diff --git a/solution/3200-3299/3262.Find Overlapping Shifts/README.md b/solution/3200-3299/3262.Find Overlapping Shifts/README.md
index d25d83995a4be..76a31ed07041c 100644
--- a/solution/3200-3299/3262.Find Overlapping Shifts/README.md
+++ b/solution/3200-3299/3262.Find Overlapping Shifts/README.md
@@ -8,7 +8,7 @@ tags:
-# [3262. Find Overlapping Shifts 🔒](https://leetcode.cn/problems/find-overlapping-shifts)
+# [3262. 查找重叠的班次 🔒](https://leetcode.cn/problems/find-overlapping-shifts)
[English Version](/solution/3200-3299/3262.Find%20Overlapping%20Shifts/README_EN.md)
@@ -16,7 +16,7 @@ tags:
-Table: EmployeeShifts
+表：EmployeeShifts
+------------------+---------+
@@ -26,23 +26,24 @@ tags:
| start_time | time |
| end_time | time |
+------------------+---------+
-(employee_id, start_time) is the unique key for this table.
-This table contains information about the shifts worked by employees, including the start and end times on a specific date.
+(employee_id, start_time) 是此表的唯一主键。
+这张表包含员工的排班工作，包括特定日期的开始和结束时间。
-Write a solution to count the number of overlapping shifts for each employee. Two shifts are considered overlapping if one shift’s end_time is later than another shift’s start_time.
+编写一个解决方案来为每个员工计算 重叠排班 的数量。如果一个排班的 end_time 比另一个排班的 start_time 更晚 则认为两个排班重叠。
-Return the result table ordered by employee_id in ascending order.
+返回结果表以 employee_id 升序 排序。
-The query result format is in the following example.
+查询结果格式如下所示。
-Example:
+
+示例：
-
Input:
+
输入：
-
EmployeeShifts table:
+
EmployeeShifts 表：
+-------------+------------+----------+
@@ -61,7 +62,7 @@ This table contains information about the shifts worked by employees, including
+-------------+------------+----------+
-
Output:
+
输出：
+-------------+--------------------+
@@ -73,38 +74,38 @@ This table contains information about the shifts worked by employees, including
+-------------+--------------------+
-
Explanation:
+
解释：
- - Employee 1 has 3 shifts:
+
- 员工 1 有 3 个排班：
- - 08:00:00 to 12:00:00
- - 11:00:00 to 15:00:00
- - 14:00:00 to 18:00:00
+ - 08:00:00 到 12:00:00
+ - 11:00:00 到 15:00:00
+ - 14:00:00 到 18:00:00
- The first shift overlaps with the second, and the second overlaps with the third, resulting in 2 overlapping shifts.
- - Employee 2 has 2 shifts:
+ 第一个排班与第二个排班重叠，第二个排班与第三个排班重叠，因此有 2 个重叠排班。
+ - 员工 2 有 2 个排班：
- - 09:00:00 to 17:00:00
- - 16:00:00 to 20:00:00
+ - 09:00:00 到 17:00:00
+ - 16:00:00 到 20:00:00
- These shifts overlap with each other, resulting in 1 overlapping shift.
- - Employee 3 has 3 shifts:
+ 这些排班彼此重叠，因此有 1 个重叠排班。
+ - 员工 3 有 3 个排班：
- - 10:00:00 to 12:00:00
- - 13:00:00 to 15:00:00
- - 16:00:00 to 18:00:00
+ - 10:00:00 到 12:00:00
+ - 13:00:00 到 15:00:00
+ - 16:00:00 到 18:00:00
- None of these shifts overlap, so Employee 3 is not included in the output.
- - Employee 4 has 2 shifts:
+ 这些排班没有重叠，所以员工 3 不包含在输出中。
+ - 员工 4 有 2 个排班：
- - 08:00:00 to 10:00:00
- - 09:00:00 to 11:00:00
+ - 08:00:00 到 10:00:00
+ - 09:00:00 到 11:00:00
- These shifts overlap with each other, resulting in 1 overlapping shift.
+ 这些排班彼此重叠，因此有 1 个重叠排班。
-
The output shows the employee_id and the count of overlapping shifts for each employee who has at least one overlapping shift, ordered by employee_id in ascending order.
+
输出展示了 employee_id 和至少有一个重叠排班的员工的重叠排班的数量，以 employee_id 升序排序。
You are given the head of a doubly linked list, which contains nodes that have a next pointer and a previous pointer.
