LeetCode - #63 Different Paths II

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Difficulty level: medium

1. Description

A robot is located in the upper left corner of a m x ngrid (the starting point is marked "Start" in the image below).

The robot can only move down or to the right one step at a time. The robot tries to reach the bottom right corner of the grid (labeled "Finish" in the image below).

Now consider that there are obstacles in the mesh. So how many different paths will there be from the top left corner to the bottom right corner?

Obstacles and empty positions in the grid are denoted by 1and 0.

2. Example

Example 1

输入：obstacleGrid = [[0,0,0],[0,1,0],[0,0,0]]
输出：2
解释：3x3 网格的正中间有一个障碍物。
从左上角到右下角一共有 2 条不同的路径：
1. 向右 -> 向右 -> 向下 -> 向下
2. 向下 -> 向下 -> 向右 -> 向右
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Example 2

输入：obstacleGrid = [[0,1],[0,0]]
输出：1
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Restrictions:

• m == obstacleGrid.length
• n == obstacleGrid[i].length
• 1 <= m, n <= 100
• obstacleGrid[i][j]for 0or1

3. Answers

class UniquePathsII {
    func uniquePathsWithObstacles(_ obstacleGrid: [[Int]]) -> Int {
        let m = obstacleGrid.count
        guard m > 0 else {
            return 0
        }
        
        let n = obstacleGrid[0].count
        guard n > 0 else {
            return 0
        }
    
        var dp = Array(repeating: Array(repeating: -1, count: n), count: m)
        
        return help(m - 1, n - 1, &dp, obstacleGrid)
    }
    
    fileprivate func help(_ m: Int, _ n: Int, _ dp: inout [[Int]], _ obstacleGrid: [[Int]]) -> Int {
        if m < 0 || n < 0 {
            return 0
        }
        if obstacleGrid[m][n] == 1 {
            return 0
        }
        if m == 0 && n == 0 {
            return 1
        }
        if dp[m][n] != -1 {
            return dp[m][n]
        }
        
        dp[m][n] = help(m - 1, n, &dp, obstacleGrid) + help(m, n - 1, &dp, obstacleGrid)
        return dp[m][n]
    }
}
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• Main idea: 2D dynamic programming, which uses 2D arrays as caches to store computational data.
• Time Complexity: O(mn)
• Space Complexity: O(mn)

Repository for the algorithm solution: LeetCode-Swift

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