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Commit 576142f

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feat: add solutions to lc problem: No.0836 (doocs#3940)
No.0836.Rectangle Overlap
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‎solution/0800-0899/0831.Masking Personal Information/README_EN.md‎

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<!-- solution:start -->
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### Solution 1
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### Solution 1: Simulation
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According to the problem description, we can first determine whether the string $s$ is an email or a phone number, and then handle it accordingly.
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The time complexity is $O(n),ドル and the space complexity is $O(n),ドル where $n$ is the length of the string $s$.
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‎solution/0800-0899/0832.Flipping an Image/README.md‎

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### 方法一:双指针
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我们可以遍历矩阵,对于遍历到的每一行 $row$:
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我们可以遍历矩阵,对于遍历到的每一行 $\textit{row},ドル我们使用双指针 $i$ 和 $j$ 分别指向该行的首尾元素,如果 $\textit{row}[i] = \textit{row}[j],ドル交换后两者的值仍然保持不变,因此,我们只需要对 $\textit{row}[i]$ 和 $\textit{row}[j]$ 进行异或反转即可,然后将 $i$ 和 $j$ 分别向中间移动一位,直到 $i \geq j$。如果 $\textit{row}[i] \neq \textit{row}[j],ドル此时交换后再反转两者的值,仍然保持不变,因此,可以不进行任何操作。
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我们使用双指针 $i$ 和 $j$ 分别指向该行的首尾元素,如果 $row[i] = row[j],ドル交换后两者的值仍然保持不变,因此,我们只需要对 $row[i]$ 和 $row[j]$ 进行异或反转即可,然后将 $i$ 和 $j$ 分别向中间移动一位,直到 $i \geq j$。如果 $row[i] \neq row[j],ドル此时交换后再反转两者的值,仍然保持不变,因此,可以不进行任何操作。
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最后,如果 $i = j,ドル我们直接对 $row[i]$ 进行反转即可。
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最后,如果 $i = j,ドル我们直接对 $\textit{row}[i]$ 进行反转即可。
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时间复杂度 $O(n^2),ドル其中 $n$ 是矩阵的行数或列数。空间复杂度 $O(1)$。
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‎solution/0800-0899/0832.Flipping an Image/README_EN.md‎

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<!-- solution:start -->
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### Solution 1
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### Solution 1: Two Pointers
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We can traverse the matrix, and for each row $\textit{row},ドル we use two pointers $i$ and $j$ pointing to the first and last elements of the row, respectively. If $\textit{row}[i] = \textit{row}[j],ドル swapping them will keep their values unchanged, so we only need to XOR invert $\textit{row}[i]$ and $\textit{row}[j],ドル then move $i$ and $j$ one position towards the center until $i \geq j$. If $\textit{row}[i] \neq \textit{row}[j],ドル swapping and then inverting their values will also keep them unchanged, so no operation is needed.
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Finally, if $i = j,ドル we directly invert $\textit{row}[i]$.
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The time complexity is $O(n^2),ドル where $n$ is the number of rows or columns in the matrix. The space complexity is $O(1)$.
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‎solution/0800-0899/0833.Find And Replace in String/README.md‎

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### 方法一:模拟
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我们遍历每个替换操作,对于当前第 $k$ 个替换操作 $(i, src),ドル如果 $s[i..i+|src|-1]$ 与 $src$ 相等,此时我们记录下标 $i$ 处需要替换的是 $targets$ 的第 $k$ 个字符串,否则不需要替换。
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我们遍历每个替换操作,对于当前第 $k$ 个替换操作 $(i, \text{src}),ドル如果 $s[i..i+|\text{src}|-1]$ 与 $\text{src}$ 相等,此时我们记录下标 $i$ 处需要替换的是 $\text{targets}$ 的第 $k$ 个字符串,否则不需要替换。
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接下来,我们只需要遍历原字符串 $s,ドル根据记录的信息进行替换即可。
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‎solution/0800-0899/0833.Find And Replace in String/README_EN.md‎

