LeetCode-in-Java

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 * 104
• 0 <= arr[i] < arr.length
• 0 <= start < arr.length

Solution

public class Solution {
    private boolean[] dp;
    private boolean found = false;

    public boolean canReach(int[] arr, int start) {
        if (arr[start] == 0) {
            return true;
        }
        dp = new boolean[arr.length];
        dp[start] = true;
        recurse(arr, start);
        return found;
    }

    private void recurse(int[] arr, int index) {
        if (found) {
            return;
        }
        if (index - arr[index] >= 0 && !dp[index - arr[index]]) {
            if (arr[index - arr[index]] == 0) {
                found = true;
                return;
            }
            dp[index - arr[index]] = true;
            recurse(arr, index - arr[index]);
        }
        if (index + arr[index] < arr.length && !dp[index + arr[index]]) {
            if (arr[index + arr[index]] == 0) {
                found = true;
                return;
            }
            dp[index + arr[index]] = true;
            recurse(arr, index + arr[index]);
        }
    }
}

This site is open source. Improve this page.