980. Unique Paths III.md

980. Unique Paths III

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leetcode Daily Challenge on September 20th, 2020.

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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 <= grid.length * grid[0].length <= 20

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Solution

• mine

  • Java

    • DFS Runtime: 0 ms, faster than 100.00%, Memory Usage: 36.7 MB, less than 33.33% of Java online submissions

      got from the most votes

      //emptyCount = 1, is mean the ending square 
      public int res = 0,sX,sY,eX,eY,emptyCount = 1;
      public int uniquePathsIII(int[][] grid) {
          calculate(grid);
          dfs(grid,sX,sY);
          return res;
      }
      
      public void calculate(int[][] grid){
          for(int i = 0; i < grid.length; i++){
              for(int j = 0; j < grid[i].length; j++){
                  if(grid[i][j] == 1){
                      sX = i;
                      sY = j;
                  }else if(grid[i][j] == 2){
                      eX = i;
                      eY = j;
                  }else if(grid[i][j] == 0){
                      emptyCount++;
                  }
              }
          }
      }
      
      public void dfs(int[][] grid, int x, int y) {
          if (x < 0 || x >= grid.length || y < 0 || y >= grid[0].length || grid[x][y] < 0)
              return;
          if (x == eX && y == eY) {
              if (emptyCount == 0) res++;
              return;
          }
          grid[x][y] = -2;// As long as it is not -1, 0, 1, 2
          emptyCount--;
          dfs(grid, x + 1, y);
          dfs(grid, x - 1, y);
          dfs(grid, x, y + 1);
          dfs(grid, x, y - 1);
          grid[x][y] = 0;
          emptyCount++;
      }
      

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• the most votes

  • DFS Runtime: 0 ms, faster than 100.00%, Memory Usage: 36.4 MB, less than 33.33% of Java online submissions

    int res = 0, empty = 1, sx, sy, ex, ey;
    public int uniquePathsIII(int[][] grid) {
        int m = grid.length, n = grid[0].length;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == 0) empty++;
                else if (grid[i][j] == 1) {
                    sx = i;
                    sy = j;
                } else if (grid[i][j] == 2) {
                    ex = i;
                    ey = j;
                }
            }
        }
        dfs(grid, sx, sy);
        return res;
    }
    
    public void dfs(int[][] grid, int x, int y) {
        if (x < 0 || x >= grid.length || y < 0 || y >= grid[0].length || grid[x][y] < 0)
            return;
        if (x == ex && y == ey) {
            if (empty == 0) res++;
            return;
        }
        grid[x][y] = -2;
        empty--;
        dfs(grid, x + 1, y);
        dfs(grid, x - 1, y);
        dfs(grid, x, y + 1);
        dfs(grid, x, y - 1);
        grid[x][y] = 0;
        empty++;
    }
    

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