Description
You are given an m x n grid. Each cell of grid represents a street. The street of grid[i][j] can be:
1 which means a street connecting the left cell and the right cell.2 which means a street connecting the upper cell and the lower cell.3 which means a street connecting the left cell and the lower cell.4 which means a street connecting the right cell and the lower cell.5 which means a street connecting the left cell and the upper cell.6 which means a street connecting the right cell and the upper cell.
You will initially start at the street of the upper-left cell (0, 0). A valid path in the grid is a path that starts from the upper left cell (0, 0) and ends at the bottom-right cell (m - 1, n - 1). The path should only follow the streets.
Notice that you are not allowed to change any street.
Return true if there is a valid path in the grid or false otherwise.
Example 1:
Input: grid = [[2,4,3],[6,5,2]]
Output: true
Explanation: As shown you can start at cell (0, 0) and visit all the cells of the grid to reach (m - 1, n - 1).
Example 2:
Input: grid = [[1,2,1],[1,2,1]]
Output: false
Explanation: As shown you the street at cell (0, 0) is not connected with any street of any other cell and you will get stuck at cell (0, 0)
Example 3:
Input: grid = [[1,1,2]]
Output: false
Explanation: You will get stuck at cell (0, 1) and you cannot reach cell (0, 2).
Constraints:
m == grid.lengthn == grid[i].length1 <= m, n <= 3001 <= grid[i][j] <= 6
Solutions
Solution 1
PythonJavaC++Go
class Solution:
def hasValidPath(self, grid: List[List[int]]) -> bool:
m, n = len(grid), len(grid[0])
p = list(range(m * n))
def find(x):
if p[x] != x:
p[x] = find(p[x])
return p[x]
def left(i, j):
if j > 0 and grid[i][j - 1] in (1, 4, 6):
p[find(i * n + j)] = find(i * n + j - 1)
def right(i, j):
if j < n - 1 and grid[i][j + 1] in (1, 3, 5):
p[find(i * n + j)] = find(i * n + j + 1)
def up(i, j):
if i > 0 and grid[i - 1][j] in (2, 3, 4):
p[find(i * n + j)] = find((i - 1) * n + j)
def down(i, j):
if i < m - 1 and grid[i + 1][j] in (2, 5, 6):
p[find(i * n + j)] = find((i + 1) * n + j)
for i in range(m):
for j in range(n):
e = grid[i][j]
if e == 1:
left(i, j)
right(i, j)
elif e == 2:
up(i, j)
down(i, j)
elif e == 3:
left(i, j)
down(i, j)
elif e == 4:
right(i, j)
down(i, j)
elif e == 5:
left(i, j)
up(i, j)
else:
right(i, j)
up(i, j)
return find(0) == find(m * n - 1)(code-box)
class Solution {
private int[] p;
private int[][] grid;
private int m;
private int n;
public boolean hasValidPath(int[][] grid) {
this.grid = grid;
m = grid.length;
n = grid[0].length;
p = new int[m * n];
for (int i = 0; i < p.length; ++i) {
p[i] = i;
}
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
int e = grid[i][j];
if (e == 1) {
left(i, j);
right(i, j);
} else if (e == 2) {
up(i, j);
down(i, j);
} else if (e == 3) {
left(i, j);
down(i, j);
} else if (e == 4) {
right(i, j);
down(i, j);
} else if (e == 5) {
left(i, j);
up(i, j);
} else {
right(i, j);
up(i, j);
}
}
}
return find(0) == find(m * n - 1);
}
private int find(int x) {
if (p[x] != x) {
p[x] = find(p[x]);
}
return p[x];
}