+
+Return an integer array which contains the elements of the linked list in order.
+
+
+Example 1:
+
+
+
Input: head = [1,2,3,4,3,2,1]
+
+
Output: [1,2,3,4,3,2,1]
+
Example 2:
+
+
+
Input: head = [2,2,2,2,2]
+
+
Output: [2,2,2,2,2]
+
Example 3:
+
+
+
Input: head = [3,2,3,2,3,2]
+
+
Output: [3,2,3,2,3,2]
+
+Constraints:
+
+
+ - The number of nodes in the given list is in the range
[1, 50].
+ 1 <= Node.val <= 50
+
+
+
+## 解法
+
+
+
+### 方法一：直接遍历
+
+我们可以直接遍历链表，将节点的值依次添加到答案数组 $\textit{ans}$ 中。
+
+遍历结束后，返回答案数组 $\textit{ans}$ 即可。
+
+时间复杂度 $O(n)$，其中 $n$ 为链表的长度。忽略答案数组的空间消耗，空间复杂度 $O(1)$。
+
+
+
+#### Python3
+
+```python
+"""
+# Definition for a Node.
+class Node:
+ def __init__(self, val, prev=None, next=None):
+ self.val = val
+ self.prev = prev
+ self.next = next
+"""
+
+
+class Solution:
+ def toArray(self, root: "Optional[Node]") -> List[int]:
+ ans = []
+ while root:
+ ans.append(root.val)
+ root = root.next
+ return ans
+```
+
+#### Java
+
+```java
+/*
+// Definition for a Node.
+class Node {
+ public int val;
+ public Node prev;
+ public Node next;
+};
+*/
+
+class Solution {
+ public int[] toArray(Node head) {
+ List ans = new ArrayList<>();
+ for (; head != null; head = head.next) {
+ ans.add(head.val);
+ }
+ return ans.stream().mapToInt(i -> i).toArray();
+ }
+}
+```
+
+#### C++
+
+```cpp
+/**
+ * Definition for doubly-linked list.
+ * class Node {
+ * int val;
+ * Node* prev;
+ * Node* next;
+ * Node() : val(0), next(nullptr), prev(nullptr) {}
+ * Node(int x) : val(x), next(nullptr), prev(nullptr) {}
+ * Node(int x, Node *prev, Node *next) : val(x), next(next), prev(prev) {}
+ * };
+ */
+class Solution {
+public:
+ vector toArray(Node* head) {
+ vector ans;
+ for (; head; head = head->next) {
+ ans.push_back(head->val);
+ }
+ return ans;
+ }
+};
+```
+
+#### Go
+
+```go
+/**
+ * Definition for a Node.
+ * type Node struct {
+ * Val int
+ * Next *Node
+ * Prev *Node
+ * }
+ */
+
+func toArray(head *Node) (ans []int) {
+ for ; head != nil; head = head.Next {
+ ans = append(ans, head.Val)
+ }
+ return
+}
+```
+
+#### TypeScript
+
+```ts
+/**
+ * Definition for _Node.
+ * class _Node {
+ * val: number
+ * prev: _Node | null
+ * next: _Node | null
+ *
+ * constructor(val?: number, prev? : _Node, next? : _Node) {
+ * this.val = (val===undefined ? 0 : val);
+ * this.prev = (prev===undefined ? null : prev);
+ * this.next = (next===undefined ? null : next);
+ * }
+ * }
+ */
+
+function toArray(head: _Node | null): number[] {
+ const ans: number[] = [];
+ for (; head; head = head.next) {
+ ans.push(head.val);
+ }
+ return ans;
+}
+```
+
+
+
+
+
+
diff --git a/solution/3200-3299/3263.Convert Doubly Linked List to Array I/README_EN.md b/solution/3200-3299/3263.Convert Doubly Linked List to Array I/README_EN.md
new file mode 100644
index 0000000000000..88319b5e7b123
--- /dev/null
+++ b/solution/3200-3299/3263.Convert Doubly Linked List to Array I/README_EN.md
@@ -0,0 +1,192 @@
+---
+comments: true
+difficulty: Easy
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3200-3299/3263.Convert%20Doubly%20Linked%20List%20to%20Array%20I/README_EN.md
+---
+
+
+
+# [3263. Convert Doubly Linked List to Array I 🔒](https://leetcode.com/problems/convert-doubly-linked-list-to-array-i)
+
+[中文文档](/solution/3200-3299/3263.Convert%20Doubly%20Linked%20List%20to%20Array%20I/README.md)
+
+## Description
+
+
+
+You are given the head of a doubly linked list, which contains nodes that have a next pointer and a previous pointer.
