feat: update solutions to lc problems (doocs#3451)

Author: yanglbme

solution/1300-1399/1326.Minimum Number of Taps to Open to Water a Garden/README.md

@@ -75,15 +75,15 @@ tags:
 
 我们注意到，对于所有能覆盖某个左端点的水龙头，选择能覆盖最远右端点的那个水龙头是最优的。
 
-因此，我们可以先预处理数组 $ranges$，对于第 $i$ 个水龙头，它能覆盖的左端点 $l = max(0, i - ranges[i])$，右端点 $r = i + ranges[i]$，我们算出所有能覆盖左端点 $l$ 的水龙头中，右端点最大的那个位置，记录在数组 $last[i]$ 中。
+因此，我们可以先预处理数组 $ranges$，对于第 $i$ 个水龙头，它能覆盖的左端点 $l = \max(0, i - ranges[i])$，右端点 $r = i + ranges[i]$，我们算出所有能覆盖左端点 $l$ 的水龙头中，右端点最大的那个位置，记录在数组 $last[i]$ 中。
 
 然后我们定义以下三个变量，其中：
 
 -   变量 $ans$ 表示最终答案，即最少水龙头数目；
 -   变量 $mx$ 表示当前能覆盖的最远右端点；
 -   变量 $pre$ 表示上一个水龙头覆盖的最远右端点。
 
-我们在 $[0,...n-1]$ 的范围内遍历所有位置，对于当前位置 $i$，我们用 $last[i]$ 更新 $mx$，即 $mx = max(mx, last[i])$。
+我们在 $[0,...n-1]$ 的范围内遍历所有位置，对于当前位置 $i$，我们用 $last[i]$ 更新 $mx$，即 $mx = \max(mx, last[i])$。
 
 -   如果 $mx \leq i$，说明无法覆盖下一个位置，返回 $-1$。
 -   如果 $pre = i$，说明需要使用一个新的子区间，因此我们将 $ans$ 加 $1$，并且更新 $pre = mx$。

solution/1300-1399/1326.Minimum Number of Taps to Open to Water a Garden/README_EN.md

@@ -66,7 +66,32 @@ Opening Only the second tap will water the whole garden [0,5]
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Greedy
+
+We note that for all taps that can cover a certain left endpoint, choosing the tap that can cover the farthest right endpoint is optimal.
+
+Therefore, we can preprocess the array $ranges$. For the $i$-th tap, it can cover the left endpoint $l = \max(0, i - ranges[i])$ and the right endpoint $r = i + ranges[i]$. We calculate the position of the tap that can cover the left endpoint $l$ with the farthest right endpoint and record it in the array $last[i]$.
+
+Then we define the following three variables:
+
+-   Variable $ans$ represents the final answer, i.e., the minimum number of taps;
+-   Variable $mx$ represents the farthest right endpoint that can currently be covered;
+-   Variable $pre$ represents the farthest right endpoint covered by the previous tap.
+
+We traverse all positions in the range $[0, \ldots, n-1]$. For the current position $i$, we use $last[i]$ to update $mx$, i.e., $mx = \max(mx, last[i])$.
+
+-   If $mx \leq i$, it means the next position cannot be covered, so we return $-1$.
+-   If $pre = i$, it means a new subinterval needs to be used, so we increment $ans$ by $1$ and update $pre = mx$.
+
+After the traversal, we return $ans$.
+
+The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the garden.
+
+Similar problems:
+
+-   [45. Jump Game II](https://github.com/doocs/leetcode/blob/main/solution/0000-0099/0045.Jump%20Game%20II/README.md)
+-   [55. Jump Game](https://github.com/doocs/leetcode/blob/main/solution/0000-0099/0055.Jump%20Game/README.md)
+-   [1024. Video Stitching](https://github.com/doocs/leetcode/blob/main/solution/1000-1099/1024.Video%20Stitching/README.md)
 
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solution/1300-1399/1328.Break a Palindrome/README_EN.md

@@ -57,7 +57,13 @@ Of all the ways, "aaccba" is the lexicographically smallest.
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Greedy
+
+First, we check if the length of the string is $1$. If it is, we directly return an empty string.
+
+Otherwise, we traverse the first half of the string from left to right, find the first character that is not `'a'`, and change it to `'a'`. If no such character exists, we change the last character to `'b'`.
+
+The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the string.
 
