diff --git a/solution/10000-10099/10035.Maximum Area of Longest Diagonal Rectangle/README.md b/solution/10000-10099/10035.Maximum Area of Longest Diagonal Rectangle/README.md new file mode 100644 index 0000000000000..572f31398e032 --- /dev/null +++ b/solution/10000-10099/10035.Maximum Area of Longest Diagonal Rectangle/README.md @@ -0,0 +1,86 @@ +# [10035. 对角线最长的矩形的面积](https://leetcode.cn/problems/maximum-area-of-longest-diagonal-rectangle) + +[English Version](/solution/10000-10099/10035.Maximum%20Area%20of%20Longest%20Diagonal%20Rectangle/README_EN.md) + +## 题目描述 + + + +

给你一个下标从 0 开始的二维整数数组 dimensions。

+ +

对于所有下标 i（0 <= i < dimensions.length），dimensions[i][0] 表示矩形 i 的长度，而 dimensions[i][1] 表示矩形 i 的宽度。

+ +

返回对角线最 长 的矩形的 面积 。如果存在多个对角线长度相同的矩形，返回面积最 大 的矩形的面积。

+ +

 

+ +

示例 1：

+ +
+输入：dimensions = [[9,3],[8,6]]
+输出：48
+解释：
+下标 = 0，长度 = 9，宽度 = 3。对角线长度 = sqrt(9 * 9 + 3 * 3) = sqrt(90) ≈ 9.487。
+下标 = 1，长度 = 8，宽度 = 6。对角线长度 = sqrt(8 * 8 + 6 * 6) = sqrt(100) = 10。
+因此，下标为 1 的矩形对角线更长，所以返回面积 = 8 * 6 = 48。
+
+ +

示例 2：

+ +
+输入：dimensions = [[3,4],[4,3]]
+输出：12
+解释：两个矩形的对角线长度相同，为 5，所以最大面积 = 12。
+
+ +

 

+ +

提示：

+ + + +## 解法 + + + + + +### **Python3** + + + +```python + +``` + +### **Java** + + + +```java + +``` + +### **C++** + +```cpp + +``` + +### **Go** + +```go + +``` + +### **...** + +``` + +``` + + diff --git a/solution/10000-10099/10035.Maximum Area of Longest Diagonal Rectangle/README_EN.md b/solution/10000-10099/10035.Maximum Area of Longest Diagonal Rectangle/README_EN.md new file mode 100644 index 0000000000000..344d43c6d78f2 --- /dev/null +++ b/solution/10000-10099/10035.Maximum Area of Longest Diagonal Rectangle/README_EN.md @@ -0,0 +1,76 @@ +# [10035. Maximum Area of Longest Diagonal Rectangle](https://leetcode.com/problems/maximum-area-of-longest-diagonal-rectangle) + +[中文文档](/solution/10000-10099/10035.Maximum%20Area%20of%20Longest%20Diagonal%20Rectangle/README.md) + +## Description + +

You are given a 2D 0-indexed integer array dimensions.

+ +

For all indices i, 0 <= i < dimensions.length, dimensions[i][0] represents the length and dimensions[i][1] represents the width of the rectangle i.

+ +

Return the area of the rectangle having the longest diagonal. If there are multiple rectangles with the longest diagonal, return the area of the rectangle having the maximum area.

+ +

 

+

Example 1:

+ +
+Input: dimensions = [[9,3],[8,6]]
+Output: 48
+Explanation: 
+For index = 0, length = 9 and width = 3. Diagonal length = sqrt(9 * 9 + 3 * 3) = sqrt(90) ≈ 9.487.
+For index = 1, length = 8 and width = 6. Diagonal length = sqrt(8 * 8 + 6 * 6) = sqrt(100) = 10.
+So, the rectangle at index 1 has a greater diagonal length therefore we return area = 8 * 6 = 48.
+
+ +

Example 2:

+ +
+Input: dimensions = [[3,4],[4,3]]
+Output: 12
+Explanation: Length of diagonal is the same for both which is 5, so maximum area = 12.
+
+ +

 

+

Constraints:

