<p>Given an array of non-negative integers <code>arr</code>, you are initially positioned at <code>start</code>&nbsp;index of the array. When you are at index <code>i</code>, you can jump&nbsp;to <code>i + arr[i]</code> or <code>i - arr[i]</code>, check if you can reach&nbsp;<strong>any</strong> index with value 0.</p>

<p>Notice that you can not jump outside of the array at any time.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> arr = [4,2,3,0,3,1,2], start = 5
<strong>Output:</strong> true
<strong>Explanation:</strong> 
All possible ways to reach at index 3 with value 0 are: 
index 5 -&gt; index 4 -&gt; index 1 -&gt; index 3 
index 5 -&gt; index 6 -&gt; index 4 -&gt; index 1 -&gt; index 3 
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> arr = [4,2,3,0,3,1,2], start = 0
<strong>Output:</strong> true 
<strong>Explanation: 
</strong>One possible way to reach at index 3 with value 0 is: 
index 0 -&gt; index 4 -&gt; index 1 -&gt; index 3
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> arr = [3,0,2,1,2], start = 2
<strong>Output:</strong> false
<strong>Explanation: </strong>There is no way to reach at index 1 with value 0.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= arr.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>0 &lt;= arr[i] &lt;&nbsp;arr.length</code></li>
	<li><code>0 &lt;= start &lt; arr.length</code></li>
</ul>

class Solution:
    def canReach(self, arr: List[int], start: int) -> bool:
        seen = set()
        L =  len(arr)
        
        def dfs(node):
            if arr[node] == 0:
                return True
            
            if node in seen:
                return False
        
            seen.add(node)
            

            left =  0 <= node + arr[node] < L and dfs(node + arr[node])
            right = 0 <= node - arr[node] < L and dfs(node - arr[node])
            
            return left or right
                
        
        return dfs(start)
                

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