63. Unique Paths II #

Problem #

A robot is located at the top-left corner of a m x n grid (marked ‘Start’ in the diagram below).

The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked ‘Finish’ in the diagram below).

Now consider if some obstacles are added to the grids. How many unique paths would there be?

An obstacle and empty space is marked as 1 and 0 respectively in the grid.

Note: m and n will be at most 100.

Example 1:

Input:
[
  [0,0,0],
  [0,1,0],
  [0,0,0]
]
Output: 2
Explanation:
There is one obstacle in the middle of the 3x3 grid above.
There are two ways to reach the bottom-right corner:
1. Right -> Right -> Down -> Down
2. Down -> Down -> Right -> Right

Problem Summary #

A robot is located at the top-left corner of an m x n grid (the starting point is marked as “Start” in the diagram below). The robot can only move one step down or right at a time. The robot is trying to reach the bottom-right corner of the grid (marked as “Finish” in the diagram below). Now consider that there are obstacles in the grid. How many different paths are there from the top-left corner to the bottom-right corner?

Solution Approach #

• This problem is an enhanced version of Problem 62. It is also a simple problem that tests DP.
• The additional condition in this problem compared to Problem 62 is that obstacles may appear in the map. The way to handle obstacles is dp[i][j]=0.
• One situation to note is that if the starting point itself is an obstacle, then directly output 0 in this case.

Code #

package leetcode

func uniquePathsWithObstacles(obstacleGrid [][]int) int {
	if len(obstacleGrid) == 0 || obstacleGrid[0][0] == 1 {
		return 0
	}
	m, n := len(obstacleGrid), len(obstacleGrid[0])
	dp := make([][]int, m)
	for i := 0; i < m; i++ {
		dp[i] = make([]int, n)
	}
	dp[0][0] = 1
	for i := 1; i < n; i++ {
		if dp[0][i-1] != 0 && obstacleGrid[0][i] != 1 {
			dp[0][i] = 1
		}
	}
	for i := 1; i < m; i++ {
		if dp[i-1][0] != 0 && obstacleGrid[i][0] != 1 {
			dp[i][0] = 1
		}
	}
	for i := 1; i < m; i++ {
		for j := 1; j < n; j++ {
			if obstacleGrid[i][j] != 1 {
				dp[i][j] = dp[i-1][j] + dp[i][j-1]
			}
		}
	}
	return dp[m-1][n-1]
}