+ + + +## Solutions + + + +### **Python3** + +```python + +``` + +### **Java** + +```java + +``` + +### **C++** + +```cpp + +``` + +### **Go** + +```go + +``` + +### **...** + +``` + +``` + + diff --git a/solution/10000-10099/10036.Minimum Moves to Capture The Queen/README.md b/solution/10000-10099/10036.Minimum Moves to Capture The Queen/README.md new file mode 100644 index 0000000000000..dbabbdf85d957 --- /dev/null +++ b/solution/10000-10099/10036.Minimum Moves to Capture The Queen/README.md @@ -0,0 +1,100 @@ +# [10036. 捕获黑皇后需要的最少移动次数](https://leetcode.cn/problems/minimum-moves-to-capture-the-queen) + +[English Version](/solution/10000-10099/10036.Minimum%20Moves%20to%20Capture%20The%20Queen/README_EN.md) + +## 题目描述 + + + +

现有一个下标从 0 开始的 8 x 8 棋盘，上面有 3 枚棋子。

+ +

给你 6 个整数 a 、b 、c 、d 、e 和 f ，其中：

+ + + +

假定你只能移动白色棋子，返回捕获黑皇后所需的最少移动次数。

+ +

请注意：

+ + + +

 

+ +

示例 1：

+ +
+输入：a = 1, b = 1, c = 8, d = 8, e = 2, f = 3
+输出：2
+解释：将白色车先移动到 (1, 3) ，然后移动到 (2, 3) 来捕获黑皇后，共需移动 2 次。
+由于起始时没有任何棋子正在攻击黑皇后，要想捕获黑皇后，移动次数不可能少于 2 次。
+
+ +

示例 2：

+ +
+输入：a = 5, b = 3, c = 3, d = 4, e = 5, f = 2
+输出：1
+解释：可以通过以下任一方式移动 1 次捕获黑皇后：
+- 将白色车移动到 (5, 2) 。
+- 将白色象移动到 (5, 2) 。
+
+ +

 

+ +

提示：

+ + + +## 解法 + + + + + +### **Python3** + + + +```python + +``` + +### **Java** + + + +```java + +``` + +### **C++** + +```cpp + +``` + +### **Go** + +```go + +``` + +### **...** + +``` + +``` + + diff --git a/solution/10000-10099/10036.Minimum Moves to Capture The Queen/README_EN.md b/solution/10000-10099/10036.Minimum Moves to Capture The Queen/README_EN.md new file mode 100644 index 0000000000000..8340757c22f7f --- /dev/null +++ b/solution/10000-10099/10036.Minimum Moves to Capture The Queen/README_EN.md @@ -0,0 +1,90 @@ +# [10036. Minimum Moves to Capture The Queen](https://leetcode.com/problems/minimum-moves-to-capture-the-queen) + +[中文文档](/solution/10000-10099/10036.Minimum%20Moves%20to%20Capture%20The%20Queen/README.md) + +## Description + +

There is a 1-indexed 8 x 8 chessboard containing 3 pieces.

+ +

You are given 6 integers a, b, c, d, e, and f where:

+ + + +

Given that you can only move the white pieces, return the minimum number of moves required to capture the black queen.

+ +

Note that:

+ + + +

 

+

Example 1:

+ +
+Input: a = 1, b = 1, c = 8, d = 8, e = 2, f = 3
+Output: 2
+Explanation: We can capture the black queen in two moves by moving the white rook to (1, 3) then to (2, 3).
+It is impossible to capture the black queen in less than two moves since it is not being attacked by any of the pieces at the beginning.
+
+ +

Example 2:

+ +
+Input: a = 5, b = 3, c = 3, d = 4, e = 5, f = 2
+Output: 1
+Explanation: We can capture the black queen in a single move by doing one of the following: 
+- Move the white rook to (5, 2).
+- Move the white bishop to (5, 2).
+
+ +

 

+

Constraints:

+ + + +## Solutions + + + +### **Python3** + +```python + +``` + +### **Java** + +```java + +``` + +### **C++** + +```cpp + +``` + +### **Go** + +```go + +``` + +### **...** + +``` + +``` + + diff --git a/solution/10000-10099/10036.Minimum Moves to Capture The Queen/images/ex1.png b/solution/10000-10099/10036.Minimum Moves to Capture The Queen/images/ex1.png new file mode 100644 index 0000000000000..7621069b3105d Binary files /dev/null and b/solution/10000-10099/10036.Minimum Moves to Capture The Queen/images/ex1.png differ diff --git a/solution/10000-10099/10036.Minimum Moves to Capture The Queen/images/ex2.png b/solution/10000-10099/10036.Minimum Moves to Capture The Queen/images/ex2.png new file mode 100644 index 0000000000000..d0f43a360a83f Binary files /dev/null and b/solution/10000-10099/10036.Minimum Moves to Capture The Queen/images/ex2.png differ diff --git a/solution/10000-10099/10037.Maximum Size of a Set After Removals/README.md b/solution/10000-10099/10037.Maximum Size of a Set After Removals/README.md new file mode 100644 index 0000000000000..1febb5498b758 --- /dev/null +++ b/solution/10000-10099/10037.Maximum Size of a Set After Removals/README.md @@ -0,0 +1,94 @@ +# [10037. 移除后集合的最多元素数](https://leetcode.cn/problems/maximum-size-of-a-set-after-removals) + +[English Version](/solution/10000-10099/10037.Maximum%20Size%20of%20a%20Set%20After%20Removals/README_EN.md) + +## 题目描述 + + + +

给你两个下标从 0 开始的整数数组 nums1 和 nums2 ，它们的长度都是偶数 n 。

+ +

你必须从 nums1 中移除 n / 2 个元素，同时从 nums2 中也移除 n / 2 个元素。移除之后，你将 nums1 和 nums2 中剩下的元素插入到集合 s 中。

+ +

返回集合 s可能的 最多 包含多少元素。

+ +

 

+ +

示例 1：

+ +
+输入：nums1 = [1,2,1,2], nums2 = [1,1,1,1]
+输出：2
+解释：从 nums1 和 nums2 中移除两个 1 。移除后，数组变为 nums1 = [2,2] 和 nums2 = [1,1] 。因此，s = {1,2} 。
+可以证明，在移除之后，集合 s 最多可以包含 2 个元素。
+
+ +

示例 2：

+ +
+输入：nums1 = [1,2,3,4,5,6], nums2 = [2,3,2,3,2,3]
+输出：5
+解释：从 nums1 中移除 2、3 和 6 ，同时从 nums2 中移除两个 3 和一个 2 。移除后，数组变为 nums1 = [1,4,5] 和 nums2 = [2,3,2] 。因此，s = {1,2,3,4,5} 。
+可以证明，在移除之后，集合 s 最多可以包含 5 个元素。 
+
+ +

示例 3：

+ +
+输入：nums1 = [1,1,2,2,3,3], nums2 = [4,4,5,5,6,6]
+输出：6
+解释：从 nums1 中移除 1、2 和 3 ，同时从 nums2 中移除 4、5 和 6 。移除后，数组变为 nums1 = [1,2,3] 和 nums2 = [4,5,6] 。因此，s = {1,2,3,4,5,6} 。
+可以证明，在移除之后，集合 s 最多可以包含 6 个元素。
+ +

 

+ +

提示：

+ + + +## 解法 + + + + + +### **Python3** + + + +```python + +``` + +### **Java** + + + +```java + +``` + +### **C++** + +```cpp + +``` + +### **Go** + +```go + +``` + +### **...** + +``` + +``` + + diff --git a/solution/10000-10099/10037.Maximum Size of a Set After Removals/README_EN.md b/solution/10000-10099/10037.Maximum Size of a Set After Removals/README_EN.md new file mode 100644 index 0000000000000..48e6ae02a235e --- /dev/null +++ b/solution/10000-10099/10037.Maximum Size of a Set After Removals/README_EN.md @@ -0,0 +1,85 @@ +# [10037. Maximum Size of a Set After Removals](https://leetcode.com/problems/maximum-size-of-a-set-after-removals) + +[中文文档](/solution/10000-10099/10037.Maximum%20Size%20of%20a%20Set%20After%20Removals/README.md) + +## Description + +

You are given two 0-indexed integer arrays nums1 and nums2 of even length n.