+
+Return an integer array which contains the elements of the linked list in order.
+
+
+Example 1:
+
+
+
Input: head = [1,2,3,4,3,2,1]
+
+
Output: [1,2,3,4,3,2,1]
+
Example 2:
+
+
+
Input: head = [2,2,2,2,2]
+
+
Output: [2,2,2,2,2]
+
Example 3:
+
+
+
Input: head = [3,2,3,2,3,2]
+
+
Output: [3,2,3,2,3,2]
+
+Constraints:
+
+
+ - The number of nodes in the given list is in the range
[1, 50].
+ 1 <= Node.val <= 50
+
+
+
+## Solutions
+
+
+
+### Solution 1: Direct Traversal
+
+We can directly traverse the linked list, adding the values of the nodes to the answer array $\textit{ans}$ one by one.
+
+After the traversal is complete, return the answer array $\textit{ans}$.
+
+The time complexity is $O(n)$, where $n$ is the length of the linked list. Ignoring the space consumption of the answer array, the space complexity is $O(1)$.
+
+
+
+#### Python3
+
+```python
+"""
+# Definition for a Node.
+class Node:
+ def __init__(self, val, prev=None, next=None):
+ self.val = val
+ self.prev = prev
+ self.next = next
+"""
+
+
+class Solution:
+ def toArray(self, root: "Optional[Node]") -> List[int]:
+ ans = []
+ while root:
+ ans.append(root.val)
+ root = root.next
+ return ans
+```
+
+#### Java
+
+```java
+/*
+// Definition for a Node.
+class Node {
+ public int val;
+ public Node prev;
+ public Node next;
+};
+*/
+
+class Solution {
+ public int[] toArray(Node head) {
+ List ans = new ArrayList<>();
+ for (; head != null; head = head.next) {
+ ans.add(head.val);
+ }
+ return ans.stream().mapToInt(i -> i).toArray();
+ }
+}
+```
+
+#### C++
+
+```cpp
+/**
+ * Definition for doubly-linked list.
+ * class Node {
+ * int val;
+ * Node* prev;
+ * Node* next;
+ * Node() : val(0), next(nullptr), prev(nullptr) {}
+ * Node(int x) : val(x), next(nullptr), prev(nullptr) {}
+ * Node(int x, Node *prev, Node *next) : val(x), next(next), prev(prev) {}
+ * };
+ */
+class Solution {
+public:
+ vector toArray(Node* head) {
+ vector ans;
+ for (; head; head = head->next) {
+ ans.push_back(head->val);
+ }
+ return ans;
+ }
+};
+```
+
+#### Go
+
+```go
+/**
+ * Definition for a Node.
+ * type Node struct {
+ * Val int
+ * Next *Node
+ * Prev *Node
+ * }
+ */
+
+func toArray(head *Node) (ans []int) {
+ for ; head != nil; head = head.Next {
+ ans = append(ans, head.Val)
+ }
+ return
+}
+```
+
+#### TypeScript
+
+```ts
+/**
+ * Definition for _Node.
+ * class _Node {
+ * val: number
+ * prev: _Node | null
+ * next: _Node | null
+ *
+ * constructor(val?: number, prev? : _Node, next? : _Node) {
+ * this.val = (val===undefined ? 0 : val);
+ * this.prev = (prev===undefined ? null : prev);
+ * this.next = (next===undefined ? null : next);
+ * }
+ * }
+ */
+
+function toArray(head: _Node | null): number[] {
+ const ans: number[] = [];
+ for (; head; head = head.next) {
+ ans.push(head.val);
+ }
+ return ans;
+}
+```
+
+
+
+
+
+
diff --git a/solution/3200-3299/3263.Convert Doubly Linked List to Array I/Solution.cpp b/solution/3200-3299/3263.Convert Doubly Linked List to Array I/Solution.cpp
new file mode 100644
index 0000000000000..c02659a356adf
--- /dev/null
+++ b/solution/3200-3299/3263.Convert Doubly Linked List to Array I/Solution.cpp
@@ -0,0 +1,21 @@
+/**
+ * Definition for doubly-linked list.
