diff --git a/solution/2800-2899/2894.Divisible and Non-divisible Sums Difference/README.md b/solution/2800-2899/2894.Divisible and Non-divisible Sums Difference/README.md new file mode 100644 index 0000000000000..3f66d34f912b7 --- /dev/null +++ b/solution/2800-2899/2894.Divisible and Non-divisible Sums Difference/README.md @@ -0,0 +1,103 @@ +# [2894. 分类求和并作差](https://leetcode.cn/problems/divisible-and-non-divisible-sums-difference) + +[English Version](/solution/2800-2899/2894.Divisible%20and%20Non-divisible%20Sums%20Difference/README_EN.md) + +## 题目描述 + + + +

给你两个正整数 n 和 m 。

+ +

现定义两个整数 num1 和 num2 ，如下所示：

+ + + +

返回整数 num1 - num2 。

+ +

 

+ +

示例 1：

+ +
+输入：n = 10, m = 3
+输出：19
+解释：在这个示例中：
+- 范围 [1, 10] 内无法被 3 整除的整数为 [1,2,4,5,7,8,10] ，num1 = 这些整数之和 = 37 。
+- 范围 [1, 10] 内能够被 3 整除的整数为 [3,6,9] ，num2 = 这些整数之和 = 18 。
+返回 37 - 18 = 19 作为答案。
+
+ +

示例 2：

+ +
+输入：n = 5, m = 6
+输出：15
+解释：在这个示例中：
+- 范围 [1, 5] 内无法被 6 整除的整数为 [1,2,3,4,5] ，num1 = 这些整数之和 =  15 。
+- 范围 [1, 5] 内能够被 6 整除的整数为 [] ，num2 = 这些整数之和 = 0 。
+返回 15 - 0 = 15 作为答案。
+
+ +

示例 3：

+ +
+输入：n = 5, m = 1
+输出：-15
+解释：在这个示例中：
+- 范围 [1, 5] 内无法被 1 整除的整数为 [] ，num1 = 这些整数之和 = 0 。 
+- 范围 [1, 5] 内能够被 1 整除的整数为 [1,2,3,4,5] ，num2 = 这些整数之和 = 15 。
+返回 0 - 15 = -15 作为答案。
+
+ +

 

+ +

提示：

+ + + +## 解法 + + + + + +### **Python3** + + + +```python + +``` + +### **Java** + + + +```java + +``` + +### **C++** + +```cpp + +``` + +### **Go** + +```go + +``` + +### **...** + +``` + +``` + + diff --git a/solution/2800-2899/2894.Divisible and Non-divisible Sums Difference/README_EN.md b/solution/2800-2899/2894.Divisible and Non-divisible Sums Difference/README_EN.md new file mode 100644 index 0000000000000..7c13e0baee5ba --- /dev/null +++ b/solution/2800-2899/2894.Divisible and Non-divisible Sums Difference/README_EN.md @@ -0,0 +1,93 @@ +# [2894. Divisible and Non-divisible Sums Difference](https://leetcode.com/problems/divisible-and-non-divisible-sums-difference) + +[中文文档](/solution/2800-2899/2894.Divisible%20and%20Non-divisible%20Sums%20Difference/README.md) + +## Description + +

You are given positive integers n and m.

+ +

Define two integers, num1 and num2, as follows:

+ + + +

Return the integer num1 - num2.

+ +

 

+

Example 1:

+ +
+Input: n = 10, m = 3
+Output: 19
+Explanation: In the given example:
+- Integers in the range [1, 10] that are not divisible by 3 are [1,2,4,5,7,8,10], num1 is the sum of those integers = 37.
+- Integers in the range [1, 10] that are divisible by 3 are [3,6,9], num2 is the sum of those integers = 18.
+We return 37 - 18 = 19 as the answer.
+
+ +

Example 2:

+ +
+Input: n = 5, m = 6
+Output: 15
+Explanation: In the given example:
+- Integers in the range [1, 5] that are not divisible by 6 are [1,2,3,4,5], num1 is the sum of those integers = 15.
+- Integers in the range [1, 5] that are divisible by 6 are [], num2 is the sum of those integers = 0.
+We return 15 - 0 = 15 as the answer.
+
+ +

Example 3:

+ +
+Input: n = 5, m = 1
+Output: -15
+Explanation: In the given example:
+- Integers in the range [1, 5] that are not divisible by 1 are [], num1 is the sum of those integers = 0.
+- Integers in the range [1, 5] that are divisible by 1 are [1,2,3,4,5], num2 is the sum of those integers = 15.
+We return 0 - 15 = -15 as the answer.
+
+ +

 

+

Constraints:

+ + + +## Solutions + + + +### **Python3** + +```python + +``` + +### **Java** + +```java + +``` + +### **C++** + +```cpp + +``` + +### **Go** + +```go + +``` + +### **...** + +``` + +``` + + diff --git a/solution/2800-2899/2895.Minimum Processing Time/README.md b/solution/2800-2899/2895.Minimum Processing Time/README.md new file mode 100644 index 0000000000000..3953037da1424 --- /dev/null +++ b/solution/2800-2899/2895.Minimum Processing Time/README.md @@ -0,0 +1,92 @@ +# [2895. 最小处理时间](https://leetcode.cn/problems/minimum-processing-time) + +[English Version](/solution/2800-2899/2895.Minimum%20Processing%20Time/README_EN.md) + +## 题目描述 + + + +

你有 n 颗处理器，每颗处理器都有 4 个核心。现有 n * 4 个待执行任务，每个核心只执行 一个 任务。

+ +

给你一个下标从 0 开始的整数数组 processorTime ，表示每颗处理器最早空闲时间。另给你一个下标从 0 开始的整数数组 tasks ，表示执行每个任务所需的时间。返回所有任务都执行完毕需要的 最小时间 。

+ +

注意：每个核心独立执行任务。

+ +

 

+ +

示例 1：

+ +
+输入：processorTime = [8,10], tasks = [2,2,3,1,8,7,4,5]
+输出：16
+解释：
+最优的方案是将下标为 4, 5, 6, 7 的任务分配给第一颗处理器（最早空闲时间 time = 8），下标为 0, 1, 2, 3 的任务分配给第二颗处理器（最早空闲时间 time = 10）。 
+第一颗处理器执行完所有任务需要花费的时间 = max(8 + 8, 8 + 7, 8 + 4, 8 + 5) = 16 。
+第二颗处理器执行完所有任务需要花费的时间 = max(10 + 2, 10 + 2, 10 + 3, 10 + 1) = 13 。
+因此，可以证明执行完所有任务需要花费的最小时间是 16 。
+ +

示例 2：

+ +
+输入：processorTime = [10,20], tasks = [2,3,1,2,5,8,4,3]
+输出：23
+解释：
+最优的方案是将下标为 1, 4, 5, 6 的任务分配给第一颗处理器（最早空闲时间 time = 10），下标为 0, 2, 3, 7 的任务分配给第二颗处理器（最早空闲时间 time = 20）。 
+第一颗处理器执行完所有任务需要花费的时间 = max(10 + 3, 10 + 5, 10 + 8, 10 + 4) = 18 。 
+第二颗处理器执行完所有任务需要花费的时间 = max(20 + 2, 20 + 1, 20 + 2, 20 + 3) = 23 。 
+因此，可以证明执行完所有任务需要花费的最小时间是 23 。
+
+ +

 

+ +

提示：

+ + + +## 解法 + + + + + +### **Python3** + + + +```python + +``` + +### **Java** + + + +```java + +``` + +### **C++** + +```cpp + +``` + +### **Go** + +```go + +``` + +### **...** + +``` + +``` + + diff --git a/solution/2800-2899/2895.Minimum Processing Time/README_EN.md b/solution/2800-2899/2895.Minimum Processing Time/README_EN.md new file mode 100644 index 0000000000000..bec0c21955f33 --- /dev/null +++ b/solution/2800-2899/2895.Minimum Processing Time/README_EN.md @@ -0,0 +1,82 @@ +# [2895. Minimum Processing Time](https://leetcode.com/problems/minimum-processing-time) + +[中文文档](/solution/2800-2899/2895.Minimum%20Processing%20Time/README.md) + +## Description + +

You have n processors each having 4 cores and n * 4 tasks that need to be executed such that each core should perform only one task.

