LeetCode Day 39 Dynamic Programming (2/17)

Second day!!

Question 1

62. Unique Paths

There is a robot on an m x n grid. The robot is 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.

Given the two integers m and n, return the number of possible unique paths that the robot can take to reach the bottom-right corner.

The test cases are generated so that the answer will be less than or equal to 2 * 10^9.

Trying to require dp[i][j] can only be derived in two directions, dp[i - 1][j] and dp[i][j - 1].

It is possible to write the code very succinctly using recursion, but it will time out.

class Solution:
    def uniquePaths(self, m: int, n: int) -> int:
        if m == 1 or n == 1:
            return 1
        return self.uniquePaths(m - 1, n) + self.uniquePaths(m, n - 1)

Using dynamic programming is written in the following code:

class Solution:
    def uniquePaths(self, m: int, n: int) -> int:
        dp = [[0] * n for _ in range(m)]
 
        for i in range(m):
            dp[i][0] = 1
        for j in range(n):
            dp[0][j] = 1
 
        for i in range(1, m):
            for j in range(1, n):
                dp[i][j] = dp[i - 1][j] + dp[i][j - 1]
        
 
        return dp[m - 1][n - 1]

Question 2

63. Unique Paths II

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 * 10^9.

The difference between this question and the previous one is that the obstacles have been added, and with the obstacles, it’s really just a matter of marking the corresponding dp table (dp array) to remain at its initial value (0).

if obstacleGrid[i][j] == 0:
    dp[i][j] = dp[i - 1][j] + dp[i][j - 1]

So the complete code is:

class Solution:
    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
        m = len(obstacleGrid)
        n = len(obstacleGrid[0])
        if obstacleGrid[m - 1][n - 1] == 1 or obstacleGrid[0][0] == 1:
            return 0
        dp = [[0] * n for _ in range(m)]
        for i in range(m):
            if obstacleGrid[i][0] == 0:  
                dp[i][0] = 1
            else:
                break
        for j in range(n):
            if obstacleGrid[0][j] == 0:
                dp[0][j] = 1
            else:
                break
        for i in range(1, m):
            for j in range(1, n):
                if obstacleGrid[i][j] == 1:
                    continue
                dp[i][j] = dp[i - 1][j] + dp[i][j - 1]
        return dp[m - 1][n - 1]