1306

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

Companies: Pinterest, Microsoft, Amazon

Related Topics:
Array, Depth-First Search, Breadth-First Search

Similar Questions:

• Jump Game II (Medium)
• Jump Game (Medium)
• Jump Game VII (Medium)
• Jump Game VIII (Medium)
• Maximum Number of Jumps to Reach the Last Index (Medium)

Solution 1. DFS

// OJ: https://leetcode.com/problems/jump-game-iii/
// Author: github.com/lzl124631x
// 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/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
public:
    bool canReach(vector<int>& A, int start) {
        queue<int> q;
        q.push(start);
        A[start] *= -1;
        while (q.size()) {
            int i = q.front();
            q.pop();
            if (A[i] == 0) return true;
            int d = -A[i];
            if (i + d < A.size() && A[i + d] >= 0) {
                A[i + d] *= -1;
                q.push(i + d);
            }
            if (i - d >= 0 && A[i - d] >= 0) {
                A[i - d] *= -1;
                q.push(i - d);
            }
        }
        return false;
    }
};