+| 103        | Math     | 90    | 2023-01-15 |
+| 104        | Physics  | 75    | 2023-01-15 |
+| 104        | Physics  | 85    | 2023-02-15 |
 +------------+----------+-------+------------+
 </pre>