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solution/1300-1399/1330.Reverse Subarray To Maximize Array Value/README_EN.md

@@ -57,7 +57,52 @@ tags:
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Classification Discussion + Enumeration
+
+According to the problem description, we need to find the maximum value of the array $\sum_{i=0}^{n-2} |a_i - a_{i+1}|$ when reversing a subarray once.
+
+Next, we discuss the following cases:
+
+1. Do not reverse the subarray.
+2. Reverse the subarray, and the subarray "includes" the first element.
+3. Reverse the subarray, and the subarray "includes" the last element.
+4. Reverse the subarray, and the subarray "does not include" the first and last elements.
+
+Let $s$ be the array value when the subarray is not reversed, then $s = \sum_{i=0}^{n-2} |a_i - a_{i+1}|$. We can initialize the answer $ans$ to $s$.
+
+If we reverse the subarray and the subarray includes the first element, we can enumerate the last element $a_i$ of the reversed subarray, where $0 \leq i < n-1$. In this case, $ans = \max(ans, s + |a_0 - a_{i+1}| - |a_i - a_{i+1}|)$.
+
+<p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1300-1399/1330.Reverse%20Subarray%20To%20Maximize%20Array%20Value/images/1-drawio.png" /></p>
+
+Similarly, if we reverse the subarray and the subarray includes the last element, we can enumerate the first element $a_{i+1}$ of the reversed subarray, where $0 \leq i < n-1$. In this case, $ans = \max(ans, s + |a_{n-1} - a_i| - |a_i - a_{i+1}|)$.
+
+<p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1300-1399/1330.Reverse%20Subarray%20To%20Maximize%20Array%20Value/images/2-drawio.png" /></p>
+
+If we reverse the subarray and the subarray does not include the first and last elements, we consider any two adjacent elements in the array as a point pair $(x, y)$. Let the first element of the reversed subarray be $y_1$, and its left adjacent element be $x_1$; let the last element of the reversed subarray be $x_2$, and its right adjacent element be $y_2$.
+
+<p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1300-1399/1330.Reverse%20Subarray%20To%20Maximize%20Array%20Value/images/3-drawio.png" /></p>
+
+At this time, compared to not reversing the subarray, the change in the array value is $|x_1 - x_2| + |y_1 - y_2| - |x_1 - y_1| - |x_2 - y_2|$, where the first two terms can be expressed as:
+
+$$
+\left | x_1 - x_2 \right |  + \left | y_1 - y_2 \right | = \max \begin{cases} (x_1 + y_1) - (x_2 + y_2) \\ (x_1 - y_1) - (x_2 - y_2) \\ (-x_1 + y_1) - (-x_2 + y_2) \\ (-x_1 - y_1) - (-x_2 - y_2) \end{cases}
+$$
+
+Then the change in the array value is:
+
+$$
+\left | x_1 - x_2 \right |  + \left | y_1 - y_2 \right | - \left | x_1 - y_1 \right | - \left | x_2 - y_2 \right |  = \max \begin{cases} (x_1 + y_1) - \left |x_1 - y_1 \right | - \left ( (x_2 + y_2) + \left |x_2 - y_2 \right | \right ) \\ (x_1 - y_1) - \left |x_1 - y_1 \right | - \left ( (x_2 - y_2) + \left |x_2 - y_2 \right | \right ) \\ (-x_1 + y_1) - \left |x_1 - y_1 \right | - \left ( (-x_2 + y_2) + \left |x_2 - y_2 \right | \right ) \\ (-x_1 - y_1) - \left |x_1 - y_1 \right | - \left ( (-x_2 - y_2) + \left |x_2 - y_2 \right | \right ) \end{cases}
+$$
+
+Therefore, we only need to find the maximum value $mx$ of $k_1 \times x + k_2 \times y$, where $k_1, k_2 \in \{-1, 1\}$, and the corresponding minimum value $mi$ of $|x - y|$. Then the maximum change in the array value is $mx - mi$. The answer is $ans = \max(ans, s + \max(0, mx - mi))$.
+
+In the code implementation, we define an array of length 5, $dirs=[1, -1, -1, 1, 1]$. Each time we take two adjacent elements of the array as the values of $k_1$ and $k_2$, which can cover all cases of $k_1, k_2 \in \{-1, 1\}$.
+
+The time complexity is $O(n)$, where $n$ is the length of the array $nums$. The space complexity is $O(1)$.
+
+Similar problems:
+
+-   [1131. Maximum of Absolute Value Expression](https://github.com/doocs/leetcode/blob/main/solution/1100-1199/1131.Maximum%20of%20Absolute%20Value%20Expression/README_EN.md)
 
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solution/1300-1399/1331.Rank Transform of an Array/README_EN.md

@@ -67,7 +67,13 @@ tags:
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Discretization
+
+First, we copy an array $t$, then sort and deduplicate it to obtain an array of length $m$ that is strictly monotonically increasing.
+
+Next, we traverse the original array $arr$. For each element $x$ in the array, we use binary search to find the position of $x$ in $t$. The position plus one is the rank of $x$.
+
+The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $arr$.
 
