diff --git a/solution/2900-2999/2959.Number of Possible Sets of Closing Branches/README.md b/solution/2900-2999/2959.Number of Possible Sets of Closing Branches/README.md
index cfda725e63bd7..2c94f18124a6a 100644
--- a/solution/2900-2999/2959.Number of Possible Sets of Closing Branches/README.md
+++ b/solution/2900-2999/2959.Number of Possible Sets of Closing Branches/README.md
@@ -123,7 +123,7 @@ class Solution:
for mask in range(1 << n):
g = [[inf] * n for _ in range(n)]
for u, v, w in roads:
- if mask >> u & 1 and mask > v & 1:
+ if mask >> u & 1 and mask >> v & 1:
g[u][v] = min(g[u][v], w)
g[v][u] = min(g[v][u], w)
for k in range(n):
diff --git a/solution/2900-2999/2959.Number of Possible Sets of Closing Branches/README_EN.md b/solution/2900-2999/2959.Number of Possible Sets of Closing Branches/README_EN.md
index 25f003b42b0a1..45f6717a7d288 100644
--- a/solution/2900-2999/2959.Number of Possible Sets of Closing Branches/README_EN.md
+++ b/solution/2900-2999/2959.Number of Possible Sets of Closing Branches/README_EN.md
@@ -117,7 +117,7 @@ class Solution:
for mask in range(1 << n):
g = [[inf] * n for _ in range(n)]
for u, v, w in roads:
- if mask >> u & 1 and mask > v & 1:
+ if mask >> u & 1 and mask >> v & 1:
g[u][v] = min(g[u][v], w)
g[v][u] = min(g[v][u], w)
for k in range(n):
diff --git a/solution/2900-2999/2959.Number of Possible Sets of Closing Branches/Solution.py b/solution/2900-2999/2959.Number of Possible Sets of Closing Branches/Solution.py
index 3ab1fdac62623..80bb9851a5dd5 100644
--- a/solution/2900-2999/2959.Number of Possible Sets of Closing Branches/Solution.py
+++ b/solution/2900-2999/2959.Number of Possible Sets of Closing Branches/Solution.py
@@ -4,7 +4,7 @@ def numberOfSets(self, n: int, maxDistance: int, roads: List[List[int]]) -> int:
for mask in range(1 << n):
g = [[inf] * n for _ in range(n)]
for u, v, w in roads:
- if mask >> u & 1 and mask > v & 1:
+ if mask >> u & 1 and mask >> v & 1:
g[u][v] = min(g[u][v], w)
g[v][u] = min(g[v][u], w)
for k in range(n):
diff --git a/solution/3100-3199/3112.Minimum Time to Visit Disappearing Nodes/README.md b/solution/3100-3199/3112.Minimum Time to Visit Disappearing Nodes/README.md
index 7ef28fbbb5218..d4b85d7db924c 100644
--- a/solution/3100-3199/3112.Minimum Time to Visit Disappearing Nodes/README.md
+++ b/solution/3100-3199/3112.Minimum Time to Visit Disappearing Nodes/README.md
@@ -133,15 +133,15 @@ class Solution:
g[v].append((u, w))
dist = [inf] * n
dist[0] = 0
- q = [(0, 0)]
- while q:
- du, u = heappop(q)
+ pq = [(0, 0)]
+ while pq:
+ du, u = heappop(pq)
if du > dist[u]:
continue
for v, w in g[u]:
if dist[v] > dist[u] + w and dist[u] + w < disappear[v]:
dist[v] = dist[u] + w
- heappush(q, (dist[v], v))
+ heappush(pq, (dist[v], v))
return [a if a < b else -1 for a, b in zip(dist, disappear)]
```
diff --git a/solution/3100-3199/3112.Minimum Time to Visit Disappearing Nodes/README_EN.md b/solution/3100-3199/3112.Minimum Time to Visit Disappearing Nodes/README_EN.md
index 08d820e5ab273..5d150f2fbba32 100644
--- a/solution/3100-3199/3112.Minimum Time to Visit Disappearing Nodes/README_EN.md
+++ b/solution/3100-3199/3112.Minimum Time to Visit Disappearing Nodes/README_EN.md
@@ -131,15 +131,15 @@ class Solution:
g[v].append((u, w))
dist = [inf] * n
dist[0] = 0
- q = [(0, 0)]
- while q:
- du, u = heappop(q)
+ pq = [(0, 0)]
+ while pq:
+ du, u = heappop(pq)
if du > dist[u]:
continue
for v, w in g[u]:
if dist[v] > dist[u] + w and dist[u] + w < disappear[v]:
dist[v] = dist[u] + w
- heappush(q, (dist[v], v))
+ heappush(pq, (dist[v], v))
return [a if a < b else -1 for a, b in zip(dist, disappear)]
```
diff --git a/solution/3100-3199/3112.Minimum Time to Visit Disappearing Nodes/Solution.py b/solution/3100-3199/3112.Minimum Time to Visit Disappearing Nodes/Solution.py
index 8695ae255103f..621600ee6e4c8 100644
--- a/solution/3100-3199/3112.Minimum Time to Visit Disappearing Nodes/Solution.py
+++ b/solution/3100-3199/3112.Minimum Time to Visit Disappearing Nodes/Solution.py
@@ -8,13 +8,13 @@ def minimumTime(
g[v].append((u, w))
dist = [inf] * n
dist[0] = 0
- q = [(0, 0)]
- while q:
- du, u = heappop(q)
+ pq = [(0, 0)]
+ while pq:
+ du, u = heappop(pq)
if du > dist[u]:
continue
for v, w in g[u]:
if dist[v] > dist[u] + w and dist[u] + w < disappear[v]:
dist[v] = dist[u] + w
- heappush(q, (dist[v], v))
+ heappush(pq, (dist[v], v))
return [a if a < b else -1 for a, b in zip(dist, disappear)]
diff --git a/solution/3200-3299/3220.Odd and Even Transactions/README.md b/solution/3200-3299/3220.Odd and Even Transactions/README.md
new file mode 100644
index 0000000000000..365fdba217daa
--- /dev/null
+++ b/solution/3200-3299/3220.Odd and Even Transactions/README.md
@@ -0,0 +1,154 @@
+---
+comments: true
+difficulty: 中等
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3200-3299/3220.Odd%20and%20Even%20Transactions/README.md
+---
+
+
+
+# [3220. Odd and Even Transactions](https://leetcode.cn/problems/odd-and-even-transactions)
+
+[English Version](/solution/3200-3299/3220.Odd%20and%20Even%20Transactions/README_EN.md)
+
+## 题目描述
+
+
+
+Table: transactions
+
+
++------------------+------+
+| Column Name | Type |
++------------------+------+
+| transaction_id | int |
+| amount | int |
+| transaction_date | date |
++------------------+------+
+The transactions_id column uniquely identifies each row in this table.
+Each row of this table contains the transaction id, amount and transaction date.
+
+
+Write a solution to find the sum of amounts for odd and even transactions for each day. If there are no odd or even transactions for a specific date, display as 0.
+
+Return the result table ordered by transaction_date in ascending order.
+
+The result format is in the following example.
+
+
+Example:
+
+
+
Input:
+
+
transactions table:
+
+
++----------------+--------+------------------+
+| transaction_id | amount | transaction_date |
++----------------+--------+------------------+
+| 1 | 150 | 2024-07-01 |
+| 2 | 200 | 2024-07-01 |
+| 3 | 75 | 2024-07-01 |
+| 4 | 300 | 2024-07-02 |
+| 5 | 50 | 2024-07-02 |
+| 6 | 120 | 2024-07-03 |
++----------------+--------+------------------+
+
+
+
Output:
+
+
++------------------+---------+----------+
+| transaction_date | odd_sum | even_sum |
++------------------+---------+----------+
+| 2024-07-01 | 75 | 350 |
+| 2024-07-02 | 0 | 350 |
+| 2024-07-03 | 0 | 120 |
++------------------+---------+----------+
+
+
+
Explanation:
+
+
+ - For transaction dates:
+
+ - 2024-07-01:
+
+ - Sum of amounts for odd transactions: 75
+ - Sum of amounts for even transactions: 150 + 200 = 350
+
+
+ - 2024-07-02:
+
+ - Sum of amounts for odd transactions: 0
+ - Sum of amounts for even transactions: 300 + 50 = 350
+
+
+ - 2024-07-03:
+
+ - Sum of amounts for odd transactions: 0
+ - Sum of amounts for even transactions: 120
+
+
+
+
+
+
+
Note: The output table is ordered by transaction_date in ascending order.
