There is an n x n
grid, with the top-left cell at (0, 0)
and the bottom-right cell at (n - 1, n - 1)
. You are given the integer n
and an integer array startPos
where startPos = [startrow, startcol]
indicates that a robot is initially at cell (startrow, startcol)
.
You are also given a 0-indexed string s
of length m
where s[i]
is the ith
instruction for the robot: 'L'
(move left), 'R'
(move right), 'U'
(move up), and 'D'
(move down).
The robot can begin executing from any ith
instruction in s
. It executes the instructions one by one towards the end of s
but it stops if either of these conditions is met:
Return an array answer
of length m
where answer[i]
is the number of instructions the robot can execute if the robot begins executing from the ith
instruction in s
.
Example 1:
Input: n = 3, startPos = [0,1], s = "RRDDLU" Output: [1,5,4,3,1,0] Explanation: Starting from startPos and beginning execution from the ith instruction: - 0th: "RRDDLU". Only one instruction "R" can be executed before it moves off the grid. - 1st: "RDDLU". All five instructions can be executed while it stays in the grid and ends at (1, 1). - 2nd: "DDLU". All four instructions can be executed while it stays in the grid and ends at (1, 0). - 3rd: "DLU". All three instructions can be executed while it stays in the grid and ends at (0, 0). - 4th: "LU". Only one instruction "L" can be executed before it moves off the grid. - 5th: "U". If moving up, it would move off the grid.
Example 2:
Input: n = 2, startPos = [1,1], s = "LURD" Output: [4,1,0,0] Explanation: - 0th: "LURD". - 1st: "URD". - 2nd: "RD". - 3rd: "D".
Example 3:
Input: n = 1, startPos = [0,0], s = "LRUD" Output: [0,0,0,0] Explanation: No matter which instruction the robot begins execution from, it would move off the grid.
Constraints:
m == s.length
1 <= n, m <= 500
startPos.length == 2
0 <= startrow, startcol < n
s
consists of 'L'
, 'R'
, 'U'
, and 'D'
.class Solution:
def executeInstructions(self, n: int, startPos: List[int], s: str) -> List[int]:
ret = []
def move(step, x, y):
if step == 'L': return (x, y - 1)
if step == 'R': return (x, y + 1)
if step == 'D': return (x + 1, y)
if step == 'U': return (x - 1, y)
for i in range(len(s)):
count = 0
nx, ny = startPos
for start in range(i, len(s)):
nx, ny = move(s[start], nx, ny)
if 0 <= nx < n and 0 <= ny < n:
count += 1
else:
break
ret.append(count)
return ret
문제의 조건에 맞게 그대로 구현하면 된다.