+ +

Given a 0-indexed integer array processorTime representing the time at which each processor becomes available for the first time and a 0-indexed integer array tasks representing the time it takes to execute each task, return the minimum time when all of the tasks have been executed by the processors.

+ +

Note: Each core executes the task independently of the others.

+ +

 

+

Example 1:

+ +
+Input: processorTime = [8,10], tasks = [2,2,3,1,8,7,4,5]
+Output: 16
+Explanation: 
+It's optimal to assign the tasks at indexes 4, 5, 6, 7 to the first processor which becomes available at time = 8, and the tasks at indexes 0, 1, 2, 3 to the second processor which becomes available at time = 10. 
+Time taken by the first processor to finish execution of all tasks = max(8 + 8, 8 + 7, 8 + 4, 8 + 5) = 16.
+Time taken by the second processor to finish execution of all tasks = max(10 + 2, 10 + 2, 10 + 3, 10 + 1) = 13.
+Hence, it can be shown that the minimum time taken to execute all the tasks is 16.
+ +

Example 2:

+ +
+Input: processorTime = [10,20], tasks = [2,3,1,2,5,8,4,3]
+Output: 23
+Explanation: 
+It's optimal to assign the tasks at indexes 1, 4, 5, 6 to the first processor which becomes available at time = 10, and the tasks at indexes 0, 2, 3, 7 to the second processor which becomes available at time = 20.
+Time taken by the first processor to finish execution of all tasks = max(10 + 3, 10 + 5, 10 + 8, 10 + 4) = 18.
+Time taken by the second processor to finish execution of all tasks = max(20 + 2, 20 + 1, 20 + 2, 20 + 3) = 23.
+Hence, it can be shown that the minimum time taken to execute all the tasks is 23.
+
+ +

 

+

Constraints:

+ + + +## Solutions + + + +### **Python3** + +```python + +``` + +### **Java** + +```java + +``` + +### **C++** + +```cpp + +``` + +### **Go** + +```go + +``` + +### **...** + +``` + +``` + + diff --git a/solution/2800-2899/2896.Apply Operations to Make Two Strings Equal/README.md b/solution/2800-2899/2896.Apply Operations to Make Two Strings Equal/README.md new file mode 100644 index 0000000000000..436910d2f9341 --- /dev/null +++ b/solution/2800-2899/2896.Apply Operations to Make Two Strings Equal/README.md @@ -0,0 +1,94 @@ +# [2896. 执行操作使两个字符串相等](https://leetcode.cn/problems/apply-operations-to-make-two-strings-equal) + +[English Version](/solution/2800-2899/2896.Apply%20Operations%20to%20Make%20Two%20Strings%20Equal/README_EN.md) + +## 题目描述 + + + +

给你两个下标从 0 开始的二进制字符串 s1 和 s2 ，两个字符串的长度都是 n ，再给你一个正整数 x 。

+ +

你可以对字符串 s1 执行以下操作 任意次 ：

+ + + +

请你返回使字符串 s1 和 s2 相等的 最小 操作代价之和，如果无法让二者相等，返回 -1 。

+ +

注意 ，反转字符的意思是将 0 变成 1 ，或者 1 变成 0 。

+ +

 

+ +

示例 1：

+ +
+输入：s1 = "1100011000", s2 = "0101001010", x = 2
+输出：4
+解释：我们可以执行以下操作：
+- 选择 i = 3 执行第二个操作。结果字符串是 s1 = "1101111000" 。
+- 选择 i = 4 执行第二个操作。结果字符串是 s1 = "1101001000" 。
+- 选择 i = 0 和 j = 8 ，执行第一个操作。结果字符串是 s1 = "0101001010" = s2 。
+总代价是 1 + 1 + 2 = 4 。这是最小代价和。
+
+ +

