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 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
class Solution:
def canReach(self, arr: List[int], start: int) -> bool:
def is_valid(index):
return 0 <= index < n
n = len(arr)
queue = [start]
visited = set()
while queue:
new_q = []
for i in queue:
visited.add(i)
if arr[i] == 0:
return True
for i in queue:
ls = i - arr[i]
rs = i + arr[i]
if is_valid(ls) and ls not in visited:
visited.add(ls)
new_q.append(ls)
if is_valid(rs) and rs not in visited:
visited.add(rs)
new_q.append(rs)
queue = new_q
return Falseclass Solution:
def canReach(self, arr: List[int], start: int) -> bool:
def is_valid(index):
return 0 <= index < n
n = len(arr)
visited = set()
def Reach(num):
if not is_valid(num) or num in visited:
return False
if arr[num] == 0:
return True
visited.add(num)
return Reach(num - arr[num]) or Reach(num + arr[num])
return Reach(start)