feat: add new lc problems and solutions: No.2773~2777 (doocs#1202)

* No.2773.Height of Special Binary Tree
* No.2774.Array Upper Bound
* No.2775.Undefined to Null
* No.2776.Convert Callback Based Function to Promise Based Function
* No.2777.Date Range Generator

Author: yanglbme

solution/0100-0199/0196.Delete Duplicate Emails/README.md

@@ -21,17 +21,19 @@ id是该表的主键列。
 
 <p>&nbsp;</p>
 
-<p>编写一个 SQL <strong>删除语句</strong>来 <strong>删除</strong> 所有重复的电子邮件，只保留一个id最小的唯一电子邮件。</p>
+<p><strong>删除</strong> 所有重复的电子邮件，只保留一个具有最小 <code>id</code> 的唯一电子邮件。</p>
 
-<p>以 <strong>任意顺序</strong> 返回结果表。 （<strong>注意</strong>： 仅需要写删除语句，将自动对剩余结果进行查询）</p>
+<p>（对于 SQL 用户，请注意你应该编写一个 <code>DELETE</code> 语句而不是 <code>SELECT</code> 语句。）</p>
 
-<p>查询结果格式如下所示。</p>
+<p>（对于 Pandas 用户，请注意你应该直接修改 <code>Person</code> 表。）</p>
 
-<p>&nbsp;</p>
+<p>运行脚本后，显示的答案是 <code>Person</code> 表。驱动程序将首先编译并运行您的代码片段，然后再显示 <code>Person</code> 表。<code>Person</code> 表的最终顺序 <strong>无关紧要</strong> 。</p>
+
+<p>返回结果格式如下示例所示。</p>
 
 <p>&nbsp;</p>
 
-<p><strong>示例 1:</strong></p>
+<p><strong class="example">示例 1:</strong></p>
 
 <pre>
 <strong>输入:</strong> 

solution/0100-0199/0196.Delete Duplicate Emails/README_EN.md

@@ -19,7 +19,11 @@ Each row of this table contains an email. The emails will not contain uppercase
 
 <p>&nbsp;</p>
 
-<p><strong>Delete</strong> all the duplicate emails, keeping only one unique email with the smallest <code>id</code>. (For SQL users, please note that you are supposed to write a <code>DELETE</code> statement and not a <code>SELECT</code> one.)</p>
+<p><strong>Delete</strong> all the duplicate emails, keeping only one unique email with the smallest <code>id</code>.</p>
+
+<p>(For SQL users, please note that you are supposed to write a <code>DELETE</code> statement and not a <code>SELECT</code> one.)</p>
+
+<p>(For Pandas users, please note that you are supposed to modify <code>Person</code> in place.)</p>
 
 <p>After running your script, the answer shown is the <code>Person</code> table. The driver will first compile and run your piece of code and then show the <code>Person</code> table. The final order of the <code>Person</code> table <strong>does not matter</strong>.</p>
 

solution/0400-0499/0460.LFU Cache/README.md

@@ -59,7 +59,7 @@ lfu.get(4);      // 返回 4
 <p><strong>提示：</strong></p>
 
 <ul>
-	<li><code>0 &lt;= capacity&nbsp;&lt;= 10<sup>4</sup></code></li>
+	<li><code>1 &lt;= capacity&nbsp;&lt;= 10<sup>4</sup></code></li>
 	<li><code>0 &lt;= key &lt;= 10<sup>5</sup></code></li>
 	<li><code>0 &lt;= value &lt;= 10<sup>9</sup></code></li>
 	<li>最多调用 <code>2 * 10<sup>5</sup></code> 次 <code>get</code> 和 <code>put</code> 方法</li>

solution/1300-1399/1357.Apply Discount Every n Orders/README_EN.md

@@ -40,7 +40,7 @@ cashier.getBill([7,3],[10,10]);                      // return 4000.0. 5<sup>th<
 cashier.getBill([7,5,3,1,6,4,2],[10,10,10,9,9,9,7]); // return 7350.0. 6<sup>th</sup> customer, 50% discount.
                                                      // Original bill = 14700, but with
                                                      // Actual bill = 14700 * ((100 - 50) / 100) = 7350.
-cashier.getBill([2,3,5],[5,3,2]);                    // return 2500.0.  6<sup>th</sup> customer, no discount.
+cashier.getBill([2,3,5],[5,3,2]);                    // return 2500.0.  7<sup>th</sup> customer, no discount.
 </pre>
 