+
Table: transactions
+
+
++------------------+------+
+| Column Name | Type |
++------------------+------+
+| transaction_id | int |
+| amount | int |
+| transaction_date | date |
++------------------+------+
+The transactions_id column uniquely identifies each row in this table.
+Each row of this table contains the transaction id, amount and transaction date.
+
+
+Write a solution to find the sum of amounts for odd and even transactions for each day. If there are no odd or even transactions for a specific date, display as 0.
+
+Return the result table ordered by transaction_date in ascending order.
+
+The result format is in the following example.
+
+
+Example:
+
+
+
Input:
+
+
transactions table:
+
+
++----------------+--------+------------------+
+| transaction_id | amount | transaction_date |
++----------------+--------+------------------+
+| 1 | 150 | 2024-07-01 |
+| 2 | 200 | 2024-07-01 |
+| 3 | 75 | 2024-07-01 |
+| 4 | 300 | 2024-07-02 |
+| 5 | 50 | 2024-07-02 |
+| 6 | 120 | 2024-07-03 |
++----------------+--------+------------------+
+
+
+
Output:
+
+
++------------------+---------+----------+
+| transaction_date | odd_sum | even_sum |
++------------------+---------+----------+
+| 2024-07-01 | 75 | 350 |
+| 2024-07-02 | 0 | 350 |
+| 2024-07-03 | 0 | 120 |
++------------------+---------+----------+
+
+
+
Explanation:
+
+
+ - For transaction dates:
+
+ - 2024-07-01:
+
+ - Sum of amounts for odd transactions: 75
+ - Sum of amounts for even transactions: 150 + 200 = 350
+
+
+ - 2024-07-02:
+
+ - Sum of amounts for odd transactions: 0
+ - Sum of amounts for even transactions: 300 + 50 = 350
+
+
+ - 2024-07-03:
+
+ - Sum of amounts for odd transactions: 0
+ - Sum of amounts for even transactions: 120
+
+
+
+
+
+
+
Note: The output table is ordered by transaction_date in ascending order.
+
Given an array nums, you have to get the maximum score starting from index 0 and hopping until you reach the last element of the array.
+
+In each hop, you can jump from index i to an index j > i, and you get a score of (j - i) * nums[j].
+
+Return the maximum score you can get.
+
+
+Example 1:
+
+
+
Input: nums = [1,5,8]
+
+
Output: 16
+
+
Explanation:
+
+
There are two possible ways to reach the last element:
+
+
+ 0 -> 1 -> 2 with a score of (1 - 0) * 5 + (2 - 1) * 8 = 13.0 -> 2 with a score of (2 - 0) * 8 = 16.
+
Example 2:
+
+
+
Input: nums = [4,5,2,8,9,1,3]
+
+
Output: 42
+
+
Explanation:
+
+
We can do the hopping 0 -> 4 -> 6 with a score of (4 - 0) * 9 + (6 - 4) * 3 = 42.
