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Nov 6, 2024
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feat: add solutions to lc problem: No.2267 #3726

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176 changes: 124 additions & 52 deletions ...tion/2200-2299/2267.Check if There Is a Valid Parentheses String Path/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -81,7 +81,17 @@ tags:

<!-- solution:start -->

### 方法一：记忆化搜索
### 方法一：DFS + 剪枝

我们记矩阵的行数为 $m$，列数为 $n$。

如果 $m + n - 1$ 为奇数，或者左上角和右下角的括号不匹配，那么一定不存在合法路径，直接返回 $\text{false}$。

否则，我们设计一个函数 $\textit{dfs}(i, j, k)$，表示从 $(i, j)$ 出发，且当前括号的平衡度为 $k$，是否存在合法路径。其中，平衡度 $k$ 的定义为：从 $(0, 0)$ 到 $(i, j)$ 的路径中，左括号的个数减去右括号的个数。

如果平衡度 $k$ 小于 $0$ 或者大于 $m + n - i - j$，那么一定不存在合法路径，直接返回 $\text{false}$。如果 $(i, j)$ 正好是右下角的格子，那么只有当 $k = 0$ 时才存在合法路径。否则，我们枚举 $(i, j)$ 的下一个格子 $(x, y)$，如果 $(x, y)$ 是合法的格子且 $\textit{dfs}(x, y, k)$ 为 $\text{true}$，那么就存在合法路径。

时间复杂度 $O(m \times n \times (m + n))$，空间复杂度 $O(m \times n \times (m + n))$。其中 $m$ 和 $n$ 分别是矩阵的行数和列数。

<!-- tabs:start -->

Expand All @@ -91,57 +101,60 @@ tags:
class Solution:
def hasValidPath(self, grid: List[List[str]]) -> bool:
@cache
def dfs(i, j, t):
if grid[i][j] == '(':
t += 1
else:
t -= 1
if t < 0:
def dfs(i: int, j: int, k: int) -> bool:
d = 1 if grid[i][j] == "(" else -1
k += d
if k < 0 or k > m - i + n - j:
return False
if i == m - 1 and j == n - 1:
return t == 0
for x, y in [(i + 1, j), (i, j + 1)]:
if x < m and y < n and dfs(x, y, t):
return k == 0
for a, b in pairwise((0, 1, 0)):
x, y = i + a, j + b
if 0 <= x < m and 0 <= y < n and dfs(x, y, k):
return True
return False

m, n = len(grid), len(grid[0])
if (m + n - 1) % 2 or grid[0][0] == ")" or grid[m - 1][n - 1] == "(":
return False
return dfs(0, 0, 0)
```

#### Java

```java
class Solution {
private boolean[][][] vis;
private int m, n;
private char[][] grid;
private int m;
private int n;
private boolean[][][] vis;

public boolean hasValidPath(char[][] grid) {
m = grid.length;
n = grid[0].length;
if ((m + n - 1) % 2 == 1 || grid[0][0] == ')' || grid[m - 1][n - 1] == '(') {
return false;
}
this.grid = grid;
vis = new boolean[m][n][m + n];
return dfs(0, 0, 0);
}

