980. Unique Paths III

Description

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.

 

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

Solution: Backtracking

• Time complexity: O(4^mn)
• Space complexity: O(mn)

 

JavaScript

/**
 * @param {number[][]} grid
 * @return {number}
 */
const uniquePathsIII = function (grid) {
  const START = 1;
  const END = 2;
  const OBSTACLE = -1;
  const m = grid.length;
  const n = grid[0].length;
  const totalCell = m * n;
  const startPoint = { row: -1, col: -1 };
  let obstacles = 0;

  const walkGrid = (row, col, step) => {
    if (row < 0 || col < 0 || row >= m || col >= n) return 0;
    const value = grid[row][col];

    if (value === END) {
      return step === totalCell - obstacles ? 1 : 0;
    }
    if (value === '.' || value === OBSTACLE) return 0;

    const nextStep = step + 1;

    grid[row][col] = '.';

    const left = walkGrid(row, col - 1, nextStep);
    const right = walkGrid(row, col + 1, nextStep);
    const up = walkGrid(row - 1, col, nextStep);
    const down = walkGrid(row + 1, col, nextStep);

    grid[row][col] = value;

    return left + right + up + down;
  };

  for (let row = 0; row < m; row++) {
    for (let col = 0; col < n; col++) {
      const value = grid[row][col];

      if (value === OBSTACLE) {
        obstacles += 1;
      }
      if (value === START) {
        startPoint.row = row;
        startPoint.col = col;
      }
    }
  }
  return walkGrid(startPoint.row, startPoint.col, 1);
};