Description
There is an m x n grid with a ball. The ball is initially at the position [startRow, startColumn]. You are allowed to move the ball to one of the four adjacent cells in the grid (possibly out of the grid crossing the grid boundary). You can apply at most maxMove moves to the ball.
Given the five integers m, n, maxMove, startRow, startColumn, return the number of paths to move the ball out of the grid boundary. Since the answer can be very large, return it modulo 10^9 + 7.
Example 1:
Input: m = 2, n = 2, maxMove = 2, startRow = 0, startColumn = 0
Output: 6
Example 2:
Input: m = 1, n = 3, maxMove = 3, startRow = 0, startColumn = 1
Output: 12
Constraints:
1 <= m, n <= 50
0 <= maxMove <= 50
0 <= startRow < m
0 <= startColumn < n
Solution
Solved after help…
Recursive, the function is f(x, y, step), and when x, y is out of boundary, return 1. If used up the step, return 0.
Time complexity:
o
(
m
∗
n
∗
m
a
x
M
o
v
e
)
o(m*n*maxMove)
o(m∗n∗maxMove)
Space complexity:
o
(
m
∗
n
∗
m
a
x
M
o
v
e
)
o(m*n*maxMove)
o(m∗n∗maxMove)
Code
class Solution:
def findPaths(self, m: int, n: int, maxMove: int, startRow: int, startColumn: int) -> int:
memo = {}
mod_val = 1000000007
def helper(x: int, y: int, step: int) -> int:
if x < 0 or x >= m or y < 0 or y >= n:
memo[(x, y, step)] = 1
return memo[(x, y, step)]
if step >= maxMove:
memo[(x, y, step)] = 0
return memo[(x, y, step)]
if (x, y, step) in memo:
return memo[(x, y, step)]
res = 0
for dz in (-1, 1):
res += helper(x + dz, y, step + 1)
res += helper(x, y + dz, step + 1)
memo[(x, y, step)] = res % mod_val
return memo[(x, y, step)]
return helper(startRow, startColumn, 0)
原文地址:https://blog.csdn.net/sinat_41679123/article/details/135858226
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