diff --git a/solution/1500-1599/1565.Unique Orders and Customers Per Month/README_EN.md b/solution/1500-1599/1565.Unique Orders and Customers Per Month/README_EN.md
index 17f2788eee346..308a4c17a7bd1 100644
--- a/solution/1500-1599/1565.Unique Orders and Customers Per Month/README_EN.md
+++ b/solution/1500-1599/1565.Unique Orders and Customers Per Month/README_EN.md
@@ -43,7 +43,7 @@ This table contains information about the orders made by customer_id.
Example 1:
-Input:
+Input:
Orders table:
+----------+------------+-------------+------------+
| order_id | order_date | customer_id | invoice |
@@ -59,7 +59,7 @@ Orders table:
| 9 | 2021-01-07 | 3 | 31 |
| 10 | 2021-01-15 | 2 | 20 |
+----------+------------+-------------+------------+
-Output:
+Output:
+---------+-------------+----------------+
| month | order_count | customer_count |
+---------+-------------+----------------+
@@ -68,7 +68,7 @@ Orders table:
| 2020-12 | 2 | 1 |
| 2021-01 | 1 | 1 |
+---------+-------------+----------------+
-Explanation:
+Explanation:
In September 2020 we have two orders from 2 different customers with invoices > $20.
In October 2020 we have two orders from 1 customer, and only one of the two orders has invoice > $20.
In November 2020 we have two orders from 2 different customers but invoices < $20, so we don't include that month.
diff --git a/solution/1500-1599/1577.Number of Ways Where Square of Number Is Equal to Product of Two Numbers/README_EN.md b/solution/1500-1599/1577.Number of Ways Where Square of Number Is Equal to Product of Two Numbers/README_EN.md
index 830fab875afd5..af21803f6266d 100644
--- a/solution/1500-1599/1577.Number of Ways Where Square of Number Is Equal to Product of Two Numbers/README_EN.md
+++ b/solution/1500-1599/1577.Number of Ways Where Square of Number Is Equal to Product of Two Numbers/README_EN.md
@@ -34,7 +34,7 @@ tags:
Input: nums1 = [7,4], nums2 = [5,2,8,9]
Output: 1
-Explanation: Type 1: (1, 1, 2), nums1[1]2 = nums2[1] * nums2[2]. (42 = 2 * 8).
+Explanation: Type 1: (1, 1, 2), nums1[1]2 = nums2[1] * nums2[2]. (42 = 2 * 8).
Example 2:
diff --git a/solution/1500-1599/1589.Maximum Sum Obtained of Any Permutation/README_EN.md b/solution/1500-1599/1589.Maximum Sum Obtained of Any Permutation/README_EN.md
index 386c07bc8ee49..f2b200a3a8618 100644
--- a/solution/1500-1599/1589.Maximum Sum Obtained of Any Permutation/README_EN.md
+++ b/solution/1500-1599/1589.Maximum Sum Obtained of Any Permutation/README_EN.md
@@ -33,7 +33,7 @@ tags:
Input: nums = [1,2,3,4,5], requests = [[1,3],[0,1]]
Output: 19
-Explanation: One permutation of nums is [2,1,3,4,5] with the following result:
+Explanation: One permutation of nums is [2,1,3,4,5] with the following result:
requests[0] -> nums[1] + nums[2] + nums[3] = 1 + 3 + 4 = 8
requests[1] -> nums[0] + nums[1] = 2 + 1 = 3
Total sum: 8 + 3 = 11.
