LeetCode [1306] Jump Game III

 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 * 10^4
• 0 <= arr[i] < arr.length
• 0 <= start < arr.length

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class Solution { int n; Set<Integer> visited = new HashSet<>(); public boolean canReach(int[] arr, int start) { this.n = arr.length; visited.add(start); return dfs(arr, start); } boolean dfs(int[] arr, int node){ if(arr[node]==0) return true; int left = node-arr[node], right = node+arr[node]; if(left>=0 && !visited.contains(left)){ visited.add(left); if(dfs(arr, left)) return true; } if(right<n && !visited.contains(right)){ visited.add(right); if(dfs(arr, right)) return true; } return false; } }