Formatted question description: https://leetcode.ca/all/1306.html

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

Related Topics:
Breadth-first Search, Graph

Similar Questions:

Solution 1. DFS

// OJ: https://leetcode.com/problems/jump-game-iii/
// Time: O(N)
// Space: O(N)
class Solution {
public:
    bool canReach(vector<int>& A, int start) {
        if (start < 0 || start >= A.size() || A[start] < 0) return false;
        if (A[start] == 0) return true;
        A[start] *= -1;
        return canReach(A, start + A[start]) || canReach(A, start - A[start]);
    }
};

Solution 2. BFS

// OJ: https://leetcode.com/problems/jump-game-iii/
// Time: O(N)
// Space: O(N)
class Solution {
public:
    bool canReach(vector<int>& A, int start) {
        queue<int> q;
        q.push(start);
        while (q.size()) {
            int i = q.front();
            q.pop();
            if (A[i] == 0) return true;
            if (A[i] < 0) continue;
            if (i + A[i] < A.size()) q.push(i + A[i]);
            if (i - A[i] >= 0) q.push(i - A[i]);
            A[i] *= -1;
        }
        return false;
    }
};

Java

class Solution {
    public boolean canReach(int[] arr, int start) {
        int length = arr.length;
        boolean[] canReach = new boolean[length];
        canReach[start] = true;
        Queue<Integer> queue = new LinkedList<Integer>();
        queue.offer(start);
        while (!queue.isEmpty()) {
            int index = queue.poll();
            int step = arr[index];
            int left = index - step, right = index + step;
            if (left >= 0) {
                if (arr[left] == 0)
                    return true;
                else if (!canReach[left]) {
                    canReach[left] = true;
                    queue.offer(left);
                }
            }
            if (right < length) {
                if (arr[right] == 0)
                    return true;
                else if (!canReach[right]) {
                    canReach[right] = true;
                    queue.offer(right);
                }
            }
        }
        return false;
    }
}

// OJ: https://leetcode.com/problems/jump-game-iii/
// Time: O(N)
// Space: O(N)
class Solution {
public:
    bool canReach(vector<int>& A, int start) {
        if (start < 0 || start >= A.size() || A[start] < 0) return false;
        if (A[start] == 0) return true;
        A[start] *= -1;
        return canReach(A, start + A[start]) || canReach(A, start - A[start]);
    }
};

```
print("Todo!")
```

1306 - Jump Game III

1306. Jump Game III (Medium)

Solution 1. DFS

Solution 2. BFS

All Problems

All Solutions