+ * class Node {
+ * int val;
+ * Node* prev;
+ * Node* next;
+ * Node() : val(0), next(nullptr), prev(nullptr) {}
+ * Node(int x) : val(x), next(nullptr), prev(nullptr) {}
+ * Node(int x, Node *prev, Node *next) : val(x), next(next), prev(prev) {}
+ * };
+ */
+class Solution {
+public:
+ vector toArray(Node* head) {
+ vector ans;
+ for (; head; head = head->next) {
+ ans.push_back(head->val);
+ }
+ return ans;
+ }
+};
diff --git a/solution/3200-3299/3263.Convert Doubly Linked List to Array I/Solution.go b/solution/3200-3299/3263.Convert Doubly Linked List to Array I/Solution.go
new file mode 100644
index 0000000000000..f7bc140b1657e
--- /dev/null
+++ b/solution/3200-3299/3263.Convert Doubly Linked List to Array I/Solution.go
@@ -0,0 +1,15 @@
+/**
+ * Definition for a Node.
+ * type Node struct {
+ * Val int
+ * Next *Node
+ * Prev *Node
+ * }
+ */
+
+func toArray(head *Node) (ans []int) {
+ for ; head != nil; head = head.Next {
+ ans = append(ans, head.Val)
+ }
+ return
+}
diff --git a/solution/3200-3299/3263.Convert Doubly Linked List to Array I/Solution.java b/solution/3200-3299/3263.Convert Doubly Linked List to Array I/Solution.java
new file mode 100644
index 0000000000000..fef1a77a1e0bf
--- /dev/null
+++ b/solution/3200-3299/3263.Convert Doubly Linked List to Array I/Solution.java
@@ -0,0 +1,18 @@
+/*
+// Definition for a Node.
+class Node {
+ public int val;
+ public Node prev;
+ public Node next;
+};
+*/
+
+class Solution {
+ public int[] toArray(Node head) {
+ List ans = new ArrayList<>();
+ for (; head != null; head = head.next) {
+ ans.add(head.val);
+ }
+ return ans.stream().mapToInt(i -> i).toArray();
+ }
+}
diff --git a/solution/3200-3299/3263.Convert Doubly Linked List to Array I/Solution.py b/solution/3200-3299/3263.Convert Doubly Linked List to Array I/Solution.py
new file mode 100644
index 0000000000000..f5498238f0900
--- /dev/null
+++ b/solution/3200-3299/3263.Convert Doubly Linked List to Array I/Solution.py
@@ -0,0 +1,17 @@
+"""
+# Definition for a Node.
+class Node:
+ def __init__(self, val, prev=None, next=None):
+ self.val = val
+ self.prev = prev
+ self.next = next
+"""
+
+
+class Solution:
+ def toArray(self, root: "Optional[Node]") -> List[int]:
+ ans = []
+ while root:
+ ans.append(root.val)
+ root = root.next
+ return ans
diff --git a/solution/3200-3299/3263.Convert Doubly Linked List to Array I/Solution.ts b/solution/3200-3299/3263.Convert Doubly Linked List to Array I/Solution.ts
new file mode 100644
index 0000000000000..af9e2027bfeb1
--- /dev/null
+++ b/solution/3200-3299/3263.Convert Doubly Linked List to Array I/Solution.ts
@@ -0,0 +1,22 @@
+/**
+ * Definition for _Node.
