1306. Jump Game III

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

Analysis

This is a plain reachability problem on an implicit graph, where each index i has up to two outgoing edges: i + arr[i] and i - arr[i]. We run a standard BFS (or DFS) starting from start, marking visited indices to avoid cycles, and stop as soon as we pop an index whose value is 0.

• Time: $O(n)$, since each index is visited at most once.
• Space: $O(n)$ for the visited array and the queue.

Code

class Solution {
public:
    bool canReach(vector<int>& arr, int start) {
        int n = arr.size();
        vector<bool> visited(n, false);
        queue<int> q;
        q.push(start);
        visited[start] = true;
        while (!q.empty()) {
            int i = q.front();
            q.pop();
            if (arr[i] == 0) return true;
            for (int next : {i + arr[i], i - arr[i]}) {
                if (next >= 0 && next < n && !visited[next]) {
                    visited[next] = true;
                    q.push(next);
                }
            }
        }
        return false;
    }
};