+ +

You must remove n / 2 elements from nums1 and n / 2 elements from nums2. After the removals, you insert the remaining elements of nums1 and nums2 into a set s.

+ +

Return the maximum possible size of the set s.

+ +

 

+

Example 1:

+ +
+Input: nums1 = [1,2,1,2], nums2 = [1,1,1,1]
+Output: 2
+Explanation: We remove two occurences of 1 from nums1 and nums2. After the removals, the arrays become equal to nums1 = [2,2] and nums2 = [1,1]. Therefore, s = {1,2}.
+It can be shown that 2 is the maximum possible size of the set s after the removals.
+
+ +

Example 2:

+ +
+Input: nums1 = [1,2,3,4,5,6], nums2 = [2,3,2,3,2,3]
+Output: 5
+Explanation: We remove 2, 3, and 6 from nums1, as well as 2 and two occurrences of 3 from nums2. After the removals, the arrays become equal to nums1 = [1,4,5] and nums2 = [2,3,2]. Therefore, s = {1,2,3,4,5}.
+It can be shown that 5 is the maximum possible size of the set s after the removals.
+
+ +

Example 3:

+ +
+Input: nums1 = [1,1,2,2,3,3], nums2 = [4,4,5,5,6,6]
+Output: 6
+Explanation: We remove 1, 2, and 3 from nums1, as well as 4, 5, and 6 from nums2. After the removals, the arrays become equal to nums1 = [1,2,3] and nums2 = [4,5,6]. Therefore, s = {1,2,3,4,5,6}.
+It can be shown that 6 is the maximum possible size of the set s after the removals.
+
+ +

 

+

Constraints:

+ + + +## Solutions + + + +### **Python3** + +```python + +``` + +### **Java** + +```java + +``` + +### **C++** + +```cpp + +``` + +### **Go** + +```go + +``` + +### **...** + +``` + +``` + + diff --git a/solution/10000-10099/10038.Maximize the Number of Partitions After Operations/README.md b/solution/10000-10099/10038.Maximize the Number of Partitions After Operations/README.md new file mode 100644 index 0000000000000..1f7e21e2497c4 --- /dev/null +++ b/solution/10000-10099/10038.Maximize the Number of Partitions After Operations/README.md @@ -0,0 +1,122 @@ +# [10038. 执行操作后的最大分割数量](https://leetcode.cn/problems/maximize-the-number-of-partitions-after-operations) + +[English Version](/solution/10000-10099/10038.Maximize%20the%20Number%20of%20Partitions%20After%20Operations/README_EN.md) + +## 题目描述 + + + +

给你一个下标从 0 开始的字符串 s 和一个整数 k。

+ +

你需要执行以下分割操作，直到字符串 s 变为 空：

+ + + +

执行操作之 前 ，你可以将 s 中 至多一处 下标的对应字符更改为另一个小写英文字母。

+ +

在最优选择情形下改变至多一处下标对应字符后，用整数表示并返回操作结束时得到的最大分割数量。

+ +

 

+ +

示例 1：

+ +
+输入：s = "accca", k = 2
+输出：3
+解释：在此示例中，为了最大化得到的分割数量，可以将 s[2] 改为 'b'。
+s 变为 "acbca"。
+按照以下方式执行操作，直到 s 变为空：
+- 选择最长且至多包含 2 个不同字符的前缀，"acbca"。
+- 删除该前缀，s 变为 "bca"。现在分割数量为 1。
+- 选择最长且至多包含 2 个不同字符的前缀，"bca"。
+- 删除该前缀，s 变为 "a"。现在分割数量为 2。
+- 选择最长且至多包含 2 个不同字符的前缀，"a"。
+- 删除该前缀，s 变为空。现在分割数量为 3。
+因此，答案是 3。
+可以证明，分割数量不可能超过 3。
+ +