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solution/1400-1499/1409.Queries on a Permutation With Key/README.md

@@ -35,13 +35,13 @@ tags:
 
 <pre>
 <strong>输入：</strong>queries = [3,1,2,1], m = 5
-<strong>输出：</strong>[2,1,2,1] 
+<strong>输出：</strong>[2,1,2,1]
 <strong>解释：处理</strong> queries 的过程如下：
 对于 i=0: queries[i]=3, P=[1,2,3,4,5], 3 在 P 中的位置是 <strong>2</strong>，然后我们把 3 移动到 P 的开头，得到 P=[3,1,2,4,5] 。
-对于 i=1: queries[i]=1, P=[3,1,2,4,5], 1 在 P 中的位置是 <strong>1</strong>，然后我们把 1 移动到 P 的开头，得到 P=[1,3,2,4,5] 。 
+对于 i=1: queries[i]=1, P=[3,1,2,4,5], 1 在 P 中的位置是 <strong>1</strong>，然后我们把 1 移动到 P 的开头，得到 P=[1,3,2,4,5] 。
 对于 i=2: queries[i]=2, P=[1,3,2,4,5], 2 在 P 中的位置是 <strong>2</strong>，然后我们把 2 移动到 P 的开头，得到 P=[2,1,3,4,5] 。
-对于 i=3: queries[i]=1, P=[2,1,3,4,5], 1 在 P 中的位置是 <strong>1</strong>，然后我们把 1 移动到 P 的开头，得到 P=[1,2,3,4,5] 。 
-因此，包含结果的数组为 [2,1,2,1] 。  
+对于 i=3: queries[i]=1, P=[2,1,3,4,5], 1 在 P 中的位置是 <strong>1</strong>，然后我们把 1 移动到 P 的开头，得到 P=[1,2,3,4,5] 。
+因此，包含结果的数组为 [2,1,2,1] 。
 </pre>
 
 <p><strong>示例 2：</strong></p>
@@ -78,21 +78,6 @@ tags:
 
 题目数据规模不大，可以直接模拟。
 
-**方法一：树状数组**
-
-树状数组，也称作“二叉索引树”（Binary Indexed Tree）或 Fenwick 树。 它可以高效地实现如下两个操作：
-
-1. **单点更新** `update(x, delta)`： 把序列 x 位置的数加上一个值 delta；
-1. **前缀和查询** `query(x)`：查询序列 `[1,...x]` 区间的区间和，即位置 x 的前缀和。
-
-这两个操作的时间复杂度均为 $O(\log n)$。
-
-树状数组最基本的功能就是求比某点 x 小的点的个数（这里的比较是抽象的概念，可以是数的大小、坐标的大小、质量的大小等等）。
-
-比如给定数组 `a[5] = {2, 5, 3, 4, 1}`，求 `b[i] = 位置 i 左边小于等于 a[i] 的数的个数`。对于此例，`b[5] = {0, 1, 1, 2, 0}`。
-
-解决方案是直接遍历数组，每个位置先求出 `query(a[i])`，然后再修改树状数组 `update(a[i], 1)` 即可。当数的范围比较大时，需要进行离散化，即先进行去重并排序，然后对每个数字进行编号。
-
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 #### Python3
@@ -189,7 +174,20 @@ func processQueries(queries []int, m int) []int {
 
