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feat: add solutions to lc problems: No.3220,3221 (#3282)
* No.3220.Odd and Even Transactions * No.3221.Maximum Array Hopping Score II
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‎solution/2900-2999/2959.Number of Possible Sets of Closing Branches/README.md

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@@ -123,7 +123,7 @@ class Solution:
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for mask in range(1 << n):
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g = [[inf] * n for _ in range(n)]
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for u, v, w in roads:
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if mask >> u & 1 and mask > v & 1:
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if mask >> u & 1 and mask >> v & 1:
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g[u][v] = min(g[u][v], w)
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g[v][u] = min(g[v][u], w)
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for k in range(n):

‎solution/2900-2999/2959.Number of Possible Sets of Closing Branches/README_EN.md

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@@ -117,7 +117,7 @@ class Solution:
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for mask in range(1 << n):
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g = [[inf] * n for _ in range(n)]
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for u, v, w in roads:
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if mask >> u & 1 and mask > v & 1:
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if mask >> u & 1 and mask >> v & 1:
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g[u][v] = min(g[u][v], w)
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g[v][u] = min(g[v][u], w)
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for k in range(n):

‎solution/2900-2999/2959.Number of Possible Sets of Closing Branches/Solution.py

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@@ -4,7 +4,7 @@ def numberOfSets(self, n: int, maxDistance: int, roads: List[List[int]]) -> int:
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for mask in range(1 << n):
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g = [[inf] * n for _ in range(n)]
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for u, v, w in roads:
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if mask >> u & 1 and mask > v & 1:
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if mask >> u & 1 and mask >> v & 1:
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g[u][v] = min(g[u][v], w)
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g[v][u] = min(g[v][u], w)
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for k in range(n):

‎solution/3100-3199/3112.Minimum Time to Visit Disappearing Nodes/README.md

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@@ -133,15 +133,15 @@ class Solution:
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g[v].append((u, w))
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dist = [inf] * n
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dist[0] = 0
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q = [(0, 0)]
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while q:
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du, u = heappop(q)
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pq = [(0, 0)]
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while pq:
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du, u = heappop(pq)
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if du > dist[u]:
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continue
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for v, w in g[u]:
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if dist[v] > dist[u] + w and dist[u] + w < disappear[v]:
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dist[v] = dist[u] + w
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heappush(q, (dist[v], v))
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heappush(pq, (dist[v], v))
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return [a if a < b else -1 for a, b in zip(dist, disappear)]
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```
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‎solution/3100-3199/3112.Minimum Time to Visit Disappearing Nodes/README_EN.md

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@@ -131,15 +131,15 @@ class Solution:
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g[v].append((u, w))
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dist = [inf] * n
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dist[0] = 0
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q = [(0, 0)]
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while q:
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du, u = heappop(q)
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pq = [(0, 0)]
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while pq:
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du, u = heappop(pq)
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if du > dist[u]:
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continue
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for v, w in g[u]:
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if dist[v] > dist[u] + w and dist[u] + w < disappear[v]:
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dist[v] = dist[u] + w
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heappush(q, (dist[v], v))
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heappush(pq, (dist[v], v))
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return [a if a < b else -1 for a, b in zip(dist, disappear)]
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```
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‎solution/3100-3199/3112.Minimum Time to Visit Disappearing Nodes/Solution.py

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@@ -8,13 +8,13 @@ def minimumTime(
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g[v].append((u, w))
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dist = [inf] * n
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dist[0] = 0
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q = [(0, 0)]
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while q:
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du, u = heappop(q)
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pq = [(0, 0)]
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while pq:
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du, u = heappop(pq)
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if du > dist[u]:
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continue
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for v, w in g[u]:
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if dist[v] > dist[u] + w and dist[u] + w < disappear[v]:
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dist[v] = dist[u] + w
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heappush(q, (dist[v], v))
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heappush(pq, (dist[v], v))
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return [a if a < b else -1 for a, b in zip(dist, disappear)]