diff --git a/solution/3400-3499/3400.Maximum Number of Matching Indices After Right Shifts/README.md b/solution/3400-3499/3400.Maximum Number of Matching Indices After Right Shifts/README.md
new file mode 100644
index 0000000000000..7e0d855f9d99f
--- /dev/null
+++ b/solution/3400-3499/3400.Maximum Number of Matching Indices After Right Shifts/README.md
@@ -0,0 +1,172 @@
+---
+comments: true
+difficulty: 中等
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3400-3499/3400.Maximum%20Number%20of%20Matching%20Indices%20After%20Right%20Shifts/README.md
+---
+
+
+
+# [3400. 右移后的最大匹配索引数 🔒](https://leetcode.cn/problems/maximum-number-of-matching-indices-after-right-shifts)
+
+[English Version](/solution/3400-3499/3400.Maximum%20Number%20of%20Matching%20Indices%20After%20Right%20Shifts/README_EN.md)
+
+## 题目描述
+
+
+
+给定两个长度相同的整数数组 nums1 和 nums2。
+
+如果 nums1[i] == nums2[i] 则认为下标 i 是 匹配 的。
+
+返回在 nums1 上进行任意次数 右移 后 最大 的 匹配 下标数量。
+
+右移 是对于所有下标，将位于下标 i 的元素移动到 (i + 1) % n。
+
+
+
+示例 1：
+
+
+
输入：nums1 = [3,1,2,3,1,2], nums2 = [1,2,3,1,2,3]
+
+
输出：6
+
+
解释：
+
+
如果我们右移 nums1 2 次，它变为 [1, 2, 3, 1, 2, 3]。每个下标都匹配，所以输出为 6。
+
示例 2：
+
+
+
输入：nums1 = [1,4,2,5,3,1], nums2 = [2,3,1,2,4,6]
+
+
输出：3
+
+
解释：
+
+
如果我们右移 nums1 3 次，它变为 [5, 3, 1, 1, 4, 2]。下标 1，2，4 匹配，所以输出为 3。
+
+
+提示：
+
+
+ nums1.length == nums2.length1 <= nums1.length, nums2.length <= 30001 <= nums1[i], nums2[i] <= 109
+
+
+
+## 解法
+
+
+
+### 方法一：枚举
+
+我们可以枚举右移的次数 $k$，其中 $0 \leq k \lt n$。对于每一个 $k$，我们可以计算出右移 $k$ 次后的数组 $\textit{nums1}$ 和 $\textit{nums2}$ 的匹配下标数量，取最大值作为答案即可。
+
+时间复杂度 $O(n^2)$，其中 $n$ 为数组 $\textit{nums1}$ 的长度。空间复杂度 $O(1)$。
+
+
+
+#### Python3
+
+```python
+class Solution:
+ def maximumMatchingIndices(self, nums1: List[int], nums2: List[int]) -> int:
+ n = len(nums1)
+ ans = 0
+ for k in range(n):
+ t = sum(nums1[(i + k) % n] == x for i, x in enumerate(nums2))
+ ans = max(ans, t)
+ return ans
+```
+
+#### Java
+
+```java
+class Solution {
+ public int maximumMatchingIndices(int[] nums1, int[] nums2) {
+ int n = nums1.length;
+ int ans = 0;
+ for (int k = 0; k < n; ++k) {
+ int t = 0;
+ for (int i = 0; i < n; ++i) {
+ if (nums1[(i + k) % n] == nums2[i]) {
+ ++t;
+ }
+ }
+ ans = Math.max(ans, t);
+ }
+ return ans;
+ }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+ int maximumMatchingIndices(vector& nums1, vector& nums2) {
+ int n = nums1.size();
+ int ans = 0;
+ for (int k = 0; k < n; ++k) {
+ int t = 0;
+ for (int i = 0; i < n; ++i) {
+ if (nums1[(i + k) % n] == nums2[i]) {
+ ++t;
+ }
+ }
+ ans = max(ans, t);
+ }
+ return ans;
+ }
+};
+```
+
+#### Go
+
+```go
+func maximumMatchingIndices(nums1 []int, nums2 []int) (ans int) {
+ n := len(nums1)
+ for k := range nums1 {
+ t := 0
