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Commit 9399197

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feat: add solutions to lc problems: No.2257,2258 (#1718)
* No.2257.Count Unguarded Cells in the Grid * No.2258.Escape the Spreading Fire
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‎solution/2200-2299/2257.Count Unguarded Cells in the Grid/README.md‎

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**方法一:模拟**
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我们创建一个 $m \times n$ 的二维数组 $g,ドル其中 $g[i][j]$ 表示第 $i$ 行第 $j$ 列的格子。初始时,$g[i][j]$ 的值为 0ドル,ドル表示该格子没有被保卫。
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我们创建一个 $m \times n$ 的二维数组 $g,ドル其中 $g[i][j]$ 表示第 $i$ 行第 $j$ 列的格子。初始时$g[i][j]$ 的值为 0ドル,ドル表示该格子没有被保卫。
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然后遍历所有的警卫和墙,将 $g[i][j]$ 的值置为 2ドル,ドル这些位置不能被访问。
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@@ -204,7 +204,34 @@ func countUnguarded(m int, n int, guards [][]int, walls [][]int) (ans int) {
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### **TypeScript**
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```ts
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function countUnguarded(m: number, n: number, guards: number[][], walls: number[][]): number {
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const g: number[][] = Array.from({ length: m }, () => Array.from({ length: n }, () => 0));
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for (const [i, j] of guards) {
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g[i][j] = 2;
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}
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for (const [i, j] of walls) {
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g[i][j] = 2;
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}
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const dirs: number[] = [-1, 0, 1, 0, -1];
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for (const [i, j] of guards) {
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for (let k = 0; k < 4; ++k) {
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let [x, y] = [i, j];
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let [a, b] = [dirs[k], dirs[k + 1]];
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while (x + a >= 0 && x + a < m && y + b >= 0 && y + b < n && g[x + a][y + b] < 2) {
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x += a;
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y += b;
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g[x][y] = 1;
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}
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}
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}
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let ans = 0;
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for (const row of g) {
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for (const v of row) {
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ans += v === 0 ? 1 : 0;
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}
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}
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return ans;
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}
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```
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### **...**

‎solution/2200-2299/2257.Count Unguarded Cells in the Grid/README_EN.md‎

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## Solutions
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**Solution 1: Simulation**
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We create a two-dimensional array $g$ of size $m \times n,ドル where $g[i][j]$ represents the cell in row $i$ and column $j$. Initially, the value of $g[i][j]$ is 0ドル,ドル indicating that the cell is not guarded.
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Then, we traverse all guards and walls, and set the value of $g[i][j]$ to 2ドル,ドル indicating that these positions cannot be accessed.
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Next, we traverse all guard positions, simulate in four directions from that position until we encounter a wall or guard, or go out of bounds. During the simulation, we set the value of the encountered cell to 1ドル,ドル indicating that the cell is guarded.
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Finally, we traverse $g$ and count the number of cells with a value of 0ドル,ドル which is the answer.
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The time complexity is $O(m \times n),ドル and the space complexity is $O(m \times n)$. Here, $m$ and $n$ are the number of rows and columns in the grid, respectively.
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<!-- tabs:start -->
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### **Python3**
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### **TypeScript**
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```ts
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function countUnguarded(m: number, n: number, guards: number[][], walls: number[][]): number {
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const g: number[][] = Array.from({ length: m }, () => Array.from({ length: n }, () => 0));
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for (const [i, j] of guards) {
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g[i][j] = 2;
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}
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for (const [i, j] of walls) {
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g[i][j] = 2;
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}
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const dirs: number[] = [-1, 0, 1, 0, -1];
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for (const [i, j] of guards) {
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for (let k = 0; k < 4; ++k) {
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let [x, y] = [i, j];
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let [a, b] = [dirs[k], dirs[k + 1]];
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while (x + a >= 0 && x + a < m && y + b >= 0 && y + b < n && g[x + a][y + b] < 2) {
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x += a;
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y += b;
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g[x][y] = 1;
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}
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}
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}
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let ans = 0;
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for (const row of g) {
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for (const v of row) {
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ans += v === 0 ? 1 : 0;
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}
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}
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return ans;
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}
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```
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### **...**
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function countUnguarded(m: number, n: number, guards: number[][], walls: number[][]): number {
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const g: number[][] = Array.from({ length: m }, () => Array.from({ length: n }, () => 0));
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for (const [i, j] of guards) {
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g[i][j] = 2;
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}
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for (const [i, j] of walls) {
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g[i][j] = 2;
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}
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const dirs: number[] = [-1, 0, 1, 0, -1];
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for (const [i, j] of guards) {
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for (let k = 0; k < 4; ++k) {
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let [x, y] = [i, j];
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let [a, b] = [dirs[k], dirs[k + 1]];
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while (x + a >= 0 && x + a < m && y + b >= 0 && y + b < n && g[x + a][y + b] < 2) {
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x += a;
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y += b;
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g[x][y] = 1;
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}
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}
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}
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let ans = 0;
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for (const row of g) {
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for (const v of row) {
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ans += v === 0 ? 1 : 0;
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}
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}
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return ans;
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}

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