Nov 2, 2021 - 5 min ' read

Unique Paths III

Tags : array, backtracking, matrix, leetcode, cpp, hard

Problem Statement - link #

You are given an m x n integer array grid where grid[i][j] could be:

1 representing the starting square. There is exactly one starting square.
2 representing the ending square. There is exactly one ending square.
0 representing empty squares we can walk over.
-1 representing 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.

Examples #

Example 1:

Input: grid = [[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: grid = [[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: grid = [[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.

Constraints #

m == grid.length
n == grid[i].length
1 <= m, n <= 20
1 <= m * n <= 20
-1 <= grid[i][j] <= 2
There is exactly one starting cell and one ending cell.

Solutions #

class Solution {
public:
    int res = 0, z = 1;
    void dfs(vector<vector<int>>& grid, int i, int j, int v) {
        if (i < 0 || i >= grid.size() || j < 0 || j >= grid[0].size() || grid[i][j] == -1) return;
        
        if (grid[i][j] == 2) {
            if(z == v) res++; 
            return;
        }
        
        grid[i][j] = -1;
        
        dfs(grid, i+1, j, v+1);
        dfs(grid, i-1, j, v+1);
        dfs(grid, i, j+1, v+1);
        dfs(grid, i, j-1, v+1);
        
        grid[i][j] = 0;
        
    }
    
    int uniquePathsIII(vector<vector<int>>& grid) {
        int x, y;
        for (int i = 0; i < grid.size(); i++) {
            for (int j = 0; j < grid[0].size(); j++) {
                if (grid[i][j] == 1) x = i, y = j;
                else if (grid[i][j] == 0) z++;
            }
        }
        
        dfs(grid, x, y, 0);
        return res;
    }
};