On a 2-dimensional grid, there are 4 types of squares:
1represents the starting square. There is exactly one starting square.2represents the ending square. There is exactly one ending square.0represents empty squares we can walk over.-1represents obstacles that we cannot walk over.
Return the number of 4-directional walks from the starting square to the ending square, that walk over every non-obstacle square exactly once.
Example 1:
Input: [[1,0,0,0],[0,0,0,0],[0,0,2,-1]] Output: 2 Explanation: We have the following two paths: 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2) 2. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2)
Example 2:
Input: [[1,0,0,0],[0,0,0,0],[0,0,0,2]] Output: 4 Explanation: We have the following four paths: 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2),(2,3) 2. (0,0),(0,1),(1,1),(1,0),(2,0),(2,1),(2,2),(1,2),(0,2),(0,3),(1,3),(2,3) 3. (0,0),(1,0),(2,0),(2,1),(2,2),(1,2),(1,1),(0,1),(0,2),(0,3),(1,3),(2,3) 4. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2),(2,3)
Example 3:
Input: [[0,1],[2,0]] Output: 0 Explanation: There is no path that walks over every empty square exactly once. Note that the starting and ending square can be anywhere in the grid.
Note:
1 <= grid.length * grid[0].length <= 20
Companies:
LimeBike
Related Topics:
Backtracking, Depth-first Search
Similar Questions:
// OJ: https://leetcode.com/problems/unique-paths-iii/
// Author: github.com/lzl124631x
// Time: O(4^(MN))
// Space: O(MN)
class Solution {
private:
int ans = 0, target = 0, visited = 0, M, N;
int dirs[4][2] = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};
void dfs(vector<vector<int>>& grid, int x, int y) {
if (grid[x][y] == 2) {
if (visited == target) ++ans;
return;
}
++visited;
grid[x][y] = -1;
for (auto dir : dirs) {
int i = x + dir[0], j = y + dir[1];
if (i < 0 || i >= M || j < 0 || j >= N || grid[i][j] == -1) continue;
dfs(grid, i, j);
}
grid[x][y] = 0;
--visited;
}
public:
int uniquePathsIII(vector<vector<int>>& grid) {
M = grid.size();
N = grid[0].size();
int x, y;
for (int i = 0; i < M; ++i) {
for (int j = 0; j < N; ++j) {
if (grid[i][j] == 1) {
x = i;
y = j;
++target;
} else if (!grid[i][j]) ++target;
}
}
dfs(grid, x, y);
return ans;
}
};