private void left(int i, int j) {
if (j > 0 && (grid[i][j - 1] == 1 || grid[i][j - 1] == 4 || grid[i][j - 1] == 6)) {
p[find(i * n + j)] = find(i * n + j - 1);
}
}
private void right(int i, int j) {
if (j < n - 1 && (grid[i][j + 1] == 1 || grid[i][j + 1] == 3 || grid[i][j + 1] == 5)) {
p[find(i * n + j)] = find(i * n + j + 1);
}
}
private void up(int i, int j) {
if (i > 0 && (grid[i - 1][j] == 2 || grid[i - 1][j] == 3 || grid[i - 1][j] == 4)) {
p[find(i * n + j)] = find((i - 1) * n + j);
}
}
private void down(int i, int j) {
if (i < m - 1 && (grid[i + 1][j] == 2 || grid[i + 1][j] == 5 || grid[i + 1][j] == 6)) {
p[find(i * n + j)] = find((i + 1) * n + j);
}
}
}(code-box)
class Solution {
public:
vector<int> p;
bool hasValidPath(vector<vector<int>>& grid) {
int m = grid.size();
int n = grid[0].size();
p.resize(m * n);
for (int i = 0; i < p.size(); ++i) p[i] = i;
auto left = [&](int i, int j) {
if (j > 0 && (grid[i][j - 1] == 1 || grid[i][j - 1] == 4 || grid[i][j - 1] == 6)) {
p[find(i * n + j)] = find(i * n + j - 1);
}
};
auto right = [&](int i, int j) {
if (j < n - 1 && (grid[i][j + 1] == 1 || grid[i][j + 1] == 3 || grid[i][j + 1] == 5)) {
p[find(i * n + j)] = find(i * n + j + 1);
}
};
auto up = [&](int i, int j) {
if (i > 0 && (grid[i - 1][j] == 2 || grid[i - 1][j] == 3 || grid[i - 1][j] == 4)) {
p[find(i * n + j)] = find((i - 1) * n + j);
}
};
auto down = [&](int i, int j) {
if (i < m - 1 && (grid[i + 1][j] == 2 || grid[i + 1][j] == 5 || grid[i + 1][j] == 6)) {
p[find(i * n + j)] = find((i + 1) * n + j);
}
};
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
int e = grid[i][j];
if (e == 1) {
left(i, j);
right(i, j);
} else if (e == 2) {
up(i, j);
down(i, j);
} else if (e == 3) {
left(i, j);
down(i, j);
} else if (e == 4) {
right(i, j);
down(i, j);
} else if (e == 5) {
left(i, j);
up(i, j);
} else {
right(i, j);
up(i, j);
}
}
}
return find(0) == find(m * n - 1);
}
int find(int x) {
if (p[x] != x) p[x] = find(p[x]);
return p[x];
}
};(code-box)
func hasValidPath(grid [][]int) bool {
m, n := len(grid), len(grid[0])
p := make([]int, m*n)
for i := range p {
p[i] = i
}
var find func(x int) int
find = func(x int) int {
if p[x] != x {
p[x] = find(p[x])
}
return p[x]
}
left := func(i, j int) {
if j > 0 && (grid[i][j-1] == 1 || grid[i][j-1] == 4 || grid[i][j-1] == 6) {
p[find(i*n+j)] = find(i*n + j - 1)
}
}
right := func(i, j int) {
if j < n-1 && (grid[i][j+1] == 1 || grid[i][j+1] == 3 || grid[i][j+1] == 5) {
p[find(i*n+j)] = find(i*n + j + 1)
}
}
up := func(i, j int) {
if i > 0 && (grid[i-1][j] == 2 || grid[i-1][j] == 3 || grid[i-1][j] == 4) {
p[find(i*n+j)] = find((i-1)*n + j)
}
}
down := func(i, j int) {
if i < m-1 && (grid[i+1][j] == 2 || grid[i+1][j] == 5 || grid[i+1][j] == 6) {
p[find(i*n+j)] = find((i+1)*n + j)
}
}
for i, row := range grid {
for j, e := range row {
if e == 1 {
left(i, j)
right(i, j)
} else if e == 2 {
up(i, j)
down(i, j)
} else if e == 3 {
left(i, j)
down(i, j)
} else if e == 4 {
right(i, j)
down(i, j)
} else if e == 5 {
left(i, j)
up(i, j)
} else {
right(i, j)
up(i, j)
}
}
}
return find(0) == find(m*n-1)
}(code-box)