+ for i, x := range nums2 {
+ if nums1[(i+k)%n] == x {
+ t++
+ }
+ }
+ ans = max(ans, t)
+ }
+ return
+}
+```
+
+#### TypeScript
+
+```ts
+function maximumMatchingIndices(nums1: number[], nums2: number[]): number {
+ const n = nums1.length;
+ let ans: number = 0;
+ for (let k = 0; k < n; ++k) {
+ let t: number = 0;
+ for (let i = 0; i < n; ++i) {
+ if (nums1[(i + k) % n] === nums2[i]) {
+ ++t;
+ }
+ }
+ ans = Math.max(ans, t);
+ }
+ return ans;
+}
+```
+
+
+
+
+
+
diff --git a/solution/3400-3499/3400.Maximum Number of Matching Indices After Right Shifts/README_EN.md b/solution/3400-3499/3400.Maximum Number of Matching Indices After Right Shifts/README_EN.md
new file mode 100644
index 0000000000000..0b88e75ddc597
--- /dev/null
+++ b/solution/3400-3499/3400.Maximum Number of Matching Indices After Right Shifts/README_EN.md
@@ -0,0 +1,170 @@
+---
+comments: true
+difficulty: Medium
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3400-3499/3400.Maximum%20Number%20of%20Matching%20Indices%20After%20Right%20Shifts/README_EN.md
+---
+
+
+
+# [3400. Maximum Number of Matching Indices After Right Shifts 🔒](https://leetcode.com/problems/maximum-number-of-matching-indices-after-right-shifts)
+
+[中文文档](/solution/3400-3499/3400.Maximum%20Number%20of%20Matching%20Indices%20After%20Right%20Shifts/README.md)
+
+## Description
+
+
+
+You are given two integer arrays, nums1 and nums2, of the same length.
+
+An index i is considered matching if nums1[i] == nums2[i].
+
+Return the maximum number of matching indices after performing any number of right shifts on nums1.
+
+A right shift is defined as shifting the element at index i to index (i + 1) % n, for all indices.
+
+
+Example 1:
+
+
+
Input: nums1 = [3,1,2,3,1,2], nums2 = [1,2,3,1,2,3]
+
+
Output: 6
+
+
Explanation:
+
+
If we right shift nums1 2 times, it becomes [1, 2, 3, 1, 2, 3]. Every index matches, so the output is 6.
+
Example 2:
+
+
+
Input: nums1 = [1,4,2,5,3,1], nums2 = [2,3,1,2,4,6]
+
+
Output: 3
+
+
Explanation:
+
+
If we right shift nums1 3 times, it becomes [5, 3, 1, 1, 4, 2]. Indices 1, 2, and 4 match, so the output is 3.
+
+Constraints:
+
+
+ nums1.length == nums2.length1 <= nums1.length, nums2.length <= 30001 <= nums1[i], nums2[i] <= 109
+
+
+
+## Solutions
+
+
+
+### Solution 1: Enumeration
+
+We can enumerate the number of right shifts $k$, where $0 \leq k < n$. For each $k$, we can calculate the number of matching indices between the array $\textit{nums1}$ after right shifting $k$ times and $\textit{nums2}$. The maximum value is taken as the answer.
+
+The time complexity is $O(n^2)$, where $n$ is the length of the array $\textit{nums1}$. The space complexity is $O(1)$.
