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feat: add new lc problems
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# [2230. The Users That Are Eligible for Discount](https://leetcode-cn.com/problems/the-users-that-are-eligible-for-discount)
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[English Version](/solution/2200-2299/2230.The%20Users%20That%20Are%20Eligible%20for%20Discount/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<p>Table: <code>Purchases</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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| user_id | int |
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| time_stamp | datetime |
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| amount | int |
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+-------------+----------+
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(user_id, time_stamp) is the primary key for this table.
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Each row contains information about the purchase time and the amount paid for the user with ID user_id.
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</pre>
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<p>&nbsp;</p>
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<p>A user is eligible for a discount if they had a purchase in the inclusive interval of time <code>[startDate, endDate]</code> with at least <code>minAmount</code> amount.</p>
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<p>Write an SQL query to report the IDs of the users that are eligible for a discount.</p>
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<p>Return the result table ordered by <code>user_id</code>.</p>
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<p>The query result format is in the following example.</p>
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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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<pre>
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<strong>Input:</strong>
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Purchases table:
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+---------+---------------------+--------+
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| user_id | time_stamp | amount |
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+---------+---------------------+--------+
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| 1 | 2022年04月20日 09:03:00 | 4416 |
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| 2 | 2022年03月19日 19:24:02 | 678 |
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| 3 | 2022年03月18日 12:03:09 | 4523 |
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| 3 | 2022年03月30日 09:43:42 | 626 |
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+---------+---------------------+--------+
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startDate = 2022年03月08日, endDate = 2022年03月20日, minAmount = 1000
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<strong>Output:</strong>
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+---------+
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| user_id |
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+---------+
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| 3 |
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+---------+
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<strong>Explanation:</strong>
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Out of the three users, only User 3 is eligible for a discount.
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- User 1 had one purchase with at least minAmount amount, but not within the time interval.
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- User 2 had one purchase within the time interval, but with less than minAmount amount.
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- User 3 is the only user who had a purchase that satisfies both conditions.
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</pre>
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<p>&nbsp;</p>
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<p><strong>Important Note:</strong> This problem is basically the same as <a href="https://leetcode.com/problems/the-number-of-users-that-are-eligible-for-discount/">The Number of Users That Are Eligible for Discount</a>.</p>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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<!-- tabs:start -->
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### **SQL**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```sql
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```
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<!-- tabs:end -->
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# [2230. The Users That Are Eligible for Discount](https://leetcode.com/problems/the-users-that-are-eligible-for-discount)
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[中文文档](/solution/2200-2299/2230.The%20Users%20That%20Are%20Eligible%20for%20Discount/README.md)
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## Description
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<p>Table: <code>Purchases</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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| user_id | int |
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| time_stamp | datetime |
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| amount | int |
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+-------------+----------+
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(user_id, time_stamp) is the primary key for this table.
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Each row contains information about the purchase time and the amount paid for the user with ID user_id.
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</pre>
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<p>&nbsp;</p>
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<p>A user is eligible for a discount if they had a purchase in the inclusive interval of time <code>[startDate, endDate]</code> with at least <code>minAmount</code> amount.</p>
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<p>Write an SQL query to report the IDs of the users that are eligible for a discount.</p>
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<p>Return the result table ordered by <code>user_id</code>.</p>
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<p>The query result format is in the following example.</p>
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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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<pre>
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<strong>Input:</strong>
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Purchases table:
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+---------+---------------------+--------+
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| user_id | time_stamp | amount |
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+---------+---------------------+--------+
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| 1 | 2022年04月20日 09:03:00 | 4416 |
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| 2 | 2022年03月19日 19:24:02 | 678 |
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| 3 | 2022年03月18日 12:03:09 | 4523 |
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| 3 | 2022年03月30日 09:43:42 | 626 |
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+---------+---------------------+--------+
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startDate = 2022年03月08日, endDate = 2022年03月20日, minAmount = 1000
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<strong>Output:</strong>
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+---------+
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| user_id |
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+---------+
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| 3 |
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+---------+
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<strong>Explanation:</strong>
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Out of the three users, only User 3 is eligible for a discount.
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- User 1 had one purchase with at least minAmount amount, but not within the time interval.
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- User 2 had one purchase within the time interval, but with less than minAmount amount.
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- User 3 is the only user who had a purchase that satisfies both conditions.