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<!-- solution:start -->
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### Solution 1
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### Solution 1: Simulation
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We iterate through each replacement operation. For the current $k$-th replacement operation $(i, \text{src}),ドル if $s[i..i+|\text{src}|-1]$ is equal to $\text{src},ドル we record that the string at index $i$ needs to be replaced with the $k$-th string in $\text{targets}$; otherwise, no replacement is needed.
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Next, we only need to iterate through the original string $s$ and perform the replacements based on the recorded information.
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The time complexity is $O(L),ドル and the space complexity is $O(n),ドル where $L$ is the sum of the lengths of all strings, and $n$ is the length of the string $s$.
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‎solution/0800-0899/0834.Sum of Distances in Tree/README_EN.md‎

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<!-- solution:start -->
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### Solution 1
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### Solution 1: Tree DP (Re-rooting)
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First, we run a DFS to calculate the size of each node's subtree, recorded in the array $size,ドル and compute the sum of distances from node 0ドル$ to all other nodes, recorded in $ans[0]$.
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Next, we run another DFS to enumerate the sum of distances from each node when it is considered as the root. Suppose the answer for the current node $i$ is $t$. When we move from node $i$ to node $j,ドル the sum of distances changes to $t - size[j] + n - size[j],ドル meaning the sum of distances to node $j$ and its subtree nodes decreases by $size[j],ドル while the sum of distances to other nodes increases by $n - size[j]$.
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The time complexity is $O(n),ドル and the space complexity is $O(n),ドル where $n$ is the number of nodes in the tree.
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Similar problems:
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- [2581. Count Number of Possible Root Nodes](https://github.com/doocs/leetcode/blob/main/solution/2500-2599/2581.Count%20Number%20of%20Possible%20Root%20Nodes/README_EN.md)
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‎solution/0800-0899/0835.Image Overlap/README.md‎

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### 方法一:枚举
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我们可以枚举 $img1$ 和 $img2$ 的每个 1ドル$ 的位置,分别记为 $(i, j)$ 和 $(h, k)$。然后我们计算得到偏移量 $(i - h, j - k),ドル记为 $(dx, dy),ドル用哈希表 $cnt$ 记录每个偏移量出现的次数。最后我们遍历哈希表 $cnt,ドル找到出现次数最多的偏移量,即为答案。
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我们可以枚举 $\textit{img1}$ 和 $\textit{img2}$ 的每个 1ドル$ 的位置,分别记为 $(i, j)$ 和 $(h, k)$。然后我们计算得到偏移量 $(i - h, j - k),ドル记为 $(dx, dy),ドル用哈希表 $\textit{cnt}$ 记录每个偏移量出现的次数。最后我们遍历哈希表 $\textit{cnt},ドル找到出现次数最多的偏移量,即为答案。
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时间复杂度 $O(n^4),ドル空间复杂度 $O(n^2)$。其中 $n$ 是 $img1$ 的边长。
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时间复杂度 $O(n^4),ドル空间复杂度 $O(n^2)$。其中 $n$ 是 $\textit{img1}$ 的边长。
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‎solution/0800-0899/0835.Image Overlap/README_EN.md‎

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<!-- solution:start -->
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### Solution 1
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### Solution 1: Enumeration
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We can enumerate each position of 1ドル$ in $\textit{img1}$ and $\textit{img2},ドル denoted as $(i, j)$ and $(h, k)$ respectively. Then we calculate the offset $(i - h, j - k),ドル denoted as $(dx, dy),ドル and use a hash table $\textit{cnt}$ to record the number of occurrences of each offset. Finally, we traverse the hash table $\textit{cnt}$ to find the offset that appears the most, which is the answer.
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The time complexity is $O(n^4),ドル and the space complexity is $O(n^2),ドル where $n$ is the side length of $\textit{img1}$.
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‎solution/0800-0899/0836.Rectangle Overlap/README.md‎

