LeetCode
LeetCode 62 Unique Paths - Medium
A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).
62. Unique Paths — Medium
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).
How many possible unique paths are there?
Above is a 7 x 3 grid. How many possible unique paths are there?
Example 1:
Input: m = 3, n = 2 Output: 3 Explanation: From the top-left corner, there are a total of 3 ways to reach the bottom-right corner:
- Right -> Right -> Down
- Right -> Down -> Right
- Down -> Right -> Right Example 2:
Input: m = 7, n = 3 Output: 28
Constraints:
1 <= m, n <= 100 It's guaranteed that the answer will be less than or equal to 2 * 10 ^ 9.
Solution
### Brute Solution Recusion
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)
### DP
class Solution:
def uniquePaths(self, m: int, n: int) -> int:
# DP problem:
# Sub question:
# robot went to (m-1, n-2), and (m-2, n-1) before last step to (m-1, n-1)
# if both have X paths, Y paths, then to the last step will have X+Y paths
# So we can define the transition function:
# f[i][j] = f[i-1][j] + f[i][j-1]
# construct a status matrix m X n
f = [[1] * n for _ in range(m) ] ### Here I just initialized all elements to be 1, but in fact, only the first row and first column should be 1
### Let us do top to bottom strategy, based on the transfer function dependency relations
for i in range(1,m):
for j in range(1,n):
f[i][j] = f[i-1][j] + f[i][j-1]
return f[m-1][n-1]