LeetCode-in-Go

63. Unique Paths II

Medium

You are given an m x n integer array grid. There is a robot initially located at the top-left corner (i.e., grid[0][0]). The robot tries to move to the bottom-right corner (i.e., grid[m-1][n-1]). The robot can only move either down or right at any point in time.

An obstacle and space are marked as 1 or 0 respectively in grid. A path that the robot takes cannot include any square that is an obstacle.

Return the number of possible unique paths that the robot can take to reach the bottom-right corner.

The testcases are generated so that the answer will be less than or equal to 2 * 109.

Example 1:

Input: obstacleGrid = [[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

Example 2:

Input: obstacleGrid = [[0,1],[0,0]]

Output: 1

Constraints:

• m == obstacleGrid.length
• n == obstacleGrid[i].length
• 1 <= m, n <= 100
• obstacleGrid[i][j] is 0 or 1.

Solution

func uniquePathsWithObstacles(obstacleGrid [][]int) int {
	if obstacleGrid[0][0] == 1 {
		return 0
	}
	obstacleGrid[0][0] = 1
	m := len(obstacleGrid)
	n := len(obstacleGrid[0])
	for i := 1; i < m; i++ {
		if obstacleGrid[i][0] == 1 {
			obstacleGrid[i][0] = 0
		} else {
			obstacleGrid[i][0] = obstacleGrid[i-1][0]
		}
	}
	for j := 1; j < n; j++ {
		if obstacleGrid[0][j] == 1 {
			obstacleGrid[0][j] = 0
		} else {
			obstacleGrid[0][j] = obstacleGrid[0][j-1]
		}
	}
	for i := 1; i < m; i++ {
		for j := 1; j < n; j++ {
			if obstacleGrid[i][j] == 1 {
				obstacleGrid[i][j] = 0
			} else {
				obstacleGrid[i][j] = obstacleGrid[i-1][j] + obstacleGrid[i][j-1]
			}
		}
	}
	return obstacleGrid[m-1][n-1]
}

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