示例 2：

+ +
+输入：s1 = "10110", s2 = "00011", x = 4
+输出：-1
+解释：无法使两个字符串相等。
+
+ +

 

+ +

提示：

+ + + +## 解法 + + + + + +### **Python3** + + + +```python + +``` + +### **Java** + + + +```java + +``` + +### **C++** + +```cpp + +``` + +### **Go** + +```go + +``` + +### **...** + +``` + +``` + + diff --git a/solution/2800-2899/2896.Apply Operations to Make Two Strings Equal/README_EN.md b/solution/2800-2899/2896.Apply Operations to Make Two Strings Equal/README_EN.md new file mode 100644 index 0000000000000..ce36722dcb2b1 --- /dev/null +++ b/solution/2800-2899/2896.Apply Operations to Make Two Strings Equal/README_EN.md @@ -0,0 +1,84 @@ +# [2896. Apply Operations to Make Two Strings Equal](https://leetcode.com/problems/apply-operations-to-make-two-strings-equal) + +[中文文档](/solution/2800-2899/2896.Apply%20Operations%20to%20Make%20Two%20Strings%20Equal/README.md) + +## Description + +

You are given two 0-indexed binary strings s1 and s2, both of length n, and a positive integer x.

+ +

You can perform any of the following operations on the string s1 any number of times:

+ + + +

Return the minimum cost needed to make the strings s1 and s2 equal, or return -1 if it is impossible.

+ +

Note that flipping a character means changing it from 0 to 1 or vice-versa.

+ +

 

+

Example 1:

+ +
+Input: s1 = "1100011000", s2 = "0101001010", x = 2
+Output: 4
+Explanation: We can do the following operations:
+- Choose i = 3 and apply the second operation. The resulting string is s1 = "1101111000".
+- Choose i = 4 and apply the second operation. The resulting string is s1 = "1101001000".
+- Choose i = 0 and j = 8 and apply the first operation. The resulting string is s1 = "0101001010" = s2.
+The total cost is 1 + 1 + 2 = 4. It can be shown that it is the minimum cost possible.
+
+ +

Example 2:

+ +
+Input: s1 = "10110", s2 = "00011", x = 4
+Output: -1
+Explanation: It is not possible to make the two strings equal.
+
+ +

 

+

Constraints:

+ + + +## Solutions + + + +### **Python3** + +```python + +``` + +### **Java** + +```java + +``` + +### **C++** + +```cpp + +``` + +### **Go** + +```go + +``` + +### **...** + +``` + +``` + + diff --git a/solution/2800-2899/2897.Apply Operations on Array to Maximize Sum of Squares/README.md b/solution/2800-2899/2897.Apply Operations on Array to Maximize Sum of Squares/README.md new file mode 100644 index 0000000000000..895ee9ef0f7dd --- /dev/null +++ b/solution/2800-2899/2897.Apply Operations on Array to Maximize Sum of Squares/README.md @@ -0,0 +1,96 @@ +# [2897. 对数组执行操作使平方和最大](https://leetcode.cn/problems/apply-operations-on-array-to-maximize-sum-of-squares) + +[English Version](/solution/2800-2899/2897.Apply%20Operations%20on%20Array%20to%20Maximize%20Sum%20of%20Squares/README_EN.md) + +## 题目描述 + + + +

给你一个下标从 0 开始的整数数组 nums 和一个 正 整数 k 。

+ +

你可以对数组执行以下操作 任意次 ：

+ + + +

你需要从最终的数组里选择 k 个元素，并计算它们的 平方 之和。

+ +

请你返回你可以得到的 最大 平方和。

+ +

由于答案可能会很大，将答案对 109 + 7 取余 后返回。

+ +

 