示例 2：

+ +
+输入：s = "aabaab", k = 3
+输出：1
+解释：在此示例中，为了最大化得到的分割数量，可以保持 s 不变。
+按照以下方式执行操作，直到 s 变为空： 
+- 选择最长且至多包含 3 个不同字符的前缀，"aabaab"。
+- 删除该前缀，s 变为空。现在分割数量为 1。
+因此，答案是 1。
+可以证明，分割数量不可能超过 1。
+ +

示例 3：

+ +
+输入：s = "xxyz", k = 1
+输出：4
+解释：在此示例中，为了最大化得到的分割数量，可以将 s[1] 改为 'a'。
+s 变为 "xayz"。
+按照以下方式执行操作，直到 s 变为空：
+- 选择最长且至多包含 1 个不同字符的前缀，"xayz"。
+- 删除该前缀，s 变为 "ayz"。现在分割数量为 1。
+- 选择最长且至多包含 1 个不同字符的前缀，"ayz"。
+- 删除该前缀，s 变为 "yz"，现在分割数量为 2。
+- 选择最长且至多包含 1 个不同字符的前缀，"yz"。
+- 删除该前缀，s 变为 "z"。现在分割数量为 3。
+- 选择最且至多包含 1 个不同字符的前缀，"z"。
+- 删除该前缀，s 变为空。现在分割数量为 4。
+因此，答案是 4。
+可以证明，分割数量不可能超过 4。
+ +

 

+ +

提示：

+ + + +## 解法 + + + + + +### **Python3** + + + +```python + +``` + +### **Java** + + + +```java + +``` + +### **C++** + +```cpp + +``` + +### **Go** + +```go + +``` + +### **...** + +``` + +``` + + diff --git a/solution/10000-10099/10038.Maximize the Number of Partitions After Operations/README_EN.md b/solution/10000-10099/10038.Maximize the Number of Partitions After Operations/README_EN.md new file mode 100644 index 0000000000000..4c356c31570e7 --- /dev/null +++ b/solution/10000-10099/10038.Maximize the Number of Partitions After Operations/README_EN.md @@ -0,0 +1,113 @@ +# [10038. Maximize the Number of Partitions After Operations](https://leetcode.com/problems/maximize-the-number-of-partitions-after-operations) + +[中文文档](/solution/10000-10099/10038.Maximize%20the%20Number%20of%20Partitions%20After%20Operations/README.md) + +## Description + +

You are given a 0-indexed string s and an integer k.

+ +

You are to perform the following partitioning operations until s is empty:

+ + + +

Before the operations, you are allowed to change at most one index in s to another lowercase English letter.

+ +

Return an integer denoting the maximum number of resulting partitions after the operations by optimally choosing at most one index to change.

+

 

+

Example 1:

+ +
+Input: s = "accca", k = 2
+Output: 3
+Explanation: In this example, to maximize the number of resulting partitions, s[2] can be changed to 'b'.
+s becomes "acbca".
+The operations can now be performed as follows until s becomes empty:
+- Choose the longest prefix containing at most 2 distinct characters, "acbca".
+- Delete the prefix, and s becomes "bca". The number of partitions is now 1.
+- Choose the longest prefix containing at most 2 distinct characters, "bca".
+- Delete the prefix, and s becomes "a". The number of partitions is now 2.
+- Choose the longest prefix containing at most 2 distinct characters, "a".
+- Delete the prefix, and s becomes empty. The number of partitions is now 3.
+Hence, the answer is 3.
+It can be shown that it is not possible to obtain more than 3 partitions.
+ +

Example 2:

+ +
+Input: s = "aabaab", k = 3
+Output: 1
+Explanation: In this example, to maximize the number of resulting partitions we can leave s as it is.
+The operations can now be performed as follows until s becomes empty: 
+- Choose the longest prefix containing at most 3 distinct characters, "aabaab".
+- Delete the prefix, and s becomes empty. The number of partitions becomes 1. 
+Hence, the answer is 1. 
+It can be shown that it is not possible to obtain more than 1 partition.
+
+ +

Example 3:

+ +
+Input: s = "xxyz", k = 1
+Output: 4
+Explanation: In this example, to maximize the number of resulting partitions, s[1] can be changed to 'a'.
+s becomes "xayz".
+The operations can now be performed as follows until s becomes empty:
+- Choose the longest prefix containing at most 1 distinct character, "xayz".
+- Delete the prefix, and s becomes "ayz". The number of partitions is now 1.
+- Choose the longest prefix containing at most 1 distinct character, "ayz".
+- Delete the prefix, and s becomes "yz". The number of partitions is now 2.
+- Choose the longest prefix containing at most 1 distinct character, "yz".
+- Delete the prefix, and s becomes "z". The number of partitions is now 3.
+- Choose the longest prefix containing at most 1 distinct character, "z".
+- Delete the prefix, and s becomes empty. The number of partitions is now 4.
+Hence, the answer is 4.
+It can be shown that it is not possible to obtain more than 4 partitions.
+
+ +

 

+

Constraints:

+ + + +## Solutions + + + +### **Python3** + +```python + +``` + +### **Java** + +```java + +``` + +### **C++** + +```cpp + +``` + +### **Go** + +```go + +``` + +### **...** + +``` + +``` + + diff --git a/solution/CONTEST_README.md b/solution/CONTEST_README.md index 2db1f3ed66f28..2c5ee4d9f3b29 100644 --- a/solution/CONTEST_README.md +++ b/solution/CONTEST_README.md @@ -22,6 +22,13 @@ ## 往期竞赛 +#### 第 379 场周赛(2024-01-07 10:30, 90 分钟) 参赛人数 3117 + +- [10035. 对角线最长的矩形的面积](/solution/10000-10099/10035.Maximum%20Area%20of%20Longest%20Diagonal%20Rectangle/README.md) +- [10036. 捕获黑皇后需要的最少移动次数](/solution/10000-10099/10036.Minimum%20Moves%20to%20Capture%20The%20Queen/README.md) +- [10037. 移除后集合的最多元素数](/solution/10000-10099/10037.Maximum%20Size%20of%20a%20Set%20After%20Removals/README.md) +- [10038. 执行操作后的最大分割数量](/solution/10000-10099/10038.Maximize%20the%20Number%20of%20Partitions%20After%20Operations/README.md) + #### 第 121 场双周赛(2024-01-06 22:30, 90 分钟) 参赛人数 2218 - [10031. 大于等于顺序前缀和的最小缺失整数](/solution/10000-10099/10031.Smallest%20Missing%20Integer%20Greater%20Than%20Sequential%20Prefix%20Sum/README.md) diff --git a/solution/CONTEST_README_EN.md b/solution/CONTEST_README_EN.md index 73e2ce765ce2f..d5474e0d4503f 100644 --- a/solution/CONTEST_README_EN.md +++ b/solution/CONTEST_README_EN.md @@ -25,6 +25,13 @@ Get your rating changes right after the completion of LeetCode contests, https:/ ## Past Contests +#### Weekly Contest 379 + +- [10035. Maximum Area of Longest Diagonal Rectangle](/solution/10000-10099/10035.Maximum%20Area%20of%20Longest%20Diagonal%20Rectangle/README_EN.md) +- [10036. Minimum Moves to Capture The Queen](/solution/10000-10099/10036.Minimum%20Moves%20to%20Capture%20The%20Queen/README_EN.md) +- [10037. Maximum Size of a Set After Removals](/solution/10000-10099/10037.Maximum%20Size%20of%20a%20Set%20After%20Removals/README_EN.md) +- [10038. Maximize the Number of Partitions After Operations](/solution/10000-10099/10038.Maximize%20the%20Number%20of%20Partitions%20After%20Operations/README_EN.md) + #### Biweekly Contest 121 - [10031. Smallest Missing Integer Greater Than Sequential Prefix