feat: add solutions to lc problems: No.3220,3221

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):

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):

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):

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)]
 ```
 

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)]
 ```
 

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)]

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
+---
+
+<!-- problem:start -->
+
+# [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
+
+<!-- description:start -->
+
+<p>Table: <code>transactions</code></p>
+
+<pre>
++------------------+------+
+| 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.
+</pre>
+
+<p>Write a solution to find the <strong>sum of amounts</strong> for <strong>odd</strong> and <strong>even</strong> transactions for each day. If there are no odd or even transactions for a specific date, display as <code>0</code>.</p>
+
+<p>Return <em>the result table ordered by</em> <code>transaction_date</code> <em>in <strong>ascending</strong> order</em>.</p>
+
+<p>The result format is in the following example.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong></p>
+
+<p><code>transactions</code> table:</p>
+
+<pre class="example-io">
++----------------+--------+------------------+
+| 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       |
++----------------+--------+------------------+
+  </pre>
+
+<p><strong>Output:</strong></p>
+
+<pre class="example-io">
++------------------+---------+----------+
+| transaction_date | odd_sum | even_sum |
++------------------+---------+----------+
+| 2024-07-01       | 75      | 350      |
+| 2024-07-02       | 0       | 350      |
+| 2024-07-03       | 0       | 120      |
++------------------+---------+----------+
+  </pre>
+
+<p><strong>Explanation:</strong></p>
+
+<ul>
+	<li>For transaction dates:
+	<ul>
+		<li>2024-07-01:
+		<ul>
+			<li>Sum of amounts for odd transactions: 75</li>
+			<li>Sum of amounts for even transactions: 150 + 200 = 350</li>
+		</ul>
+		</li>
+		<li>2024-07-02:
+		<ul>
+			<li>Sum of amounts for odd transactions: 0</li>
+			<li>Sum of amounts for even transactions: 300 + 50 = 350</li>
+		</ul>
+		</li>
+		<li>2024-07-03:
+		<ul>
+			<li>Sum of amounts for odd transactions: 0</li>
+			<li>Sum of amounts for even transactions: 120</li>
+		</ul>
+		</li>
+	</ul>
+	</li>
+</ul>
+
+<p><strong>Note:</strong> The output table is ordered by <code>transaction_date</code> in ascending order.</p>
+</div>
+
+<!-- description:end -->
+
+## 解法
+
+<!-- solution:start -->
+
+### 方法一：分组求和
+
+我们可以将数据按照 `transaction_date` 进行分组，然后分别计算奇数和偶数的交易金额之和。最后按照 `transaction_date` 升序排序。
+
+<!-- tabs:start -->
+
+#### MySQL
+
+```sql
+# Write your MySQL query statement below
+SELECT
+    transaction_date,
+    SUM(IF(amount % 2 = 1, amount, 0)) AS odd_sum,
+    SUM(IF(amount % 2 = 0, amount, 0)) AS even_sum
+FROM transactions
+GROUP BY 1
+ORDER BY 1;
+```
+
+#### Pandas
+
+```python
+import pandas as pd
+
+
+def sum_daily_odd_even(transactions: pd.DataFrame) -> pd.DataFrame:
+    transactions["odd_sum"] = transactions["amount"].where(
+        transactions["amount"] % 2 == 1, 0
+    )
+    transactions["even_sum"] = transactions["amount"].where(
+        transactions["amount"] % 2 == 0, 0
+    )
+
+    result = (
+        transactions.groupby("transaction_date")
+        .agg(odd_sum=("odd_sum", "sum"), even_sum=("even_sum", "sum"))
+        .reset_index()
+    )
+
+    result = result.sort_values("transaction_date")
+
+    return result
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->

solution/3200-3299/3220.Odd and Even Transactions/README_EN.md

@@ -0,0 +1,154 @@
+---
+comments: true
+difficulty: Medium
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3200-3299/3220.Odd%20and%20Even%20Transactions/README_EN.md
+---
+
+<!-- problem:start -->
+
+# [3220. Odd and Even Transactions](https://leetcode.com/problems/odd-and-even-transactions)
+
+[中文文档](/solution/3200-3299/3220.Odd%20and%20Even%20Transactions/README.md)
+
+## Description
+
+<!-- description:start -->
+
+<p>Table: <code>transactions</code></p>
+
+<pre>
++------------------+------+
+| 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.
+</pre>
+
+<p>Write a solution to find the <strong>sum of amounts</strong> for <strong>odd</strong> and <strong>even</strong> transactions for each day. If there are no odd or even transactions for a specific date, display as <code>0</code>.</p>
+
+<p>Return <em>the result table ordered by</em> <code>transaction_date</code> <em>in <strong>ascending</strong> order</em>.</p>
+
+<p>The result format is in the following example.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong></p>
+
+<p><code>transactions</code> table:</p>
+
+<pre class="example-io">
++----------------+--------+------------------+
+| 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       |
++----------------+--------+------------------+
+  </pre>
+
+<p><strong>Output:</strong></p>
+
+<pre class="example-io">
++------------------+---------+----------+
+| transaction_date | odd_sum | even_sum |
++------------------+---------+----------+
+| 2024-07-01       | 75      | 350      |
+| 2024-07-02       | 0       | 350      |
+| 2024-07-03       | 0       | 120      |
++------------------+---------+----------+
+  </pre>
+
+<p><strong>Explanation:</strong></p>
+
+<ul>
+	<li>For transaction dates:
+	<ul>
+		<li>2024-07-01:
+		<ul>
+			<li>Sum of amounts for odd transactions: 75</li>
+			<li>Sum of amounts for even transactions: 150 + 200 = 350</li>
+		</ul>
+		</li>
+		<li>2024-07-02:
+		<ul>
+			<li>Sum of amounts for odd transactions: 0</li>
+			<li>Sum of amounts for even transactions: 300 + 50 = 350</li>
+		</ul>
+		</li>
+		<li>2024-07-03:
+		<ul>
+			<li>Sum of amounts for odd transactions: 0</li>
+			<li>Sum of amounts for even transactions: 120</li>
+		</ul>
+		</li>
+	</ul>
+	</li>
+</ul>
+
+<p><strong>Note:</strong> The output table is ordered by <code>transaction_date</code> in ascending order.</p>
+</div>
+
+<!-- description:end -->
+
+## Solutions
+
+<!-- solution:start -->
+
+### Solution 1: Grouping and Summing
+
+We can group the data by `transaction_date`, and then calculate the sum of transaction amounts for odd and even dates separately. Finally, sort by `transaction_date` in ascending order.
+
+<!-- tabs:start -->
+
+#### MySQL
+
+```sql
+# Write your MySQL query statement below
+SELECT
+    transaction_date,
+    SUM(IF(amount % 2 = 1, amount, 0)) AS odd_sum,
+    SUM(IF(amount % 2 = 0, amount, 0)) AS even_sum
+FROM transactions
+GROUP BY 1
+ORDER BY 1;
+```
+
+#### Pandas
+
+```python
+import pandas as pd
+
+
+def sum_daily_odd_even(transactions: pd.DataFrame) -> pd.DataFrame:
+    transactions["odd_sum"] = transactions["amount"].where(
+        transactions["amount"] % 2 == 1, 0
+    )
+    transactions["even_sum"] = transactions["amount"].where(
+        transactions["amount"] % 2 == 0, 0
+    )
+
+    result = (
+        transactions.groupby("transaction_date")
+        .agg(odd_sum=("odd_sum", "sum"), even_sum=("even_sum", "sum"))
+        .reset_index()
+    )
+
+    result = result.sort_values("transaction_date")
+
+    return result
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->