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 * 109.
Example 1:
Input: m = 3, n = 7 Output: 28
Example 2: 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: 1. Right -> Down -> Down 2. Down -> Down -> Right 3. Down -> Right -> Down
Constraints:
1 <= m, n <= 100
- code time limit exceeded
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)
- code
class Solution:
def uniquePaths(self, m, n):
dp = {}
for row in range(m):
for col in range(n):
if row == 0 or col == 0:
dp[(row, col)] = 1
else:
dp[(row, col)] = dp[(row-1, col)] + dp[(row, col-1)]
return dp[(m-1, n-1)]
- code
class Solution:
def uniquePaths(self, m: int, n: int) -> int:
grid = [[1 for _ in range(n)] for _ in range(m)]
for i in range(1, m):
for j in range(1, n):
grid[i][j] = grid[i][j-1]+grid[i-1][j]
return grid[-1][-1]
- code
class Solution:
def uniquePaths(self, m, n):
cur = [1] * n
for _ in range(1, m):
for j in range(1, n):
cur[j] += cur[j-1]
return cur[-1]
- code
class Solution:
def uniquePaths(self, m: int, n: int) -> int:
if m > n: n, m = m, n
dp = [0] * m
dp[0] = 1 # row
for _ in range(n):
for row in range(1, m):
dp[row] += dp[row - 1]
return dp[-1]
- code
class Solution:
def uniquePaths(self, m, n):
return math.factorial(m+n-2)//(math.factorial(n-1) * math.factorial(m-1))