private boolean dfs(int i, int j, int t) {
if (vis[i][j][t]) {
private boolean dfs(int i, int j, int k) {
if (vis[i][j][k]) {
return false;
}
vis[i][j][t] = true;
t += grid[i][j] == '(' ? 1 : -1;
if (t < 0) {
vis[i][j][k] = true;
k += grid[i][j] == '(' ? 1 : -1;
if (k < 0 || k > m - i + n - j) {
return false;
}
if (i == m - 1 && j == n - 1) {
return t == 0;
return k == 0;
}
int[] dirs = {0, 1, 0};
for (int k = 0; k < 2; ++k) {
int x = i + dirs[k], y = j + dirs[k + 1];
if (x < m && y < n && dfs(x, y, t)) {
final int[] dirs = {1, 0, 1};
for (int d = 0; d < 2; ++d) {
int x = i + dirs[d], y = j + dirs[d + 1];
if (x >= 0 && x < m && y >= 0 && y < n && dfs(x, y, k)) {
return true;
}
}
Expand All @@ -153,28 +166,37 @@ class Solution {
#### C++

```cpp
bool vis[100][100][200];
int dirs[3] = {1, 0, 1};

class Solution {
public:
bool hasValidPath(vector<vector<char>>& grid) {
memset(vis, 0, sizeof(vis));
return dfs(0, 0, 0, grid);
}

bool dfs(int i, int j, int t, vector<vector<char>>& grid) {
if (vis[i][j][t]) return false;
vis[i][j][t] = true;
t += grid[i][j] == '(' ? 1 : -1;
if (t < 0) return false;
int m = grid.size(), n = grid[0].size();
if (i == m - 1 && j == n - 1) return t == 0;
for (int k = 0; k < 2; ++k) {
int x = i + dirs[k], y = j + dirs[k + 1];
if (x < m && y < n && dfs(x, y, t, grid)) return true;
if ((m + n - 1) % 2 || grid[0][0] == ')' || grid[m - 1][n - 1] == '(') {
return false;
}
return false;
bool vis[m][n][m + n];
memset(vis, false, sizeof(vis));
int dirs[3] = {1, 0, 1};
auto dfs = [&](auto&& dfs, int i, int j, int k) -> bool {
if (vis[i][j][k]) {
return false;
}
vis[i][j][k] = true;
k += grid[i][j] == '(' ? 1 : -1;
if (k < 0 || k > m - i + n - j) {
return false;
}
if (i == m - 1 && j == n - 1) {
return k == 0;
}
for (int d = 0; d < 2; ++d) {
int x = i + dirs[d], y = j + dirs[d + 1];
if (x >= 0 && x < m && y >= 0 && y < n && dfs(dfs, x, y, k)) {
return true;
}
}
return false;
};
return dfs(dfs, 0, 0, 0);
}
};
```
Expand All @@ -184,34 +206,37 @@ public:
```go
func hasValidPath(grid [][]byte) bool {
m, n := len(grid), len(grid[0])
if (m+n-1)%2 == 1 || grid[0][0] == ')' || grid[m-1][n-1] == '(' {
return false
}
vis := make([][][]bool, m)
for i := range vis {
vis[i] = make([][]bool, n)
for j := range vis[i] {
vis[i][j] = make([]bool, m+n)
}
}
var dfs func(int, int, int) bool
dfs = func(i, j, t int) bool {
if vis[i][j][t] {
dirs := [3]int{1, 0, 1}
var dfs func(i, j, k int) bool
dfs = func(i, j, k int) bool {
if vis[i][j][k] {
return false
}
vis[i][j][t] = true
vis[i][j][k] = true
if grid[i][j] == '(' {
t += 1
k++
} else {
t -= 1
k--
}
if t < 0 {
if k < 0 || k > m-i+n-j {
return false
}
if i == m-1 && j == n-1 {
return t == 0
return k == 0
}
dirs := []int{1, 0, 1}
for k := 0; k < 2; k++ {
x, y := i+dirs[k], j+dirs[k+1]
if x < m && y < n && dfs(x, y, t) {
for d := 0; d < 2; d++ {
x, y := i+dirs[d], j+dirs[d+1]
if x >= 0 && x < m && y >= 0 && y < n && dfs(x, y, k) {
return true
}
}
Expand All @@ -221,6 +246,53 @@ func hasValidPath(grid [][]byte) bool {
}
```

#### TypeScript

```ts
function hasValidPath(grid: string[][]): boolean {
const m = grid.length,
n = grid[0].length;

if ((m + n - 1) % 2 || grid[0][0] === ')' || grid[m - 1][n - 1] === '(') {
return false;
}

const vis: boolean[][][] = Array.from({ length: m }, () =>
Array.from({ length: n }, () => Array(m + n).fill(false)),
);
const dirs = [1, 0, 1];

const dfs = (i: number, j: number, k: number): boolean => {
if (vis[i][j][k]) {
return false;
}

vis[i][j][k] = true;
k += grid[i][j] === '(' ? 1 : -1;

if (k < 0 || k > m - i + n - j) {
return false;
}

if (i === m - 1 && j === n - 1) {
return k === 0;
}

for (let d = 0; d < 2; ++d) {
const x = i + dirs[d],
y = j + dirs[d + 1];
if (x >= 0 && x < m && y >= 0 && y < n && dfs(x, y, k)) {
return true;
}
}

return false;
};

return dfs(0, 0, 0);
}
```

<!-- tabs:end -->

<!-- solution:end -->
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