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### 方法一:判断不重叠的情况
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我们记矩形 $rec1$ 的坐标点为 $(x_1, y_1, x_2, y_2),ドル矩形 $rec2$ 的坐标点为 $(x_3, y_3, x_4, y_4)$。
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我们记矩形 $\text{rec1}$ 的坐标点为 $(x_1, y_1, x_2, y_2),ドル矩形 $\text{rec2}$ 的坐标点为 $(x_3, y_3, x_4, y_4)$。
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那么当满足以下任一条件时,矩形 $rec1$ 和 $rec2$ 不重叠:
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那么当满足以下任一条件时,矩形 $\text{rec1}$ 和 $\text{rec2}$ 不重叠:
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- 满足 $y_3 \geq y_2,ドル即 $rec2$ 在 $rec1$ 的上方;
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- 满足 $y_4 \leq y_1,ドル即 $rec2$ 在 $rec1$ 的下方;
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- 满足 $x_3 \geq x_2,ドル即 $rec2$ 在 $rec1$ 的右方;
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- 满足 $x_4 \leq x_1,ドル即 $rec2$ 在 $rec1$ 的左方。
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- 满足 $y_3 \geq y_2,ドル即 $\text{rec2}$ 在 $\text{rec1}$ 的上方;
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- 满足 $y_4 \leq y_1,ドル即 $\text{rec2}$ 在 $\text{rec1}$ 的下方;
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- 满足 $x_3 \geq x_2,ドル即 $\text{rec2}$ 在 $\text{rec1}$ 的右方;
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- 满足 $x_4 \leq x_1,ドル即 $\text{rec2}$ 在 $\text{rec1}$ 的左方。
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当以上条件都不满足时,矩形 $rec1$ 和 $rec2$ 重叠。
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当以上条件都不满足时,矩形 $\text{rec1}$ 和 $\text{rec2}$ 重叠。
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时间复杂度 $O(1),ドル空间复杂度 $O(1)$。
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#### TypeScript
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```ts
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function isRectangleOverlap(rec1: number[], rec2: number[]): boolean {
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const [x1, y1, x2, y2] = rec1;
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const [x3, y3, x4, y4] = rec2;
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return !(y3 >= y2 || y4 <= y1 || x3 >= x2 || x4 <= x1);
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}
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```
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‎solution/0800-0899/0836.Rectangle Overlap/README_EN.md‎

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<!-- solution:start -->
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### Solution 1
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### Solution 1: Determine Non-Overlap Cases
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Let the coordinates of rectangle $\text{rec1}$ be $(x_1, y_1, x_2, y_2),ドル and the coordinates of rectangle $\text{rec2}$ be $(x_3, y_3, x_4, y_4)$.
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The rectangles $\text{rec1}$ and $\text{rec2}$ do not overlap if any of the following conditions are met:
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- $y_3 \geq y_2$: $\text{rec2}$ is above $\text{rec1}$;
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- $y_4 \leq y_1$: $\text{rec2}$ is below $\text{rec1}$;
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- $x_3 \geq x_2$: $\text{rec2}$ is to the right of $\text{rec1}$;
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- $x_4 \leq x_1$: $\text{rec2}$ is to the left of $\text{rec1}$.
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If none of the above conditions are met, the rectangles $\text{rec1}$ and $\text{rec2}$ overlap.
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The time complexity is $O(1),ドル and the space complexity is $O(1)$.
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#### TypeScript
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```ts
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function isRectangleOverlap(rec1: number[], rec2: number[]): boolean {
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const [x1, y1, x2, y2] = rec1;
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const [x3, y3, x4, y4] = rec2;
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return !(y3 >= y2 || y4 <= y1 || x3 >= x2 || x4 <= x1);
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}
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```
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<!-- solution:end -->

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