Problem
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 * 10^4
0 <= arr[i] < arr.length
0 <= start < arr.length
Notes
Use dfs to search for 0. Mark the visited place to trigger stop.
Important:
Remember to reset the mark if can not find along the path, so that it can search into another path.
Solution
class Solution:
def canReach(self, arr, start):
if start >= len(arr) or start < 0:
return False
if arr[start] == 0:
return True
if arr[start] == -1:
return False
step = arr[start]
arr[start] = -1
if self.canReach(arr, start - step) or self.canReach(arr, start + step):
return True
else:
arr[start] = step
return False