 <p>&nbsp;</p>

solution/1300-1399/1378.Replace Employee ID With The Unique Identifier/README.md

@@ -36,15 +36,15 @@ id 是这张表的主键。
 
 <p>&nbsp;</p>
 
-<p>写一段SQL查询来展示每位用户的<strong> 唯一标识码（unique ID ）</strong>；如果某位员工没有唯一标识码，使用 null 填充即可。</p>
+<p>展示每位用户的<strong> 唯一标识码（unique ID ）</strong>；如果某位员工没有唯一标识码，使用 null 填充即可。</p>
 
 <p>你可以以<strong> 任意</strong> 顺序返回结果表。</p>
 
-<p>查询结果的格式如下例所示。</p>
+<p>返回结果的格式如下例所示。</p>
 
 <p>&nbsp;</p>
 
-<p><strong>示例 1：</strong></p>
+<p><strong class="example">示例 1：</strong></p>
 
 <pre>
 <code><strong>输入：</strong>

solution/1600-1699/1667.Fix Names in a Table/README.md

@@ -21,15 +21,15 @@ user_id 是该表的主键。
 
 <p>&nbsp;</p>
 
-<p>编写一个 SQL 查询来修复名字，使得只有第一个字符是大写的，其余都是小写的。</p>
+<p>修复名字，使得只有第一个字符是大写的，其余都是小写的。</p>
 
 <p>返回按 <code>user_id</code> 排序的结果表。</p>
 
-<p>查询结果格式示例如下。</p>
+<p>返回结果格式示例如下。</p>
 
 <p>&nbsp;</p>
 
-<p><strong>示例 1：</strong></p>
+<p><strong class="example">示例 1：</strong></p>
 
 <pre>
 <strong>输入：</strong>

solution/1700-1799/1741.Find Total Time Spent by Each Employee/README.md

@@ -8,7 +8,8 @@
 
 <p>表: <code>Employees</code></p>
 
-<pre>+-------------+------+
+<pre>
++-------------+------+
 | Column Name | Type |
 +-------------+------+
 | emp_id      | int  |
@@ -23,14 +24,20 @@ in_time 和 out_time 的取值在1到1440之间。
 题目保证同一天没有两个事件在时间上是相交的，并且保证 in_time 小于 out_time。
 </pre>
 
-<p> </p>
+<p>&nbsp;</p>
 
-<p>编写一个SQL查询以计算每位员工每天在办公室花费的总时间（以分钟为单位）。 请注意，在一天之内，同一员工是可以多次进入和离开办公室的。 在办公室里一次进出所花费的时间为out_time 减去 in_time。</p>
+<p>计算每位员工每天在办公室花费的总时间（以分钟为单位）。 请注意，在一天之内，同一员工是可以多次进入和离开办公室的。 在办公室里一次进出所花费的时间为out_time 减去 in_time。</p>
 
-<p>返回结果表单的顺序无要求。<br>
+<p>返回结果表单的顺序无要求。<br />
 查询结果的格式如下：</p>
 
-<pre>Employees table:
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1：</strong></p>
+
+<pre>
+<strong>输入：</strong>
+Employees table:
 +--------+------------+---------+----------+
 | emp_id | event_day  | in_time | out_time |
 +--------+------------+---------+----------+
@@ -40,7 +47,7 @@ in_time 和 out_time 的取值在1到1440之间。
 | 2      | 2020-11-28 | 3       | 33       |
 | 2      | 2020-12-09 | 47      | 74       |
 +--------+------------+---------+----------+
-Result table:
+<strong>输出：</strong>
 +------------+--------+------------+
 | day        | emp_id | total_time |
 +------------+--------+------------+
@@ -49,6 +56,7 @@ Result table:
 | 2020-12-03 | 1      | 41         |
 | 2020-12-09 | 2      | 27         |
 +------------+--------+------------+
+<strong>解释：</strong>
 雇员 1 有三次进出: 有两次发生在 2020-11-28 花费的时间为 (32 - 4) + (200 - 55) = 173, 有一次发生在 2020-12-03 花费的时间为 (42 - 1) = 41。
 雇员 2 有两次进出: 有一次发生在 2020-11-28 花费的时间为 (33 - 3) = 30,  有一次发生在 2020-12-09 花费的时间为 (74 - 47) = 27。
 </pre>

