# 1306. Jump Game III

## Medium

***

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`

```python
class Solution:
    def __init__(self):
        self.visited = set()
    def canReach(self, arr: List[int], start: int) -> bool:
        if start < 0 or start >= len(arr):
            return False
        # print(start)
        self.visited.add(start)
        if arr[start] == 0:
            return True
        left = False
        right = False
        if (start-arr[start]) not in self.visited:
            left = self.canReach(arr, start - arr[start])
        if (start + arr[start]) not in self.visited:
            right = self.canReach(arr, start + arr[start])
        return left or right
```