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

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---
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comments: true
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difficulty: 中等
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edit_url: https://github.com/doocs/leetcode/edit/main/solution/3200-3299/3220.Odd%20and%20Even%20Transactions/README.md
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---
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<!-- problem:start -->
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# [3220. Odd and Even Transactions](https://leetcode.cn/problems/odd-and-even-transactions)
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[English Version](/solution/3200-3299/3220.Odd%20and%20Even%20Transactions/README_EN.md)
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## 题目描述
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<!-- description:start -->
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<p>Table: <code>transactions</code></p>
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<pre>
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+------------------+------+
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| Column Name | Type |
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+------------------+------+
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| transaction_id | int |
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| amount | int |
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| transaction_date | date |
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+------------------+------+
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The transactions_id column uniquely identifies each row in this table.
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Each row of this table contains the transaction id, amount and transaction date.
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</pre>
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<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>
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<p>Return <em>the result table ordered by</em> <code>transaction_date</code> <em>in <strong>ascending</strong> order</em>.</p>
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<p>The result format is in the following example.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example:</strong></p>
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<div class="example-block">
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<p><strong>Input:</strong></p>
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<p><code>transactions</code> table:</p>
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<pre class="example-io">
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+----------------+--------+------------------+
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| transaction_id | amount | transaction_date |
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+----------------+--------+------------------+
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| 1 | 150 | 2024-07-01 |
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| 2 | 200 | 2024-07-01 |
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| 3 | 75 | 2024-07-01 |
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| 4 | 300 | 2024-07-02 |
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| 5 | 50 | 2024-07-02 |
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| 6 | 120 | 2024-07-03 |
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+----------------+--------+------------------+
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</pre>
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<p><strong>Output:</strong></p>
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<pre class="example-io">
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+------------------+---------+----------+
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| transaction_date | odd_sum | even_sum |
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+------------------+---------+----------+
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| 2024-07-01 | 75 | 350 |
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| 2024-07-02 | 0 | 350 |
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| 2024-07-03 | 0 | 120 |
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+------------------+---------+----------+
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</pre>
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<p><strong>Explanation:</strong></p>
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<ul>
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<li>For transaction dates:
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<ul>
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<li>2024-07-01:
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<ul>
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<li>Sum of amounts for odd transactions: 75</li>
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<li>Sum of amounts for even transactions: 150 + 200 = 350</li>
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</ul>
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</li>
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<li>2024-07-02:
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<ul>
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<li>Sum of amounts for odd transactions: 0</li>
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<li>Sum of amounts for even transactions: 300 + 50 = 350</li>
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</ul>
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</li>
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<li>2024-07-03:
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<ul>
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<li>Sum of amounts for odd transactions: 0</li>
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<li>Sum of amounts for even transactions: 120</li>
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</ul>
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</li>
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</ul>
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</li>
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</ul>
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<p><strong>Note:</strong> The output table is ordered by <code>transaction_date</code> in ascending order.</p>
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</div>
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<!-- description:end -->
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## 解法
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<!-- solution:start -->
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### 方法一：分组求和
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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。
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<!-- tabs:start -->
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#### MySQL
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```sql
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# Write your MySQL query statement below
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SELECT
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transaction_date,
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SUM(IF(amount % 2 = 1, amount, 0)) AS odd_sum,
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SUM(IF(amount % 2 = 0, amount, 0)) AS even_sum
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FROM transactions
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GROUP BY 1
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ORDER BY 1;
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```
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#### Pandas
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```python
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import pandas as pd
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def sum_daily_odd_even(transactions: pd.DataFrame) -> pd.DataFrame:
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transactions["odd_sum"] = transactions["amount"].where(
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transactions["amount"] % 2 == 1, 0
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)
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transactions["even_sum"] = transactions["amount"].where(
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transactions["amount"] % 2 == 0, 0
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)
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result = (
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transactions.groupby("transaction_date")
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.agg(odd_sum=("odd_sum", "sum"), even_sum=("even_sum", "sum"))
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.reset_index()
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)
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result = result.sort_values("transaction_date")
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return result
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->