Sum](/solution/10000-10099/10031.Smallest%20Missing%20Integer%20Greater%20Than%20Sequential%20Prefix%20Sum/README_EN.md) diff --git a/solution/README.md b/solution/README.md index 406559bf185f7..c8d48881372f0 100644 --- a/solution/README.md +++ b/solution/README.md @@ -1018,6 +1018,10 @@ | 10032 | [使数组异或和等于 K 的最少操作次数](/solution/10000-10099/10032.Minimum%20Number%20of%20Operations%20to%20Make%20Array%20XOR%20Equal%20to%20K/README.md) | | 中等 | 第 121 场双周赛 | | 10033 | [使 X 和 Y 相等的最少操作次数](/solution/10000-10099/10033.Minimum%20Number%20of%20Operations%20to%20Make%20X%20and%20Y%20Equal/README.md) | | 中等 | 第 121 场双周赛 | | 10034 | [统计强大整数的数目](/solution/10000-10099/10034.Count%20the%20Number%20of%20Powerful%20Integers/README.md) | | 困难 | 第 121 场双周赛 | +| 10035 | [对角线最长的矩形的面积](/solution/10000-10099/10035.Maximum%20Area%20of%20Longest%20Diagonal%20Rectangle/README.md) | | 简单 | 第 379 场周赛 | +| 10036 | [捕获黑皇后需要的最少移动次数](/solution/10000-10099/10036.Minimum%20Moves%20to%20Capture%20The%20Queen/README.md) | | 中等 | 第 379 场周赛 | +| 10037 | [移除后集合的最多元素数](/solution/10000-10099/10037.Maximum%20Size%20of%20a%20Set%20After%20Removals/README.md) | | 中等 | 第 379 场周赛 | +| 10038 | [执行操作后的最大分割数量](/solution/10000-10099/10038.Maximize%20the%20Number%20of%20Partitions%20After%20Operations/README.md) | | 困难 | 第 379 场周赛 | | 1004 | [最大连续1的个数 III](/solution/1000-1099/1004.Max%20Consecutive%20Ones%20III/README.md) | `数组`,`二分查找`,`前缀和`,`滑动窗口` | 中等 | 第 126 场周赛 | | 1005 | [K 次取反后最大化的数组和](/solution/1000-1099/1005.Maximize%20Sum%20Of%20Array%20After%20K%20Negations/README.md) | `贪心`,`数组`,`排序` | 简单 | 第 127 场周赛 | | 1006 | [笨阶乘](/solution/1000-1099/1006.Clumsy%20Factorial/README.md) | `栈`,`数学`,`模拟` | 中等 | 第 127 场周赛 | diff --git a/solution/README_EN.md b/solution/README_EN.md index 9277b30a025e8..1b50bfa70d655 100644 --- a/solution/README_EN.md +++ b/solution/README_EN.md @@ -1016,6 +1016,10 @@ Press Control + F(or Command + F on | 10032 | [Minimum Number of Operations to Make Array XOR Equal to K](/solution/10000-10099/10032.Minimum%20Number%20of%20Operations%20to%20Make%20Array%20XOR%20Equal%20to%20K/README_EN.md) | | Medium | Biweekly Contest 121 | | 10033 | [Minimum Number of Operations to Make X and Y Equal](/solution/10000-10099/10033.Minimum%20Number%20of%20Operations%20to%20Make%20X%20and%20Y%20Equal/README_EN.md) | | Medium | Biweekly Contest 121 | | 10034 | [Count the Number of Powerful Integers](/solution/10000-10099/10034.Count%20the%20Number%20of%20Powerful%20Integers/README_EN.md) | | Hard | Biweekly Contest 121 | +| 10035 | [Maximum Area of Longest Diagonal Rectangle](/solution/10000-10099/10035.Maximum%20Area%20of%20Longest%20Diagonal%20Rectangle/README_EN.md) | | Easy | Weekly Contest 379 | +| 10036 | [Minimum Moves to Capture The Queen](/solution/10000-10099/10036.Minimum%20Moves%20to%20Capture%20The%20Queen/README_EN.md) | | Medium | Weekly Contest 379 | +| 10037 | [Maximum Size of a Set After Removals](/solution/10000-10099/10037.Maximum%20Size%20of%20a%20Set%20After%20Removals/README_EN.md) | | Medium | Weekly Contest 379 | +| 10038 | [Maximize the Number of Partitions After Operations](/solution/10000-10099/10038.Maximize%20the%20Number%20of%20Partitions%20After%20Operations/README_EN.md) | | Hard | Weekly Contest 379 | | 1004 | [Max Consecutive Ones III](/solution/1000-1099/1004.Max%20Consecutive%20Ones%20III/README_EN.md) | `Array`,`Binary Search`,`Prefix Sum`,`Sliding Window` | Medium | Weekly Contest 126 | | 1005 | [Maximize Sum Of Array After K Negations](/solution/1000-1099/1005.Maximize%20Sum%20Of%20Array%20After%20K%20Negations/README_EN.md) | `Greedy`,`Array`,`Sorting` | Easy | Weekly Contest 127 | | 1006 | [Clumsy Factorial](/solution/1000-1099/1006.Clumsy%20Factorial/README_EN.md) | `Stack`,`Math`,`Simulation` | Medium | Weekly Contest 127 | diff --git a/solution/summary.md b/solution/summary.md index 24748a8f8e242..42bc814454b43 100644 --- a/solution/summary.md +++ b/solution/summary.md @@ -1125,6 +1125,10 @@ - [10032检查替换后的词是否有效](/solution/10000-10099/10032.Minimum%20Number%20of%20Operations%20to%20Make%20Array%20XOR%20Equal%20to%20K/README.md) - [10033检查替换后的词是否有效](/solution/10000-10099/10033.Minimum%20Number%20of%20Operations%20to%20Make%20X%20and%20Y%20Equal/README.md) - [10034检查替换后的词是否有效](/solution/10000-10099/10034.Count%20the%20Number%20of%20Powerful%20Integers/README.md) + - [10035检查替换后的词是否有效](/solution/10000-10099/10035.Maximum%20Area%20of%20Longest%20Diagonal%20Rectangle/README.md) + - [10036检查替换后的词是否有效](/solution/10000-10099/10036.Minimum%20Moves%20to%20Capture%20The%20Queen/README.md) + - [10037检查替换后的词是否有效](/solution/10000-10099/10037.Maximum%20Size%20of%20a%20Set%20After%20Removals/README.md) + - [10038检查替换后的词是否有效](/solution/10000-10099/10038.Maximize%20the%20Number%20of%20Partitions%20After%20Operations/README.md) - 1100-1199 - [1100.长度为 K 的无重复字符子串](/solution/1100-1199/1100.Find%20K-Length%20Substrings%20With%20No%20Repeated%20Characters/README.md) diff --git a/solution/summary_en.md b/solution/summary_en.md index bbd2e3f04acf3..543e358aa639e 100644 --- a/solution/summary_en.md +++ b/solution/summary_en.md @@ -1125,6 +1125,10 @@ - [10032.Minimum Number of Operations to Make Array XOR Equal to K](/solution/10000-10099/10032.Minimum%20Number%20of%20Operations%20to%20Make%20Array%20XOR%20Equal%20to%20K/README_EN.md) - [10033.Minimum Number of Operations to Make X and Y Equal](/solution/10000-10099/10033.Minimum%20Number%20of%20Operations%20to%20Make%20X%20and%20Y%20Equal/README_EN.md) - [10034.Count the Number of Powerful Integers](/solution/10000-10099/10034.Count%20the%20Number%20of%20Powerful%20Integers/README_EN.md) + - [10035.Maximum Area of Longest Diagonal Rectangle](/solution/10000-10099/10035.Maximum%20Area%20of%20Longest%20Diagonal%20Rectangle/README_EN.md) + - [10036.Minimum Moves to Capture The Queen](/solution/10000-10099/10036.Minimum%20Moves%20to%20Capture%20The%20Queen/README_EN.md) + - [10037.Maximum Size of a Set After Removals](/solution/10000-10099/10037.Maximum%20Size%20of%20a%20Set%20After%20Removals/README_EN.md) + - [10038.Maximize the Number of Partitions After Operations](/solution/10000-10099/10038.Maximize%20the%20Number%20of%20Partitions%20After%20Operations/README_EN.md) - 1100-1199 - [1100.Find K-Length Substrings With No Repeated Characters](/solution/1100-1199/1100.Find%20K-Length%20Substrings%20With%20No%20Repeated%20Characters/README_EN.md)