+ * class _Node {
+ * val: number
+ * prev: _Node | null
+ * next: _Node | null
+ *
+ * constructor(val?: number, prev? : _Node, next? : _Node) {
+ * this.val = (val===undefined ? 0 : val);
+ * this.prev = (prev===undefined ? null : prev);
+ * this.next = (next===undefined ? null : next);
+ * }
+ * }
+ */
+
+function toArray(head: _Node | null): number[] {
+ const ans: number[] = [];
+ for (; head; head = head.next) {
+ ans.push(head.val);
+ }
+ return ans;
+}
diff --git a/solution/CONTEST_README.md b/solution/CONTEST_README.md
index 7d35115f978c7..ca8a7957c7b8c 100644
--- a/solution/CONTEST_README.md
+++ b/solution/CONTEST_README.md
@@ -2730,6 +2730,7 @@ comments: true
#### 第 154 场周赛(2019-09-15 10:30, 90 分钟) 参赛人数 1299
+- [1189. “气球” 的最大数量](/solution/1100-1199/1189.Maximum%20Number%20of%20Balloons/README.md)
- [1190. 反转每对括号间的子串](/solution/1100-1199/1190.Reverse%20Substrings%20Between%20Each%20Pair%20of%20Parentheses/README.md)
- [1191. K 次串联后最大子数组之和](/solution/1100-1199/1191.K-Concatenation%20Maximum%20Sum/README.md)
- [1192. 查找集群内的关键连接](/solution/1100-1199/1192.Critical%20Connections%20in%20a%20Network/README.md)
diff --git a/solution/CONTEST_README_EN.md b/solution/CONTEST_README_EN.md
index 13ae499d2057a..4d933a032a96a 100644
--- a/solution/CONTEST_README_EN.md
+++ b/solution/CONTEST_README_EN.md
@@ -2733,6 +2733,7 @@ If you want to estimate your score changes after the contest ends, you can visit
#### Weekly Contest 154
+- [1189. Maximum Number of Balloons](/solution/1100-1199/1189.Maximum%20Number%20of%20Balloons/README_EN.md)
- [1190. Reverse Substrings Between Each Pair of Parentheses](/solution/1100-1199/1190.Reverse%20Substrings%20Between%20Each%20Pair%20of%20Parentheses/README_EN.md)
- [1191. K-Concatenation Maximum Sum](/solution/1100-1199/1191.K-Concatenation%20Maximum%20Sum/README_EN.md)
- [1192. Critical Connections in a Network](/solution/1100-1199/1192.Critical%20Connections%20in%20a%20Network/README_EN.md)
diff --git a/solution/DATABASE_README.md b/solution/DATABASE_README.md
index f2bdefbd30147..24636f820ef0a 100644
--- a/solution/DATABASE_README.md
+++ b/solution/DATABASE_README.md
@@ -292,7 +292,7 @@
| 3236 | [首席执行官下属层级](/solution/3200-3299/3236.CEO%20Subordinate%20Hierarchy/README.md) | `数据库` | 困难 | 🔒 |
| 3246 | [英超积分榜排名](/solution/3200-3299/3246.Premier%20League%20Table%20Ranking/README.md) | `数据库` | 简单 | 🔒 |
| 3252 | [英超积分榜排名 II](/solution/3200-3299/3252.Premier%20League%20Table%20Ranking%20II/README.md) | `数据库` | 中等 | 🔒 |
-| 3262 | [Find Overlapping Shifts](/solution/3200-3299/3262.Find%20Overlapping%20Shifts/README.md) | | 中等 | 🔒 |
+| 3262 | [查找重叠的班次](/solution/3200-3299/3262.Find%20Overlapping%20Shifts/README.md) | | 中等 | 🔒 |
## 版权
diff --git a/solution/README.md b/solution/README.md
index d98f0ecb11f7b..3739034b935a7 100644
--- a/solution/README.md
+++ b/solution/README.md
@@ -1199,6 +1199,7 @@
| 1186 | [删除一次得到子数组最大和](/solution/1100-1199/1186.Maximum%20Subarray%20Sum%20with%20One%20Deletion/README.md) | `数组`,`动态规划` | 中等 | 第 153 场周赛 |
| 1187 | [使数组严格递增](/solution/1100-1199/1187.Make%20Array%20Strictly%20Increasing/README.md) | `数组`,`二分查找`,`动态规划`,`排序` | 困难 | 第 153 场周赛 |
| 1188 | [设计有限阻塞队列](/solution/1100-1199/1188.Design%20Bounded%20Blocking%20Queue/README.md) | `多线程` | 中等 | 🔒 |
+| 1189 | [“气球” 的最大数量](/solution/1100-1199/1189.Maximum%20Number%20of%20Balloons/README.md) | `哈希表`,`字符串`,`计数` | 简单 | 第 154 场周赛 |
| 1190 | [反转每对括号间的子串](/solution/1100-1199/1190.Reverse%20Substrings%20Between%20Each%20Pair%20of%20Parentheses/README.md) | `栈`,`字符串` | 中等 | 第 154 场周赛 |
