1306. Jump Game III (M)

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

Solution:

Version 1: BFS

class Solution {
    public boolean canReach(int[] arr, int start) {
        
        if(arr[start] == 0) return true;
        Queue<Integer> queue = new LinkedList<Integer>();
        boolean[] visited = new boolean[arr.length];
        
        queue.offer(start);
        visited[start] = true;
        
        while(!queue.isEmpty())
        {
            int currentIndex = queue.poll();
            if(arr[currentIndex] == 0)
            {
                return true;
            }
            
            int dir1 = currentIndex + arr[currentIndex];
            int dir2 = currentIndex - arr[currentIndex];
            
            if(dir1 < arr.length && !visited[dir1])
            {
                queue.offer(dir1);
                visited[dir1] = true;
            }
            if(dir2 >= 0 && !visited[dir2])
            {
                queue.offer(dir2);
                visited[dir2] = true;
            }
        }

        return false;         
    }
}