solution/1800-1899/1873.Calculate Special Bonus/README.md

@@ -22,17 +22,17 @@ employee_id 是这个表的主键。
 
 <p>&nbsp;</p>
 
-<p>写出一个SQL 查询语句，计算每个雇员的奖金。如果一个雇员的id是奇数并且他的名字不是以'M'开头，那么他的奖金是他工资的100%，否则奖金为0。</p>
+<p>计算每个雇员的奖金。如果一个雇员的id是奇数并且他的名字不是以'M'开头，那么他的奖金是他工资的100%，否则奖金为0。</p>
 
-<p>Return the result table ordered by <code>employee_id</code>.</p>
+<p>&nbsp;</p>
 
 <p>返回的结果集请按照<code>employee_id</code>排序。</p>
 
-<p>查询结果格式如下面的例子所示。</p>
+<p>返回结果格式如下面的例子所示。</p>
 
 <p>&nbsp;</p>
 
-<p><strong>示例 1:</strong></p>
+<p><strong class="example">示例 1:</strong></p>
 
 <pre>
 <strong>输入：</strong>

solution/2500-2599/2582.Pass the Pillow/README.md

@@ -204,7 +204,7 @@ impl Solution {
     pub fn pass_the_pillow(n: i32, time: i32) -> i32 {
         let mut ans = 1;
         let mut k = 1;
-        
+
         for i in 1..=time {
             ans += k;
 

solution/2500-2599/2582.Pass the Pillow/README_EN.md

@@ -191,7 +191,7 @@ impl Solution {
     pub fn pass_the_pillow(n: i32, time: i32) -> i32 {
         let mut ans = 1;
         let mut k = 1;
-        
+
         for i in 1..=time {
             ans += k;
 

solution/2500-2599/2586.Count the Number of Vowel Strings in Range/README.md

@@ -155,7 +155,7 @@ function vowelStrings(words: string[], left: number, right: number): number {
 ```rust
 impl Solution {
     pub fn vowel_strings(words: Vec<String>, left: i32, right: i32) -> i32 {
-        let check = |c: u8| -> bool { 
+        let check = |c: u8| -> bool {
             c == b'a' || c == b'e' || c == b'i' || c == b'o' || c == b'u'
         };
 

solution/2500-2599/2586.Count the Number of Vowel Strings in Range/README_EN.md

@@ -145,7 +145,7 @@ function vowelStrings(words: string[], left: number, right: number): number {
 ```rust
 impl Solution {
     pub fn vowel_strings(words: Vec<String>, left: i32, right: i32) -> i32 {
-        let check = |c: u8| -> bool { 
+        let check = |c: u8| -> bool {
             c == b'a' || c == b'e' || c == b'i' || c == b'o' || c == b'u'
         };
 

solution/2700-2799/2715.Execute Cancellable Function With Delay/README.md

@@ -15,10 +15,11 @@
 <p><strong class="example">示例 1:</strong></p>
 
 <pre>
-<b>输入：</b>fn = (x) =&gt; x * 5, args = [2], t = 20, cancelTime = 50
+<b>输入：</b>fn = (x) =&gt; x * 5, args = [2], t = 20
 <b>输出：</b>[{"time": 20, "returned": 10}]
 <b>解释：</b>
-const cancel = cancellable(fn, [2], 20); // // 在 t=20ms 时调用 fn(2)
+const cancelTime = 50
+const cancel = cancellable((x) =&gt; x * 5, [2], 20); // 在 t=20ms 时调用 fn(2)
 setTimeout(cancel, 50);
 
 cancelTime（50ms）在延迟时间（20ms）之后，所以 fn(2) 应该在 t=20ms 时调用。fn 的返回值是 10。
@@ -27,17 +28,27 @@ cancelTime（50ms）在延迟时间（20ms）之后，所以 fn(2) 应该在 t=2
 <p><strong class="example">示例 2：</strong></p>
 