| 1191 | [K 次串联后最大子数组之和](/solution/1100-1199/1191.K-Concatenation%20Maximum%20Sum/README.md) | `数组`,`动态规划` | 中等 | 第 154 场周赛 |
| 1192 | [查找集群内的关键连接](/solution/1100-1199/1192.Critical%20Connections%20in%20a%20Network/README.md) | `深度优先搜索`,`图`,`双连通分量` | 困难 | 第 154 场周赛 |
@@ -3271,7 +3272,8 @@
| 3259 | [超级饮料的最大强化能量](/solution/3200-3299/3259.Maximum%20Energy%20Boost%20From%20Two%20Drinks/README.md) | `数组`,`动态规划` | 中等 | 第 411 场周赛 |
| 3260 | [找出最大的 N 位 K 回文数](/solution/3200-3299/3260.Find%20the%20Largest%20Palindrome%20Divisible%20by%20K/README.md) | `贪心`,`数学`,`字符串`,`动态规划`,`数论` | 困难 | 第 411 场周赛 |
| 3261 | [统计满足 K 约束的子字符串数量 II](/solution/3200-3299/3261.Count%20Substrings%20That%20Satisfy%20K-Constraint%20II/README.md) | `数组`,`字符串`,`二分查找`,`前缀和`,`滑动窗口` | 困难 | 第 411 场周赛 |
-| 3262 | [Find Overlapping Shifts](/solution/3200-3299/3262.Find%20Overlapping%20Shifts/README.md) | | 中等 | 🔒 |
+| 3262 | [查找重叠的班次](/solution/3200-3299/3262.Find%20Overlapping%20Shifts/README.md) | | 中等 | 🔒 |
+| 3263 | [Convert Doubly Linked List to Array I](/solution/3200-3299/3263.Convert%20Doubly%20Linked%20List%20to%20Array%20I/README.md) | | 简单 | 🔒 |
## 版权
diff --git a/solution/README_EN.md b/solution/README_EN.md
index 4ff2f3aefd7c6..44acf5de94cd0 100644
--- a/solution/README_EN.md
+++ b/solution/README_EN.md
@@ -1197,6 +1197,7 @@ Press Control + F(or Command + F on
| 1186 | [Maximum Subarray Sum with One Deletion](/solution/1100-1199/1186.Maximum%20Subarray%20Sum%20with%20One%20Deletion/README_EN.md) | `Array`,`Dynamic Programming` | Medium | Weekly Contest 153 |
| 1187 | [Make Array Strictly Increasing](/solution/1100-1199/1187.Make%20Array%20Strictly%20Increasing/README_EN.md) | `Array`,`Binary Search`,`Dynamic Programming`,`Sorting` | Hard | Weekly Contest 153 |
| 1188 | [Design Bounded Blocking Queue](/solution/1100-1199/1188.Design%20Bounded%20Blocking%20Queue/README_EN.md) | `Concurrency` | Medium | 🔒 |
+| 1189 | [Maximum Number of Balloons](/solution/1100-1199/1189.Maximum%20Number%20of%20Balloons/README_EN.md) | `Hash Table`,`String`,`Counting` | Easy | Weekly Contest 154 |
| 1190 | [Reverse Substrings Between Each Pair of Parentheses](/solution/1100-1199/1190.Reverse%20Substrings%20Between%20Each%20Pair%20of%20Parentheses/README_EN.md) | `Stack`,`String` | Medium | Weekly Contest 154 |
| 1191 | [K-Concatenation Maximum Sum](/solution/1100-1199/1191.K-Concatenation%20Maximum%20Sum/README_EN.md) | `Array`,`Dynamic Programming` | Medium | Weekly Contest 154 |
| 1192 | [Critical Connections in a Network](/solution/1100-1199/1192.Critical%20Connections%20in%20a%20Network/README_EN.md) | `Depth-First Search`,`Graph`,`Biconnected Component` | Hard | Weekly Contest 154 |
@@ -3270,6 +3271,7 @@ Press Control + F(or Command + F on
| 3260 | [Find the Largest Palindrome Divisible by K](/solution/3200-3299/3260.Find%20the%20Largest%20Palindrome%20Divisible%20by%20K/README_EN.md) | `Greedy`,`Math`,`String`,`Dynamic Programming`,`Number Theory` | Hard | Weekly Contest 411 |
| 3261 | [Count Substrings That Satisfy K-Constraint II](/solution/3200-3299/3261.Count%20Substrings%20That%20Satisfy%20K-Constraint%20II/README_EN.md) | `Array`,`String`,`Binary Search`,`Prefix Sum`,`Sliding Window` | Hard | Weekly Contest 411 |
| 3262 | [Find Overlapping Shifts](/solution/3200-3299/3262.Find%20Overlapping%20Shifts/README_EN.md) | | Medium | 🔒 |
+| 3263 | [Convert Doubly Linked List to Array I](/solution/3200-3299/3263.Convert%20Doubly%20Linked%20List%20to%20Array%20I/README_EN.md) | | Easy | 🔒 |
## Copyright