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

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---
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comments: true
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difficulty: Medium
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edit_url: https://github.com/doocs/leetcode/edit/main/solution/3200-3299/3220.Odd%20and%20Even%20Transactions/README_EN.md
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---
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<!-- problem:start -->
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# [3220. Odd and Even Transactions](https://leetcode.com/problems/odd-and-even-transactions)
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[中文文档](/solution/3200-3299/3220.Odd%20and%20Even%20Transactions/README.md)
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## Description
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<!-- description:start -->
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<p>Table: <code>transactions</code></p>
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<pre>
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+------------------+------+
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| Column Name | Type |
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+------------------+------+
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| transaction_id | int |
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| amount | int |
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| transaction_date | date |
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+------------------+------+
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The transactions_id column uniquely identifies each row in this table.
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Each row of this table contains the transaction id, amount and transaction date.
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</pre>
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<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>
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<p>Return <em>the result table ordered by</em> <code>transaction_date</code> <em>in <strong>ascending</strong> order</em>.</p>
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<p>The result format is in the following example.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example:</strong></p>
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<div class="example-block">
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<p><strong>Input:</strong></p>
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<p><code>transactions</code> table:</p>
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<pre class="example-io">
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+----------------+--------+------------------+
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| transaction_id | amount | transaction_date |
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+----------------+--------+------------------+
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| 1 | 150 | 2024-07-01 |
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| 2 | 200 | 2024-07-01 |
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| 3 | 75 | 2024-07-01 |
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| 4 | 300 | 2024-07-02 |
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| 5 | 50 | 2024-07-02 |
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| 6 | 120 | 2024-07-03 |
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+----------------+--------+------------------+
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</pre>
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<p><strong>Output:</strong></p>
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<pre class="example-io">
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+------------------+---------+----------+
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| transaction_date | odd_sum | even_sum |
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+------------------+---------+----------+
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| 2024-07-01 | 75 | 350 |
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| 2024-07-02 | 0 | 350 |
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| 2024-07-03 | 0 | 120 |
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+------------------+---------+----------+
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</pre>
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<p><strong>Explanation:</strong></p>
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<ul>
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<li>For transaction dates:
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<ul>
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<li>2024-07-01:
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<ul>
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<li>Sum of amounts for odd transactions: 75</li>
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<li>Sum of amounts for even transactions: 150 + 200 = 350</li>
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</ul>
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</li>
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<li>2024-07-02:
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<ul>
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<li>Sum of amounts for odd transactions: 0</li>
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<li>Sum of amounts for even transactions: 300 + 50 = 350</li>
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</ul>
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</li>
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<li>2024-07-03:
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<ul>
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<li>Sum of amounts for odd transactions: 0</li>
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<li>Sum of amounts for even transactions: 120</li>
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</ul>
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</li>
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</ul>
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</li>
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</ul>
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<p><strong>Note:</strong> The output table is ordered by <code>transaction_date</code> in ascending order.</p>
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</div>
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<!-- description:end -->
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## Solutions
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<!-- solution:start -->
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### Solution 1: Grouping and Summing
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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.
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<!-- tabs:start -->
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#### MySQL
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```sql
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# Write your MySQL query statement below
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SELECT
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transaction_date,
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SUM(IF(amount % 2 = 1, amount, 0)) AS odd_sum,
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SUM(IF(amount % 2 = 0, amount, 0)) AS even_sum
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FROM transactions
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GROUP BY 1
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ORDER BY 1;
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```
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#### Pandas
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```python
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import pandas as pd
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def sum_daily_odd_even(transactions: pd.DataFrame) -> pd.DataFrame:
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transactions["odd_sum"] = transactions["amount"].where(
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transactions["amount"] % 2 == 1, 0
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)
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transactions["even_sum"] = transactions["amount"].where(
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transactions["amount"] % 2 == 0, 0
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)
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result = (
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transactions.groupby("transaction_date")
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.agg(odd_sum=("odd_sum", "sum"), even_sum=("even_sum", "sum"))
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.reset_index()
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)
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result = result.sort_values("transaction_date")
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return result
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->

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