+
+
+
+#### Python3
+
+```python
+class Solution:
+ def maximumMatchingIndices(self, nums1: List[int], nums2: List[int]) -> int:
+ n = len(nums1)
+ ans = 0
+ for k in range(n):
+ t = sum(nums1[(i + k) % n] == x for i, x in enumerate(nums2))
+ ans = max(ans, t)
+ return ans
+```
+
+#### Java
+
+```java
+class Solution {
+ public int maximumMatchingIndices(int[] nums1, int[] nums2) {
+ int n = nums1.length;
+ int ans = 0;
+ for (int k = 0; k < n; ++k) {
+ int t = 0;
+ for (int i = 0; i < n; ++i) {
+ if (nums1[(i + k) % n] == nums2[i]) {
+ ++t;
+ }
+ }
+ ans = Math.max(ans, t);
+ }
+ return ans;
+ }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+ int maximumMatchingIndices(vector& nums1, vector& nums2) {
+ int n = nums1.size();
+ int ans = 0;
+ for (int k = 0; k < n; ++k) {
+ int t = 0;
+ for (int i = 0; i < n; ++i) {
+ if (nums1[(i + k) % n] == nums2[i]) {
+ ++t;
+ }
+ }
+ ans = max(ans, t);
+ }
+ return ans;
+ }
+};
+```
+
+#### Go
+
+```go
+func maximumMatchingIndices(nums1 []int, nums2 []int) (ans int) {
+ n := len(nums1)
+ for k := range nums1 {
+ t := 0
+ for i, x := range nums2 {
+ if nums1[(i+k)%n] == x {
+ t++
+ }
+ }
+ ans = max(ans, t)
+ }
+ return
+}
+```
+
+#### TypeScript
+
+```ts
+function maximumMatchingIndices(nums1: number[], nums2: number[]): number {
+ const n = nums1.length;
+ let ans: number = 0;
+ for (let k = 0; k < n; ++k) {
+ let t: number = 0;
+ for (let i = 0; i < n; ++i) {
+ if (nums1[(i + k) % n] === nums2[i]) {
+ ++t;
+ }
+ }
+ ans = Math.max(ans, t);
+ }
+ return ans;
+}
+```
+
+
+
+
+
+
diff --git a/solution/3400-3499/3400.Maximum Number of Matching Indices After Right Shifts/Solution.cpp b/solution/3400-3499/3400.Maximum Number of Matching Indices After Right Shifts/Solution.cpp
new file mode 100644
index 0000000000000..2832ec0b23e3c
--- /dev/null
+++ b/solution/3400-3499/3400.Maximum Number of Matching Indices After Right Shifts/Solution.cpp
@@ -0,0 +1,17 @@
+class Solution {
+public:
+ int maximumMatchingIndices(vector& nums1, vector& nums2) {
+ int n = nums1.size();
+ int ans = 0;
+ for (int k = 0; k < n; ++k) {
+ int t = 0;
+ for (int i = 0; i < n; ++i) {
+ if (nums1[(i + k) % n] == nums2[i]) {
+ ++t;
+ }
+ }
+ ans = max(ans, t);
+ }
+ return ans;
+ }
+};
diff --git a/solution/3400-3499/3400.Maximum Number of Matching Indices After Right Shifts/Solution.go b/solution/3400-3499/3400.Maximum Number of Matching Indices After Right Shifts/Solution.go
new file mode 100644
index 0000000000000..c2d173c2ce12c
--- /dev/null
+++ b/solution/3400-3499/3400.Maximum Number of Matching Indices After Right Shifts/Solution.go
@@ -0,0 +1,13 @@
+func maximumMatchingIndices(nums1 []int, nums2 []int) (ans int) {
+ n := len(nums1)
+ for k := range nums1 {
+ t := 0
+ for i, x := range nums2 {
+ if nums1[(i+k)%n] == x {
+ t++
+ }
+ }
+ ans = max(ans, t)
+ }
+ return
+}
diff --git a/solution/3400-3499/3400.Maximum Number of Matching Indices After Right Shifts/Solution.java b/solution/3400-3499/3400.Maximum Number of Matching Indices After Right Shifts/Solution.java
new file mode 100644
index 0000000000000..301cba253c1d4
--- /dev/null
+++ b/solution/3400-3499/3400.Maximum Number of Matching Indices After Right Shifts/Solution.java
@@ -0,0 +1,16 @@
+class Solution {
+ public int maximumMatchingIndices(int[] nums1, int[] nums2) {