+
+Constraints:
+
+
+ 2 <= nums.length <= 1051 <= nums[i] <= 105
+
+
+
+## 解法
+
+
+
+### 方法一：单调栈
+
+我们观察发现，对于当前位置 $i$，我们应该跳到下一个值最大的位置 $j$，这样才能获得最大的分数。
+
+因此，我们遍历数组 $\text{nums}$，维护一个从栈底到栈顶单调递减的栈 $\text{stk}$。对于当前遍历到的位置 $i$，如果栈顶元素对应的值小于等于 $\text{nums}[i]$，我们就不断地弹出栈顶元素，直到栈为空或者栈顶元素对应的值大于 $\text{nums}[i]$，然后将 $i$ 入栈。
+
+然后，我们初始化答案 $\text{ans}$ 和当前位置 $i = 0$，遍历栈中的元素，每次取出栈顶元素 $j$，更新答案 $\text{ans} += \text{nums}[j] \times (j - i)$，然后更新 $i = j$。
+
+最后返回答案 $\text{ans}$。
+
+时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是数组的长度。
+
+
+
+#### Python3
+
+```python
+class Solution:
+ def maxScore(self, nums: List[int]) -> int:
+ stk = []
+ for i, x in enumerate(nums):
+ while stk and nums[stk[-1]] <= x:
+ stk.pop()
+ stk.append(i)
+ ans = i = 0
+ for j in stk:
+ ans += nums[j] * (j - i)
+ i = j
+ return ans
+```
+
+#### Java
+
+```java
+class Solution {
+ public long maxScore(int[] nums) {
+ Deque stk = new ArrayDeque<>();
+ for (int i = 0; i < nums.length; ++i) {
+ while (!stk.isEmpty() && nums[stk.peek()] <= nums[i]) {
+ stk.pop();
+ }
+ stk.push(i);
+ }
+ long ans = 0, i = 0;
+ while (!stk.isEmpty()) {
+ int j = stk.pollLast();
+ ans += (j - i) * nums[j];
+ i = j;
+ }
+ return ans;
+ }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+ long long maxScore(vector& nums) {
+ vector stk;
+ for (int i = 0; i < nums.size(); ++i) {
+ while (stk.size() && nums[stk.back()] <= nums[i]) {
+ stk.pop_back();
+ }
+ stk.push_back(i);
+ }
+ long long ans = 0, i = 0;
+ for (int j : stk) {
+ ans += (j - i) * nums[j];
+ i = j;
+ }
+ return ans;
+ }
+};
+```
+
+#### Go
+
+```go
+func maxScore(nums []int) (ans int64) {
+ stk := []int{}
+ for i, x := range nums {
+ for len(stk) > 0 && nums[stk[len(stk)-1]] <= x {
+ stk = stk[:len(stk)-1]
+ }
+ stk = append(stk, i)
+ }
+ i := 0
+ for _, j := range stk {
+ ans += int64((j - i) * nums[j])
+ i = j
+ }
+ return
+}
+```
+
+#### TypeScript
+
+```ts
+function maxScore(nums: number[]): number {
+ const stk: number[] = [];
+ for (let i = 0; i < nums.length; ++i) {
+ while (stk.length && nums[stk.at(-1)!] <= nums[i]) {
+ stk.pop();
+ }
+ stk.push(i);
+ }
+ let ans = 0;
+ let i = 0;
+ for (const j of stk) {
+ ans += (j - i) * nums[j];
+ i = j;
+ }
+ return ans;
+}
+```
+
+
+
+
+
+
diff --git a/solution/3200-3299/3221.Maximum Array Hopping Score II/README_EN.md b/solution/3200-3299/3221.Maximum Array Hopping Score II/README_EN.md
new file mode 100644
index 0000000000000..d1b3ab984a648
--- /dev/null
+++ b/solution/3200-3299/3221.Maximum Array Hopping Score II/README_EN.md
@@ -0,0 +1,185 @@
+---
+comments: true
+difficulty: Medium
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3200-3299/3221.Maximum%20Array%20Hopping%20Score%20II/README_EN.md
+---
+
+
+
+# [3221. Maximum Array Hopping Score II 🔒](https://leetcode.com/problems/maximum-array-hopping-score-ii)
+
+[中文文档](/solution/3200-3299/3221.Maximum%20Array%20Hopping%20Score%20II/README.md)
+
+## Description
+
+
+
+Given an array nums, you have to get the maximum score starting from index 0 and hopping until you reach the last element of the array.
+
+In each hop, you can jump from index i to an index j > i, and you get a score of (j - i) * nums[j].
+
+Return the maximum score you can get.
+
+
+Example 1:
+
+
+
Input: nums = [1,5,8]
+
+
Output: 16
+
+
Explanation:
+
+
There are two possible ways to reach the last element:
+
+
+ 0 -> 1 -> 2 with a score of (1 - 0) * 5 + (2 - 1) * 8 = 13.0 -> 2 with a score of (2 - 0) * 8 = 16.
+
Example 2:
+
+
+
Input: nums = [4,5,2,8,9,1,3]
+
+
Output: 42
+
+
Explanation:
+
+
We can do the hopping 0 -> 4 -> 6 with a score of (4 - 0) * 9 + (6 - 4) * 3 = 42.