 <pre>
-<b>输入：</b>fn = (x) =&gt; x**2, args = [2], t = 100, cancelTime = 50
+<b>输入：</b>fn = (x) =&gt; x**2, args = [2], t = 100
 <b>输出：</b>[]
-<b>解释：</b>fn(2) 从未被调用，因为 cancelTime（50ms）在延迟时间（100ms）之前。
+<b>解释：</b>
+const cancelTime = 50 
+const cancel = cancellable((x) =&gt; x**2, [2], 100); // fn(2) 没被调用
+setTimeout(cancel, cancelTime);
+
+fn(2) 从未被调用，因为 cancelTime（50ms）在延迟时间（100ms）之前。
 </pre>
 
 <p><strong class="example">示例 3：</strong></p>
 
 <pre>
 <b>输入：</b>fn = (x1, x2) =&gt; x1 * x2, args = [2,4], t = 30, cancelTime = 100
 <b>输出：</b>[{"time": 30, "returned": 8}]
-<b>解释：</b>fn(2) 从未被调用，因为 cancelTime（50ms）在延迟时间（100ms）之前。
+<b>解释：</b>
+const cancelTime = 100
+const cancel = cancellable((x1, x2) =&gt; x1 * x2, [2,4], 30); // fn(2,4) 在 t=30ms 时被调用
+setTimeout(cancel, cancelTime);
+
+fn(2) 从未被调用，因为 cancelTime（50ms）在延迟时间（100ms）之前。
 </pre>
 
 <p>&nbsp;</p>

solution/2700-2799/2715.Execute Cancellable Function With Delay/README_EN.md

@@ -12,29 +12,40 @@
 <p><strong class="example">Example 1:</strong></p>
 
 <pre>
-<strong>Input:</strong> fn = (x) =&gt; x * 5, args = [2], t = 20, cancelTime = 50
+<strong>Input:</strong> fn = (x) =&gt; x * 5, args = [2], t = 20
 <strong>Output:</strong> [{&quot;time&quot;: 20, &quot;returned&quot;: 10}]
 <strong>Explanation:</strong> 
-const cancel = cancellable(fn, [2], 20); // fn(2) called at t=20ms
-setTimeout(cancel, 50);
+const cancelTime = 50
+const cancel = cancellable((x) =&gt; x * 5, [2], 20); // fn(2) called at t=20ms
+setTimeout(cancel, cancelTime);
 
-the cancelTime (50ms) is after the delay time (20ms), so fn(2) should be called at t=20ms. The value returned from fn is 10.
+The cancellation was scheduled to occur after a delay of cancelTime (50ms), which happened after the execution of fn(2) at 20ms.
 </pre>
 
 <p><strong class="example">Example 2:</strong></p>
 
 <pre>
-<strong>Input:</strong> fn = (x) =&gt; x**2, args = [2], t = 100, cancelTime = 50
+<strong>Input:</strong> fn = (x) =&gt; x**2, args = [2], t = 100
 <strong>Output:</strong> []
-<strong>Explanation:</strong> fn(2) was never called because cancelTime (50ms) is before the delay time (100ms).
+<strong>Explanation:</strong> 
+const cancelTime = 50 
+const cancel = cancellable((x) =&gt; x**2, [2], 100); // fn(2) not called
+setTimeout(cancel, cancelTime);
+
+The cancellation was scheduled to occur after a delay of cancelTime (50ms), which happened before the execution of fn(2) at 100ms, resulting in fn(2) never being called.
 </pre>
 
 <p><strong class="example">Example 3:</strong></p>
 
 <pre>
-<strong>Input:</strong> fn = (x1, x2) =&gt; x1 * x2, args = [2,4], t = 30, cancelTime = 100
+<strong>Input:</strong> fn = (x1, x2) =&gt; x1 * x2, args = [2,4], t = 30
 <strong>Output:</strong> [{&quot;time&quot;: 30, &quot;returned&quot;: 8}]
-<strong>Explanation:</strong> fn(2, 4) was called at t=30ms because cancelTime &gt; t.
+<strong>Explanation:</strong>
+const cancelTime = 100
+const cancel = cancellable((x1, x2) =&gt; x1 * x2, [2,4], 30); // fn(2,4) called at t=30ms
+setTimeout(cancel, cancelTime);
+
+The cancellation was scheduled to occur after a delay of cancelTime (100ms), which happened after the execution of fn(2,4) at 30ms.
 </pre>
 
