<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 {
public:
    bool canReach(vector<int>& arr, int start) {
     int n=arr.size();
    queue<int>q;
    vector<int>vis(n,0);
   q.push(start);
    vis[start]=1;
    
   while(!q.empty()){
      
    int node =q.front();
       cout<<node<<endl;
     q.pop();
     if(arr[node]==0) return true;
     int x=node+arr[node];
     int y=node-arr[node];
       
     if(x>=0&&x<n&&vis[x]==0){
         q.push(x);
          vis[x]=1;
         }
    if(y>=0&&y<n&&vis[y]==0){

         q.push(y);

          vis[y]=1;

         }
       }

       
    return false;   
       
        
    }
};
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