+
+Constraints:
+
+
+ 2 <= nums.length <= 1051 <= nums[i] <= 105
+
+
+
+## Solutions
+
+
+
+### Solution 1: Monotonic Stack
+
+We observe that for the current position $i$, we should jump to the next position $j$ with the maximum value to obtain the maximum score.
+
+Therefore, we traverse the array $\text{nums}$, maintaining a stack $\text{stk}$ that is monotonically decreasing from the bottom to the top of the stack. For the current position $i$ being traversed, if the value corresponding to the top element of the stack is less than or equal to $\text{nums}[i]$, we continuously pop the top element of the stack until the stack is empty or the value corresponding to the top element of the stack is greater than $\text{nums}[i]$, and then push $i$ into the stack.
+
+Next, we initialize the answer $\text{ans}$ and the current position $i = 0$, traverse the elements in the stack, each time taking out the top element $j$, updating the answer $\text{ans} += \text{nums}[j] \times (j - i)$, and then updating $i = j$.
+
+
+
+#### Python3
+
+```python
+class Solution:
+ def maxScore(self, nums: List[int]) -> int:
+ stk = []
+ for i, x in enumerate(nums):
+ while stk and nums[stk[-1]] <= x:
+ stk.pop()
+ stk.append(i)
+ ans = i = 0
+ for j in stk:
+ ans += nums[j] * (j - i)
+ i = j
+ return ans
+```
+
+#### Java
+
+```java
+class Solution {
+ public long maxScore(int[] nums) {
+ Deque stk = new ArrayDeque<>();
+ for (int i = 0; i < nums.length; ++i) {
+ while (!stk.isEmpty() && nums[stk.peek()] <= nums[i]) {
+ stk.pop();
+ }
+ stk.push(i);
+ }
+ long ans = 0, i = 0;
+ while (!stk.isEmpty()) {
+ int j = stk.pollLast();
+ ans += (j - i) * nums[j];
+ i = j;
+ }
+ return ans;
+ }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+ long long maxScore(vector& nums) {
+ vector stk;
+ for (int i = 0; i < nums.size(); ++i) {
+ while (stk.size() && nums[stk.back()] <= nums[i]) {
+ stk.pop_back();
+ }
+ stk.push_back(i);
+ }
+ long long ans = 0, i = 0;
+ for (int j : stk) {
+ ans += (j - i) * nums[j];
+ i = j;
+ }
+ return ans;
+ }
+};
+```
+
+#### Go
+
+```go
+func maxScore(nums []int) (ans int64) {
+ stk := []int{}
+ for i, x := range nums {
+ for len(stk) > 0 && nums[stk[len(stk)-1]] <= x {
+ stk = stk[:len(stk)-1]
+ }
+ stk = append(stk, i)
+ }
+ i := 0
+ for _, j := range stk {
+ ans += int64((j - i) * nums[j])
+ i = j
+ }
+ return
+}
+```
+
+#### TypeScript
+
+```ts
+function maxScore(nums: number[]): number {
+ const stk: number[] = [];
+ for (let i = 0; i < nums.length; ++i) {
+ while (stk.length && nums[stk.at(-1)!] <= nums[i]) {
+ stk.pop();
+ }
+ stk.push(i);
+ }
+ let ans = 0;
+ let i = 0;
+ for (const j of stk) {
+ ans += (j - i) * nums[j];
+ i = j;
+ }
+ return ans;
+}
+```
+
+
+
+
+
+
diff --git a/solution/3200-3299/3221.Maximum Array Hopping Score II/Solution.cpp b/solution/3200-3299/3221.Maximum Array Hopping Score II/Solution.cpp
new file mode 100644
index 0000000000000..7365af25c1e7f
--- /dev/null
+++ b/solution/3200-3299/3221.Maximum Array Hopping Score II/Solution.cpp
@@ -0,0 +1,18 @@
+class Solution {
+public:
+ long long maxScore(vector& nums) {
+ vector stk;
+ for (int i = 0; i < nums.size(); ++i) {
