980. Unique Paths III

Posted 2021-02-08 ruruozhenhao

On a 2-dimensional grid, there are 4 types of squares:

• 1 represents the starting square.  There is exactly one starting square.
• 2 represents the ending square.  There is exactly one ending square.
• 0 represents empty squares we can walk over.
• -1 represents 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. 1 <= grid.length * grid[0].length <= 20

 

My code:

class Solution {
public:
    int uniquePathsIII(vector<vector<int>>& grid) {
        int ans;
        pair<int, int> start, end;
        for (int i = 0; i < grid.size(); ++i) {
            for (int j = 0; j < grid[0].size(); ++j) {
                if (grid[i][j] == 1)
                    start = {i, j};
                if (grid[i][j] == 2)
                    end = {i, j};
            }
        }
        
        helper(grid, start, end);
        
        return ans;
    }
    
private:
    int ans = 0;
    
    void helper(const vector<vector<int>>& grid, const pair<int, int>& start, const pair<int, int>& end) {
        vector<vector<int>> temp = grid;
        int startX = start.first;
        int startY = start.second;
        // cout << start.first << " " << start.second << endl;
        // cout << end.first << " " << end.second << endl;
        dfs(temp, startX+1, startY, end);
        dfs(temp, startX-1, startY, end);
        dfs(temp, startX, startY+1, end);
        dfs(temp, startX, startY-1, end);
    }
    
    void dfs(vector<vector<int>> temp, int curX, int curY, const pair<int, int>& end) {
        if (curX == end.first && curY == end.second && check(temp)) ans++, return; 
        if (curX < 0 || curX >= temp.size() || curY < 0 || curY >= temp[0].size() || temp[curX][curY] != 0) return;
        
        // cout << curX << " " << curY << endl;
        
        temp[curX][curY] = 3;
        dfs(temp, curX+1, curY, end);
        dfs(temp, curX-1, curY, end);
        dfs(temp, curX, curY+1, end);
        dfs(temp, curX, curY-1, end);
        
    }
    
    bool check(vector<vector<int>>& temp) {
        for (int i = 0; i < temp.size(); ++i) {
            for (int j = 0; j < temp[0].size(); ++j) {
                if (temp[i][j] == 0) return false;
            }
        }
        return true;
    }
};

/*

[[1,0,0,0],
 [0,0,0,0],
 [0,0,2,-1]]

*/

　　

Approach #2: DFS. [C++]

class Solution {
public:
    int uniquePathsIII(vector<vector<int>>& grid) {
        int num = 1;
        int sx = -1, sy = -1;
        for (int i = 0; i < grid.size(); ++i) 
            for (int j = 0; j < grid[0].size(); ++j) 
                if (grid[i][j] == 0) num++;
                else if (grid[i][j] == 1) sx = i, sy = j;
        return dfs(grid, sx, sy, num);
    }
    
private:
    int dfs(vector<vector<int>>& grid, int cx, int cy, int num) {
        if (cx < 0 || cx >= grid.size() || cy < 0 || cy >= grid[0].size() || grid[cx][cy] == -1) return 0;
        if (grid[cx][cy] == 2) return num == 0;
        grid[cx][cy] = -1;
        int path = dfs(grid, cx+1, cy, num-1)+
                   dfs(grid, cx-1, cy, num-1)+
                   dfs(grid, cx, cy+1, num-1)+
                   dfs(grid, cx, cy-1, num-1);
        grid[cx][cy] = 0;
        return path;
    }
};

　　

Analysis:

In the first code. I don‘t get the correct answer. Maybe my thinking is wrong in this problem.