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 * 1040 <= arr[i] < arr.length0 <= start < arr.lengthclass Solution {
int[] arr;
boolean[] vis;
public boolean canReach(int[] arr, int start) {
this.arr = arr;
int n = arr.length;
this.vis = new boolean[n];
return move(start);
}
private boolean move(int i) {
if (i < 0 || i >= arr.length || vis[i])
return false;
if (arr[i] == 0)
return true;
vis[i] = true;
return move(i + arr[i]) || move(i - arr[i]);
}
}