+ while (stk.size() && nums[stk.back()] <= nums[i]) {
+ stk.pop_back();
+ }
+ stk.push_back(i);
+ }
+ long long ans = 0, i = 0;
+ for (int j : stk) {
+ ans += (j - i) * nums[j];
+ i = j;
+ }
+ return ans;
+ }
+};
\ No newline at end of file
diff --git a/solution/3200-3299/3221.Maximum Array Hopping Score II/Solution.go b/solution/3200-3299/3221.Maximum Array Hopping Score II/Solution.go
new file mode 100644
index 0000000000000..34ecd330347d6
--- /dev/null
+++ b/solution/3200-3299/3221.Maximum Array Hopping Score II/Solution.go
@@ -0,0 +1,15 @@
+func maxScore(nums []int) (ans int64) {
+ stk := []int{}
+ for i, x := range nums {
+ for len(stk) > 0 && nums[stk[len(stk)-1]] <= x {
+ stk = stk[:len(stk)-1]
+ }
+ stk = append(stk, i)
+ }
+ i := 0
+ for _, j := range stk {
+ ans += int64((j - i) * nums[j])
+ i = j
+ }
+ return
+}
\ No newline at end of file
diff --git a/solution/3200-3299/3221.Maximum Array Hopping Score II/Solution.java b/solution/3200-3299/3221.Maximum Array Hopping Score II/Solution.java
new file mode 100644
index 0000000000000..a862b3028d4fa
--- /dev/null
+++ b/solution/3200-3299/3221.Maximum Array Hopping Score II/Solution.java
@@ -0,0 +1,18 @@
+class Solution {
+ public long maxScore(int[] nums) {
+ Deque stk = new ArrayDeque<>();
+ for (int i = 0; i < nums.length; ++i) {
+ while (!stk.isEmpty() && nums[stk.peek()] <= nums[i]) {
+ stk.pop();
+ }
+ stk.push(i);
+ }
+ long ans = 0, i = 0;
+ while (!stk.isEmpty()) {
+ int j = stk.pollLast();
+ ans += (j - i) * nums[j];
+ i = j;
+ }
+ return ans;
+ }
+}
\ No newline at end of file
diff --git a/solution/3200-3299/3221.Maximum Array Hopping Score II/Solution.py b/solution/3200-3299/3221.Maximum Array Hopping Score II/Solution.py
new file mode 100644
index 0000000000000..02a773ddd21f7
--- /dev/null
+++ b/solution/3200-3299/3221.Maximum Array Hopping Score II/Solution.py
@@ -0,0 +1,12 @@
+class Solution:
+ def maxScore(self, nums: List[int]) -> int:
+ stk = []
+ for i, x in enumerate(nums):
+ while stk and nums[stk[-1]] <= x:
+ stk.pop()
+ stk.append(i)
+ ans = i = 0
+ for j in stk:
+ ans += nums[j] * (j - i)
+ i = j
+ return ans
diff --git a/solution/3200-3299/3221.Maximum Array Hopping Score II/Solution.ts b/solution/3200-3299/3221.Maximum Array Hopping Score II/Solution.ts
new file mode 100644
index 0000000000000..a5411d1b49c0a
--- /dev/null
+++ b/solution/3200-3299/3221.Maximum Array Hopping Score II/Solution.ts
@@ -0,0 +1,16 @@
+function maxScore(nums: number[]): number {
+ const stk: number[] = [];
+ for (let i = 0; i < nums.length; ++i) {
+ while (stk.length && nums[stk.at(-1)!] <= nums[i]) {
+ stk.pop();
+ }
+ stk.push(i);
+ }
+ let ans = 0;
+ let i = 0;
+ for (const j of stk) {
+ ans += (j - i) * nums[j];
+ i = j;
+ }
+ return ans;
+}
diff --git a/solution/DATABASE_README.md b/solution/DATABASE_README.md
index 9552987e6c05e..5e07c7b269103 100644
--- a/solution/DATABASE_README.md
+++ b/solution/DATABASE_README.md
@@ -287,6 +287,7 @@
| 3198 | [查找每个州的城市](/solution/3100-3199/3198.Find%20Cities%20in%20Each%20State/README.md) | `数据库` | 简单 | 🔒 |
| 3204 | [按位用户权限分析](/solution/3200-3299/3204.Bitwise%20User%20Permissions%20Analysis/README.md) | `数据库` | 中等 | 🔒 |