+ +

示例 1：

+ +
+输入：nums = [2,6,5,8], k = 2
+输出：261
+解释：我们可以对数组执行以下操作：
+- 选择 i = 0 和 j = 3 ，同时将 nums[0] 变为 (2 AND 8) = 0 且 nums[3] 变为 (2 OR 8) = 10 ，结果数组为 nums = [0,6,5,10] 。
+- 选择 i = 2 和 j = 3 ，同时将 nums[2] 变为 (5 AND 10) = 0 且 nums[3] 变为 (5 OR 10) = 15 ，结果数组为 nums = [0,6,0,15] 。
+从最终数组里选择元素 15 和 6 ，平方和为 152 + 62 = 261 。
+261 是可以得到的最大结果。
+
+ +

示例 2：

+ +
+输入：nums = [4,5,4,7], k = 3
+输出：90
+解释：不需要执行任何操作。
+选择元素 7 ，5 和 4 ，平方和为 72 + 52 + 42 = 90 。
+90 是可以得到的最大结果。
+
+ +

 

+ +

提示：

+ + + +## 解法 + + + + + +### **Python3** + + + +```python + +``` + +### **Java** + + + +```java + +``` + +### **C++** + +```cpp + +``` + +### **Go** + +```go + +``` + +### **...** + +``` + +``` + + diff --git a/solution/2800-2899/2897.Apply Operations on Array to Maximize Sum of Squares/README_EN.md b/solution/2800-2899/2897.Apply Operations on Array to Maximize Sum of Squares/README_EN.md new file mode 100644 index 0000000000000..590360f0bda48 --- /dev/null +++ b/solution/2800-2899/2897.Apply Operations on Array to Maximize Sum of Squares/README_EN.md @@ -0,0 +1,86 @@ +# [2897. Apply Operations on Array to Maximize Sum of Squares](https://leetcode.com/problems/apply-operations-on-array-to-maximize-sum-of-squares) + +[中文文档](/solution/2800-2899/2897.Apply%20Operations%20on%20Array%20to%20Maximize%20Sum%20of%20Squares/README.md) + +## Description + +

You are given a 0-indexed integer array nums and a positive integer k.

+ +

You can do the following operation on the array any number of times:

+ + + +

You have to choose k elements from the final array and calculate the sum of their squares.

+ +

Return the maximum sum of squares you can achieve.

+ +

Since the answer can be very large, return it modulo 109 + 7.

+ +

 

+

Example 1:

+ +
+Input: nums = [2,6,5,8], k = 2
+Output: 261
+Explanation: We can do the following operations on the array:
+- Choose i = 0 and j = 3, then change nums[0] to (2 AND 8) = 0 and nums[3] to (2 OR 8) = 10. The resulting array is nums = [0,6,5,10].
+- Choose i = 2 and j = 3, then change nums[2] to (5 AND 10) = 0 and nums[3] to (5 OR 10) = 15. The resulting array is nums = [0,6,0,15].
+We can choose the elements 15 and 6 from the final array. The sum of squares is 152 + 62 = 261.
+It can be shown that this is the maximum value we can get.
+
+ +

Example 2:

+ +
+Input: nums = [4,5,4,7], k = 3
+Output: 90
+Explanation: We do not need to apply any operations.
+We can choose the elements 7, 5, and 4 with a sum of squares: 72 + 52 + 42 = 90.
+It can be shown that this is the maximum value we can get.
+
+ +

 

+

Constraints:

+ + + +## Solutions + + + +### **Python3** + +```python + +``` + +### **Java** + +```java + +``` + +### **C++** + +```cpp + +``` + +### **Go** + +```go + +``` + +### **...** + +``` + +``` + + diff --git a/solution/CONTEST_README.md b/solution/CONTEST_README.md index 28f2e07820fbc..8c239d0bd0053 100644 --- a/solution/CONTEST_README.md +++ b/solution/CONTEST_README.md @@ -22,6 +22,13 @@ ## 往期竞赛 +#### 第 366 场周赛(2023-10-08 10:30, 90 分钟) 参赛人数 2790 + +- [2894. 分类求和并作差](/solution/2800-2899/2894.Divisible%20and%20Non-divisible%20Sums%20Difference/README.md) +- [2895. 最小处理时间](/solution/2800-2899/2895.Minimum%20Processing%20Time/README.md) +- [2896. 执行操作使两个字符串相等](/solution/2800-2899/2896.Apply%20Operations%20to%20Make%20Two%20Strings%20Equal/README.md) +- [2897. 对数组执行操作使平方和最大](/solution/2800-2899/2897.Apply%20Operations%20on%20Array%20to%20Maximize%20Sum%20of%20Squares/README.md) + #### 第 365 场周赛(2023-10-01 10:30, 90 分钟) 参赛人数 2909 - [2873. 有序三元组中的最大值 I](/solution/2800-2899/2873.Maximum%20Value%20of%20an%20Ordered%20Triplet%20I/README.md) diff --git a/solution/CONTEST_README_EN.md b/solution/CONTEST_README_EN.md index 792ec1703868c..fbba212a38f33 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 366 + +- [2894. Divisible and Non-divisible Sums Difference](/solution/2800-2899/2894.Divisible%20and%20Non-divisible%20Sums%20Difference/README_EN.md) +- [2895. Minimum Processing Time](/solution/2800-2899/2895.Minimum%20Processing%20Time/README_EN.md) +- [2896. Apply Operations to Make Two Strings Equal](/solution/2800-2899/2896.Apply%20Operations%20to%20Make%20Two%20Strings%20Equal/README_EN.md) +- [2897. Apply Operations on Array to Maximize Sum of Squares](/solution/2800-2899/2897.Apply%20Operations%20on%20Array%20to%20Maximize%20Sum%20of%20Squares/README_EN.md) + #### Weekly Contest 365 - [2873. Maximum Value of an Ordered Triplet I](/solution/2800-2899/2873.Maximum%20Value%20of%20an%20Ordered%20Triplet%20I/README_EN.md) diff --git a/solution/README.md b/solution/README.md index 988c0c1082dd2..2328ae0a5357e 100644 --- a/solution/README.md +++ b/solution/README.md @@ -2904,6 +2904,10 @@ | 2891 | [Method Chaining](/solution/2800-2899/2891.Method%20Chaining/README.md) | | 简单 | | | 2892 | [Minimizing Array After Replacing Pairs With Their Product](/solution/2800-2899/2892.Minimizing%20Array%20After%20Replacing%20Pairs%20With%20Their%20Product/README.md) | | 中等 | 🔒 | | 2893 | [Calculate Orders Within Each Interval](/solution/2800-2899/2893.Calculate%20Orders%20Within%20Each%20Interval/README.md) | | 中等 | 🔒 | +| 2894 | [分类求和并作差](/solution/2800-2899/2894.Divisible%20and%20Non-divisible%20Sums%20Difference/README.md) | | 简单 | 第 366 场周赛 | +| 2895 | [最小处理时间](/solution/2800-2899/2895.Minimum%20Processing%20Time/README.md) | | 中等 | 第 366 场周赛 | +| 2896 | [执行操作使两个字符串相等](/solution/2800-2899/2896.Apply%20Operations%20to%20Make%20Two%20Strings%20Equal/README.md) | | 中等 | 第 366 场周赛 | +| 2897 | [对数组执行操作使平方和最大](/solution/2800-2899/2897.Apply%20Operations%20on%20Array%20to%20Maximize%20Sum%20of%20Squares/README.md) | | 困难 | 第 366 场周赛 | ## 版权 diff --git a/solution/README_EN.md b/solution/README_EN.md index 040c0fa3e8ba5..43ee56d3f9858 100644 --- a/solution/README_EN.md +++ b/solution/README_EN.md @@ -2902,6 +2902,10 @@ Press Control + F(or Command + F on | 