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</pre>
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<p>&nbsp;</p>
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<p><strong>Important Note:</strong> This problem is basically the same as <a href="https://leetcode.com/problems/the-number-of-users-that-are-eligible-for-discount/">The Number of Users That Are Eligible for Discount</a>.</p>
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## Solutions
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<!-- tabs:start -->
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### **SQL**
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```sql
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```
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<!-- tabs:end -->
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# [2231. 按奇偶性交换后的最大数字](https://leetcode-cn.com/problems/largest-number-after-digit-swaps-by-parity)
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[English Version](/solution/2200-2299/2231.Largest%20Number%20After%20Digit%20Swaps%20by%20Parity/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<p>给你一个正整数 <code>num</code> 。你可以交换 <code>num</code> 中 <strong>奇偶性</strong> 相同的任意两位数字(即,都是奇数或者偶数)。</p>
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<p>返回交换 <strong>任意</strong> 次之后 <code>num</code> 的 <strong>最大</strong> 可能值<em>。</em></p>
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<p>&nbsp;</p>
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<p><strong>示例 1:</strong></p>
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<pre><strong>输入:</strong>num = 1234
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<strong>输出:</strong>3412
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<strong>解释:</strong>交换数字 3 和数字 1 ,结果得到 3214 。
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交换数字 2 和数字 4 ,结果得到 3412 。
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注意,可能存在其他交换序列,但是可以证明 3412 是最大可能值。
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注意,不能交换数字 4 和数字 1 ,因为它们奇偶性不同。
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</pre>
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<p><strong>示例 2:</strong></p>
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<pre><strong>输入:</strong>num = 65875
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<strong>输出:</strong>87655
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<strong>解释:</strong>交换数字 8 和数字 6 ,结果得到 85675 。
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交换数字 5 和数字 7 ,结果得到 87655 。
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注意,可能存在其他交换序列,但是可以证明 87655 是最大可能值。
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</pre>
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<p>&nbsp;</p>
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<p><strong>提示:</strong></p>
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<ul>
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<li><code>1 &lt;= num &lt;= 10<sup>9</sup></code></li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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```
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### **TypeScript**
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```ts
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```
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### **...**
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```
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```
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<!-- tabs:end -->
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# [2231. Largest Number After Digit Swaps by Parity](https://leetcode.com/problems/largest-number-after-digit-swaps-by-parity)
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[中文文档](/solution/2200-2299/2231.Largest%20Number%20After%20Digit%20Swaps%20by%20Parity/README.md)
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## Description
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<p>You are given a positive integer <code>num</code>. You may swap any two digits of <code>num</code> that have the same <strong>parity</strong> (i.e. both odd digits or both even digits).</p>
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<p>Return<em> the <strong>largest</strong> possible value of </em><code>num</code><em> after <strong>any</strong> number of swaps.</em></p>
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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> num = 1234
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<strong>Output:</strong> 3412
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<strong>Explanation:</strong> Swap the digit 3 with the digit 1, this results in the number 3214.
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Swap the digit 2 with the digit 4, this results in the number 3412.
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Note that there may be other sequences of swaps but it can be shown that 3412 is the largest possible number.
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Also note that we may not swap the digit 4 with the digit 1 since they are of different parities.
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</pre>
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<p><strong>Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> num = 65875
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<strong>Output:</strong> 87655
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<strong>Explanation:</strong> Swap the digit 8 with the digit 6, this results in the number 85675.
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Swap the first digit 5 with the digit 7, this results in the number 87655.
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Note that there may be other sequences of swaps but it can be shown that 87655 is the largest possible number.
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt;= num &lt;= 10<sup>9</sup></code></li>
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</ul>
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## Solutions
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<!-- tabs:start -->
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### **Python3**
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```python
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```
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### **Java**
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```java
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```
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### **TypeScript**
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```ts
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```
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### **...**
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```
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```
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<!-- tabs:end -->
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# [2232. 向表达式添加括号后的最小结果](https://leetcode-cn.com/problems/minimize-result-by-adding-parentheses-to-expression)
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[English Version](/solution/2200-2299/2232.Minimize%20Result%20by%20Adding%20Parentheses%20to%20Expression/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<p>给你一个下标从 <strong>0</strong> 开始的字符串 <code>expression</code> ,格式为 <code>"&lt;num1&gt;+&lt;num2&gt;"</code> ,其中 <code>&lt;num1&gt;</code> 和 <code>&lt;num2&gt;</code> 表示正整数。</p>
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<p>请你向 <code>expression</code> 中添加一对括号,使得在添加之后, <code>expression</code> 仍然是一个有效的数学表达式,并且计算后可以得到 <strong>最小</strong> 可能值。左括号 <strong>必须</strong> 添加在 <code>'+'</code> 的左侧,而右括号必须添加在 <code>'+'</code> 的右侧。</p>
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<p>返回添加一对括号后形成的表达式&nbsp;<code>expression</code> ,且满足<em> </em><code>expression</code><em> </em>计算得到 <strong>最小</strong> 可能值<em>。</em>如果存在多个答案都能产生相同结果,返回任意一个答案。</p>
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<p>生成的输入满足:<code>expression</code> 的原始值和添加满足要求的任一对括号之后 <code>expression</code> 的值,都符合 32-bit 带符号整数范围。</p>
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<p>&nbsp;</p>
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<p><strong>示例 1:</strong></p>
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<pre><strong>输入:</strong>expression = "247+38"
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<strong>输出:</strong>"2(47+38)"
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<strong>解释:</strong>表达式计算得到 2 * (47 + 38) = 2 * 85 = 170 。
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注意 "2(4)7+38" 不是有效的结果,因为右括号必须添加在 <code>'+' 的右侧。</code>
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可以证明 170 是最小可能值。
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</pre>
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<p><strong>示例 2:</strong></p>
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<pre><strong>输入:</strong>expression = "12+34"
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<strong>输出:</strong>"1(2+3)4"
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<strong>解释:</strong>表达式计算得到 1 * (2 + 3) * 4 = 1 * 5 * 4 = 20 。
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</pre>
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<p><strong>示例 3:</strong></p>
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<pre><strong>输入:</strong>expression = "999+999"
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<strong>输出:</strong>"(999+999)"
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<strong>解释:</strong>表达式计算得到 999 + 999 = 1998 。
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</pre>
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<p>&nbsp;</p>
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<p><strong>提示:</strong></p>
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<ul>
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<li><code>3 &lt;= expression.length &lt;= 10</code></li>
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<li><code>expression</code> 仅由数字 <code>'1'</code> 到 <code>'9'</code> 和 <code>'+'</code> 组成</li>
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<li><code>expression</code> 由数字开始和结束</li>
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<li><code>expression</code> 恰好仅含有一个 <code>'+'</code>.</li>
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<li><code>expression</code> 的原始值和添加满足要求的任一对括号之后 <code>expression</code> 的值,都符合 32-bit 带符号整数范围</li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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```
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### **TypeScript**
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```ts
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```
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### **...**
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```
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```
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<!-- tabs:end -->

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