| 3214 | [同比增长率](/solution/3200-3299/3214.Year%20on%20Year%20Growth%20Rate/README.md) | `数据库` | 困难 | 🔒 |
+| 3220 | [Odd and Even Transactions](/solution/3200-3299/3220.Odd%20and%20Even%20Transactions/README.md) | | 中等 | |
## 版权
diff --git a/solution/DATABASE_README_EN.md b/solution/DATABASE_README_EN.md
index 18d7f63a90c8e..498c7ac873005 100644
--- a/solution/DATABASE_README_EN.md
+++ b/solution/DATABASE_README_EN.md
@@ -285,6 +285,7 @@ Press Control + F(or Command + F on
| 3198 | [Find Cities in Each State](/solution/3100-3199/3198.Find%20Cities%20in%20Each%20State/README_EN.md) | `Database` | Easy | 🔒 |
| 3204 | [Bitwise User Permissions Analysis](/solution/3200-3299/3204.Bitwise%20User%20Permissions%20Analysis/README_EN.md) | `Database` | Medium | 🔒 |
| 3214 | [Year on Year Growth Rate](/solution/3200-3299/3214.Year%20on%20Year%20Growth%20Rate/README_EN.md) | `Database` | Hard | 🔒 |
+| 3220 | [Odd and Even Transactions](/solution/3200-3299/3220.Odd%20and%20Even%20Transactions/README_EN.md) | | Medium | |
## Copyright
diff --git a/solution/README.md b/solution/README.md
index 0a4c9bc30b211..973b4707f8e17 100644
--- a/solution/README.md
+++ b/solution/README.md
@@ -3230,6 +3230,8 @@
| 3217 | [从链表中移除在数组中存在的节点](/solution/3200-3299/3217.Delete%20Nodes%20From%20Linked%20List%20Present%20in%20Array/README.md) | | 中等 | 第 406 场周赛 |
| 3218 | [切蛋糕的最小总开销 I](/solution/3200-3299/3218.Minimum%20Cost%20for%20Cutting%20Cake%20I/README.md) | | 中等 | 第 406 场周赛 |
| 3219 | [切蛋糕的最小总开销 II](/solution/3200-3299/3219.Minimum%20Cost%20for%20Cutting%20Cake%20II/README.md) | | 困难 | 第 406 场周赛 |
+| 3220 | [Odd and Even Transactions](/solution/3200-3299/3220.Odd%20and%20Even%20Transactions/README.md) | | 中等 | |
+| 3221 | [Maximum Array Hopping Score II](/solution/3200-3299/3221.Maximum%20Array%20Hopping%20Score%20II/README.md) | | 中等 | 🔒 |
## 版权
diff --git a/solution/README_EN.md b/solution/README_EN.md
index 1e25425af4baa..79c33c379eb5a 100644
--- a/solution/README_EN.md
+++ b/solution/README_EN.md
@@ -3228,6 +3228,8 @@ Press Control + F(or Command + F on
| 3217 | [Delete Nodes From Linked List Present in Array](/solution/3200-3299/3217.Delete%20Nodes%20From%20Linked%20List%20Present%20in%20Array/README_EN.md) | | Medium | Weekly Contest 406 |
| 3218 | [Minimum Cost for Cutting Cake I](/solution/3200-3299/3218.Minimum%20Cost%20for%20Cutting%20Cake%20I/README_EN.md) | | Medium | Weekly Contest 406 |
| 3219 | [Minimum Cost for Cutting Cake II](/solution/3200-3299/3219.Minimum%20Cost%20for%20Cutting%20Cake%20II/README_EN.md) | | Hard | Weekly Contest 406 |
+| 3220 | [Odd and Even Transactions](/solution/3200-3299/3220.Odd%20and%20Even%20Transactions/README_EN.md) | | Medium | |
+| 3221 | [Maximum Array Hopping Score II](/solution/3200-3299/3221.Maximum%20Array%20Hopping%20Score%20II/README_EN.md) | | Medium | 🔒 |
## Copyright
diff --git a/solution/main.py b/solution/main.py
index 366e4aa9decf8..e6b60b0a9aa9f 100644
--- a/solution/main.py
+++ b/solution/main.py
@@ -453,7 +453,7 @@ def run():
) or spider.get_question_detail_en(slug, retry=8)
if not detail:
continue
- time.sleep(0.3)
+ time.sleep(1)
question_details[slug] = Spider.format_question_detail(
detail, str(qid).zfill(4)
)