Given an array of non-negative integers arr
, you are initially positioned at start
index of the array. When you are at index i
, you can jump to i + arr[i]
or i - arr[i]
, check if you can reach to any index with value 0.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [4,2,3,0,3,1,2], start = 5
Output: true
Explanation:
All possible ways to reach at index 3 with value 0 are:
index 5 -> index 4 -> index 1 -> index 3
index 5 -> index 6 -> index 4 -> index 1 -> index 3
Example 2:
Input: arr = [4,2,3,0,3,1,2], start = 0
Output: true
Explanation:
One possible way to reach at index 3 with value 0 is:
index 0 -> index 4 -> index 1 -> index 3
Example 3:
Input: arr = [3,0,2,1,2], start = 2
Output: false
Explanation: There is no way to reach at index 1 with value 0.
Constraints:
-
1 <= arr.length <= 5 * 104
-
0 <= arr[i] < arr.length
-
0 <= start < arr.length
SOLUTION:
class Solution:
def canReach(self, arr: List[int], start: int) -> bool:
n = len(arr)
paths = [[start]]
visited = set()
while len(paths) > 0:
curr = paths.pop(0)
currIndex = curr[-1]
jump = arr[currIndex]
if jump == 0:
return True
visited.add(currIndex)
if currIndex - jump >= 0 and (currIndex - jump) not in visited:
paths.append(path + [currIndex - jump])
if currIndex + jump <= n - 1 and (currIndex + jump) not in visited:
paths.append(path + [currIndex + jump])
return False
Top comments (0)