Unique Paths

Info

📖 Leetcode-63. Unique Paths II

Reference: 📖 Leetcode-62. Unique Paths

Description

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:

img

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:

img

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.

Thinking

• If one stop is blocked, then it’s unreachable.
• How to calculate avaliable paths counts of reachable stops? Still use DP

Solution

DP

Basic idea follows same with 📖 Leetcode-62. Unique Paths, but need to do some actions for extreme situations(when facing blocked stop)

So the basic dp formula remains the same:

$$F[i][j]=F[i-1][j]+F[i][j-1]$$

But notice that: F[i - 1][j] is blocked, then the value it’s 0.(which means this stop is unreachable and can’t get the (i, j) of course)

class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        if (obstacleGrid[0][0] == 1) {
            return 0;
        }
        int[][] dp = new int[obstacleGrid.length][obstacleGrid[0].length];
        dp[0][0] = 1;
        for (int i = 0; i < dp.length; i++) {
            for (int j = 0; j < dp[0].length; j++) {
                if (i == 0 && j == 0 || obstacleGrid[i][j] == 1) {
                    continue;
                }
                
                dp[i][j] = getAvailablePathNum(obstacleGrid, dp, i - 1, j) + getAvailablePathNum(obstacleGrid, dp, i, j - 1);
            }
        }
        return dp[dp.length - 1][dp[0].length - 1];
    }

    public int getAvailablePathNum(int[][] obstacleGrid, int[][] dp, int i, int j) {
        // Error / Extreme Situation
        if (i < 0 || j < 0 || obstacleGrid[i][j] == 1) {
            return 0;
        }
        return dp[i][j];
    }
}

Optimization

The getAvailablePathNum is useless, so we wrap it in the main func.

class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        if (obstacleGrid[0][0] == 1) {
            return 0;
        }
        int[][] dp = new int[obstacleGrid.length][obstacleGrid[0].length];
        dp[0][0] = 1;
        for (int i = 0; i < dp.length; i++) {
            for (int j = 0; j < dp[0].length; j++) {
                if (i == 0 && j == 0 || obstacleGrid[i][j] == 1) {
                    continue;
                }
                int leftStop = i - 1 >=0 ? dp[i - 1][j] : 0;
                int topStop = j - 1 >=0 ? dp[i][j - 1] : 0;
                dp[i][j] = leftStop + topStop;
            }
        }
        return dp[dp.length - 1][dp[0].length - 1];
    }
}

Final Optimization

**Rolling Array**

When handling DP Problems, idea of Rolling Array will helps us optimize the performacne. Reduce space complexity effectively

class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int n = obstacleGrid.length, m = obstacleGrid[0].length;
        int[] f = new int[m];

        f[0] = obstacleGrid[0][0] == 0 ? 1 : 0;
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < m; ++j) {
                if (obstacleGrid[i][j] == 1) {
                    f[j] = 0;
                    continue;
                }
                if (j - 1 >= 0 && obstacleGrid[i][j - 1] == 0) {
                    // That's where the thought works
                    // Only do the leftStop calculate, `case we've already done the topStop calculate through '+='
                    f[j] += f[j - 1];
                }
            }
        }
        
        return f[m - 1];
    }
}

Reflection

• Remember the idea of Rolling Array when doing dp optimization.