2891 | [Method Chaining](/solution/2800-2899/2891.Method%20Chaining/README_EN.md) | | Easy | | | 2892 | [Minimizing Array After Replacing Pairs With Their Product](/solution/2800-2899/2892.Minimizing%20Array%20After%20Replacing%20Pairs%20With%20Their%20Product/README_EN.md) | | Medium | 🔒 | | 2893 | [Calculate Orders Within Each Interval](/solution/2800-2899/2893.Calculate%20Orders%20Within%20Each%20Interval/README_EN.md) | | Medium | 🔒 | +| 2894 | [Divisible and Non-divisible Sums Difference](/solution/2800-2899/2894.Divisible%20and%20Non-divisible%20Sums%20Difference/README_EN.md) | | Easy | Weekly Contest 366 | +| 2895 | [Minimum Processing Time](/solution/2800-2899/2895.Minimum%20Processing%20Time/README_EN.md) | | Medium | Weekly Contest 366 | +| 2896 | [Apply Operations to Make Two Strings Equal](/solution/2800-2899/2896.Apply%20Operations%20to%20Make%20Two%20Strings%20Equal/README_EN.md) | | Medium | Weekly Contest 366 | +| 2897 | [Apply Operations on Array to Maximize Sum of Squares](/solution/2800-2899/2897.Apply%20Operations%20on%20Array%20to%20Maximize%20Sum%20of%20Squares/README_EN.md) | | Hard | Weekly Contest 366 | ## Copyright diff --git a/solution/summary.md b/solution/summary.md index 664db13ecb20d..60cef39e98a14 100644 --- a/solution/summary.md +++ b/solution/summary.md @@ -2949,3 +2949,7 @@ - [2891.Method Chaining](/solution/2800-2899/2891.Method%20Chaining/README.md) - [2892.Minimizing Array After Replacing Pairs With Their Product](/solution/2800-2899/2892.Minimizing%20Array%20After%20Replacing%20Pairs%20With%20Their%20Product/README.md) - [2893.Calculate Orders Within Each Interval](/solution/2800-2899/2893.Calculate%20Orders%20Within%20Each%20Interval/README.md) + - [2894.分类求和并作差](/solution/2800-2899/2894.Divisible%20and%20Non-divisible%20Sums%20Difference/README.md) + - [2895.最小处理时间](/solution/2800-2899/2895.Minimum%20Processing%20Time/README.md) + - [2896.执行操作使两个字符串相等](/solution/2800-2899/2896.Apply%20Operations%20to%20Make%20Two%20Strings%20Equal/README.md) + - [2897.对数组执行操作使平方和最大](/solution/2800-2899/2897.Apply%20Operations%20on%20Array%20to%20Maximize%20Sum%20of%20Squares/README.md) diff --git a/solution/summary_en.md b/solution/summary_en.md index 6200030e654ae..fa7e7f91712ba 100644 --- a/solution/summary_en.md +++ b/solution/summary_en.md @@ -2949,3 +2949,7 @@ - [2891.Method Chaining](/solution/2800-2899/2891.Method%20Chaining/README_EN.md) - [2892.Minimizing Array After Replacing Pairs With Their Product](/solution/2800-2899/2892.Minimizing%20Array%20After%20Replacing%20Pairs%20With%20Their%20Product/README_EN.md) - [2893.Calculate Orders Within Each Interval](/solution/2800-2899/2893.Calculate%20Orders%20Within%20Each%20Interval/README_EN.md) + - [2894.Divisible and Non-divisible Sums Difference](/solution/2800-2899/2894.Divisible%20and%20Non-divisible%20Sums%20Difference/README_EN.md) + - [2895.Minimum Processing Time](/solution/2800-2899/2895.Minimum%20Processing%20Time/README_EN.md) + - [2896.Apply Operations to Make Two Strings Equal](/solution/2800-2899/2896.Apply%20Operations%20to%20Make%20Two%20Strings%20Equal/README_EN.md) + - [2897.Apply Operations on Array to Maximize Sum of Squares](/solution/2800-2899/2897.Apply%20Operations%20on%20Array%20to%20Maximize%20Sum%20of%20Squares/README_EN.md)