+ int n = nums1.length;
+ int ans = 0;
+ for (int k = 0; k < n; ++k) {
+ int t = 0;
+ for (int i = 0; i < n; ++i) {
+ if (nums1[(i + k) % n] == nums2[i]) {
+ ++t;
+ }
+ }
+ ans = Math.max(ans, t);
+ }
+ return ans;
+ }
+}
diff --git a/solution/3400-3499/3400.Maximum Number of Matching Indices After Right Shifts/Solution.py b/solution/3400-3499/3400.Maximum Number of Matching Indices After Right Shifts/Solution.py
new file mode 100644
index 0000000000000..ced0cc1c3ef6e
--- /dev/null
+++ b/solution/3400-3499/3400.Maximum Number of Matching Indices After Right Shifts/Solution.py
@@ -0,0 +1,8 @@
+class Solution:
+ def maximumMatchingIndices(self, nums1: List[int], nums2: List[int]) -> int:
+ n = len(nums1)
+ ans = 0
+ for k in range(n):
+ t = sum(nums1[(i + k) % n] == x for i, x in enumerate(nums2))
+ ans = max(ans, t)
+ return ans
diff --git a/solution/3400-3499/3400.Maximum Number of Matching Indices After Right Shifts/Solution.ts b/solution/3400-3499/3400.Maximum Number of Matching Indices After Right Shifts/Solution.ts
new file mode 100644
index 0000000000000..b0a09fe147751
--- /dev/null
+++ b/solution/3400-3499/3400.Maximum Number of Matching Indices After Right Shifts/Solution.ts
@@ -0,0 +1,14 @@
+function maximumMatchingIndices(nums1: number[], nums2: number[]): number {
+ const n = nums1.length;
+ let ans: number = 0;
+ for (let k = 0; k < n; ++k) {
+ let t: number = 0;
+ for (let i = 0; i < n; ++i) {
+ if (nums1[(i + k) % n] === nums2[i]) {
+ ++t;
+ }
+ }
+ ans = Math.max(ans, t);
+ }
+ return ans;
+}
diff --git a/solution/README.md b/solution/README.md
index d73f2b1355251..753f3ac5cc3d3 100644
--- a/solution/README.md
+++ b/solution/README.md
@@ -3410,6 +3410,7 @@
| 3397 | [执行操作后不同元素的最大数量](/solution/3300-3399/3397.Maximum%20Number%20of%20Distinct%20Elements%20After%20Operations/README.md) | | 中等 | 第 429 场周赛 |
| 3398 | [字符相同的最短子字符串 I](/solution/3300-3399/3398.Smallest%20Substring%20With%20Identical%20Characters%20I/README.md) | | 困难 | 第 429 场周赛 |
| 3399 | [字符相同的最短子字符串 II](/solution/3300-3399/3399.Smallest%20Substring%20With%20Identical%20Characters%20II/README.md) | | 困难 | 第 429 场周赛 |
+| 3400 | [右移后的最大匹配索引数](/solution/3400-3499/3400.Maximum%20Number%20of%20Matching%20Indices%20After%20Right%20Shifts/README.md) | | 中等 | 🔒 |
## 版权
diff --git a/solution/README_EN.md b/solution/README_EN.md
index d2a51356a556b..5ce0855910a6a 100644
--- a/solution/README_EN.md
+++ b/solution/README_EN.md
@@ -3408,6 +3408,7 @@ Press Control + F(or Command + F on
| 3397 | [Maximum Number of Distinct Elements After Operations](/solution/3300-3399/3397.Maximum%20Number%20of%20Distinct%20Elements%20After%20Operations/README_EN.md) | | Medium | Weekly Contest 429 |
| 3398 | [Smallest Substring With Identical Characters I](/solution/3300-3399/3398.Smallest%20Substring%20With%20Identical%20Characters%20I/README_EN.md) | | Hard | Weekly Contest 429 |
| 3399 | [Smallest Substring With Identical Characters II](/solution/3300-3399/3399.Smallest%20Substring%20With%20Identical%20Characters%20II/README_EN.md) | | Hard | Weekly Contest 429 |
+| 3400 | [Maximum Number of Matching Indices After Right Shifts](/solution/3400-3499/3400.Maximum%20Number%20of%20Matching%20Indices%20After%20Right%20Shifts/README_EN.md) | | Medium | 🔒 |
## Copyright