 <p>&nbsp;</p>

solution/2700-2799/2721.Execute Asynchronous Functions in Parallel/README.md

@@ -65,7 +65,7 @@ promiseAll(functions).then(console.log); // [5]
 <p><strong>提示：</strong></p>
 
 <ul>
-	<li><code>functions&nbsp;is an array of functions that returns promises</code></li>
+	<li><code>函数 functions 是一个返回 Promise 的函数数组</code></li>
 	<li><code>1 &lt;= functions.length &lt;= 10</code></li>
 </ul>
 

solution/2700-2799/2721.Execute Asynchronous Functions in Parallel/README_EN.md

@@ -9,7 +9,7 @@
 <p><code>promise</code> resolves:</p>
 
 <ul>
-	<li>When all the promises returned from&nbsp;<code>functions</code>&nbsp;were resolved successfully.&nbsp;The resolved&nbsp;value of&nbsp;<code>promise</code> should be an array of all the resolved values of promises in the same order as they were in the&nbsp;<code>functions.</code></li>
+	<li>When all the promises returned from&nbsp;<code>functions</code>&nbsp;were resolved successfully.&nbsp;The resolved&nbsp;value of&nbsp;<code>promise</code> should be an array of all the resolved values of promises in the same order as they were in the&nbsp;<code>functions</code>.</li>
 </ul>
 
 <p><code>promise</code> rejects:</p>

solution/2700-2799/2725.Interval Cancellation/README.md

@@ -23,15 +23,19 @@
    {"time": 80, "returned": 8},
    {"time": 100, "returned": 8}
 ]
-<strong>解释：</strong> 每隔 20ms，调用 fn(4)。直到 t=110ms，然后取消。
+<strong>解释：</strong> 
+const cancelT = 110
 const cancel = cancellable(x =&gt; x * 2, [4], 20);
-setTimeout(cancel, 110);
+setTimeout(cancel, cancelT);
+每隔 20ms，调用 fn(4)。
 第一次调用 fn 是在 0ms。fn(4) 返回 8。
 第二次调用 fn 是在 20ms。fn(4) 返回 8。
 第三次调用 fn 是在 40ms。fn(4) 返回 8。
 第四次调用 fn 是在&nbsp;60ms。fn(4) 返回 8。
 第五次调用 fn 是在 80ms。fn(4) 返回 8。
-第六次调用 fn 是在 100ms。fn(4) 返回 8。</pre>
+第六次调用 fn 是在 100ms。fn(4) 返回 8。
+在 t=110ms 时取消。
+</pre>
 
 <p><strong class="example">示例 2：</strong></p>
 
@@ -46,7 +50,12 @@ setTimeout(cancel, 110);
    {"time": 100, "returned": 10},
    {"time": 125, "returned": 10}
 ]
-<strong>解释：</strong>每隔 25ms，调用 fn(2, 5)。直到 t=140ms，然后取消。
+<strong>解释：
+</strong>const cancelT = 140
+const cancel = cancellable((x1, x2) =&gt; (x1 * x2), [2, 5], 25); 
+setTimeout(cancel, cancelT);
+
+每隔 25ms，调用 fn(2, 5)。直到 t=140ms，然后取消。
 第一次调用 fn 是在 0ms
 第二次调用 fn 是在 25ms
 第三次调用 fn 是在 50ms
@@ -67,7 +76,12 @@ setTimeout(cancel, 110);
    {"time": 100, "returned": 9},
    {"time": 150, "returned": 9}
 ]
-<b>解释：</b>每隔 50ms，调用 fn(5, 1, 3)。直到 t=180ms，然后取消。
+<b>解释：</b>
+const cancelT = 180
+const cancel = cancellable((x1, x2, x3) =&gt; (x1 + x2 + x3), [5, 1, 3], 50);
+setTimeout(cancel, cancelT);
+
+每隔 50ms，调用 fn(5, 1, 3)。直到 t=180ms，然后取消。
 第一次调用 fn 是在 0ms
 第二次调用 fn 是在 50ms
 第三次调用 fn 是在 100ms