 <!-- solution:start -->
 
-### 方法二
+### 方法二：树状数组
+
+树状数组，也称作“二叉索引树”（Binary Indexed Tree）或 Fenwick 树。 它可以高效地实现如下两个操作：
+
+1. **单点更新** `update(x, delta)`： 把序列 x 位置的数加上一个值 delta；
+1. **前缀和查询** `query(x)`：查询序列 `[1,...x]` 区间的区间和，即位置 x 的前缀和。
+
+这两个操作的时间复杂度均为 $O(\log n)$。
+
+树状数组最基本的功能就是求比某点 x 小的点的个数（这里的比较是抽象的概念，可以是数的大小、坐标的大小、质量的大小等等）。
+
+比如给定数组 `a[5] = {2, 5, 3, 4, 1}`，求 `b[i] = 位置 i 左边小于等于 a[i] 的数的个数`。对于此例，`b[5] = {0, 1, 1, 2, 0}`。
+
+解决方案是直接遍历数组，每个位置先求出 `query(a[i])`，然后再修改树状数组 `update(a[i], 1)` 即可。当数的范围比较大时，需要进行离散化，即先进行去重并排序，然后对每个数字进行编号。
 
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solution/1400-1499/1409.Queries on a Permutation With Key/README_EN.md

@@ -34,13 +34,13 @@ tags:
 
 <pre>
 <strong>Input:</strong> queries = [3,1,2,1], m = 5
-<strong>Output:</strong> [2,1,2,1] 
-<strong>Explanation:</strong> The queries are processed as follow: 
-For i=0: queries[i]=3, P=[1,2,3,4,5], position of 3 in P is <strong>2</strong>, then we move 3 to the beginning of P resulting in P=[3,1,2,4,5]. 
-For i=1: queries[i]=1, P=[3,1,2,4,5], position of 1 in P is <strong>1</strong>, then we move 1 to the beginning of P resulting in P=[1,3,2,4,5]. 
-For i=2: queries[i]=2, P=[1,3,2,4,5], position of 2 in P is <strong>2</strong>, then we move 2 to the beginning of P resulting in P=[2,1,3,4,5]. 
-For i=3: queries[i]=1, P=[2,1,3,4,5], position of 1 in P is <strong>1</strong>, then we move 1 to the beginning of P resulting in P=[1,2,3,4,5]. 
-Therefore, the array containing the result is [2,1,2,1].  
+<strong>Output:</strong> [2,1,2,1]
+<strong>Explanation:</strong> The queries are processed as follow:
+For i=0: queries[i]=3, P=[1,2,3,4,5], position of 3 in P is <strong>2</strong>, then we move 3 to the beginning of P resulting in P=[3,1,2,4,5].
+For i=1: queries[i]=1, P=[3,1,2,4,5], position of 1 in P is <strong>1</strong>, then we move 1 to the beginning of P resulting in P=[1,3,2,4,5].
+For i=2: queries[i]=2, P=[1,3,2,4,5], position of 2 in P is <strong>2</strong>, then we move 2 to the beginning of P resulting in P=[2,1,3,4,5].
+For i=3: queries[i]=1, P=[2,1,3,4,5], position of 1 in P is <strong>1</strong>, then we move 1 to the beginning of P resulting in P=[1,2,3,4,5].
+Therefore, the array containing the result is [2,1,2,1].
 </pre>
 
 <p><strong class="example">Example 2:</strong></p>
@@ -72,7 +72,9 @@ Therefore, the array containing the result is [2,1,2,1].
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Simulation
+
+The problem's data scale is not large, so we can directly simulate it.
 
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@@ -170,7 +172,20 @@ func processQueries(queries []int, m int) []int {
 
 <!-- solution:start -->
 
-### Solution 2
+### Solution 2: Binary Indexed Tree
+
+The Binary Indexed Tree (BIT), also known as the Fenwick Tree, efficiently supports the following two operations:
+
+1. **Point Update** `update(x, delta)`: Adds a value `delta` to the element at position `x` in the sequence.
+2. **Prefix Sum Query** `query(x)`: Queries the sum of the sequence over the interval `[1,...,x]`, i.e., the prefix sum at position `x`.
+
+Both operations have a time complexity of $O(\log n)$.
+
+The fundamental functionality of the Binary Indexed Tree is to count the number of elements smaller than a given element `x`. This comparison is abstract and can refer to size, coordinate, mass, etc.
+
+For example, given the array `a[5] = {2, 5, 3, 4, 1}`, the task is to compute `b[i] = the number of elements to the left of position i that are less than or equal to a[i]`. For this example, `b[5] = {0, 1, 1, 2, 0}`.
+
+The solution is to traverse the array, first calculating `query(a[i])` for each position, and then updating the Binary Indexed Tree with `update(a[i], 1)`. When the range of numbers is large, discretization is necessary, which involves removing duplicates, sorting, and then assigning an index to each number.
 
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