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

// BFS
bool canReach(vector<int>& arr, int start) { // time: O(n); space: O(n)
    int n = arr.size();
    queue<int> q({start});
    // unordered_set<int> visited;
    vector<bool> visited(n, false); // since 0 <= arr[i] < arr.length
    while(!q.empty()) {
        int idx = q.front(); q.pop();
        if (arr[idx] == 0) return true;
        // if (visited.count(idx)) continue;
        // visited.insert(idx);
        if (visited[idx]) continue;
        visited[idx] = true;
        int backward_jump = idx - arr[idx];
        int forward_jump = idx + arr[idx];
        // if (backward_jump >= 0 && !visited.count(backward_jump)) q.push(backward_jump);
        // if (forward_jump < n && !visited.count(forward_jump)) q.push(forward_jump);
        if (backward_jump >= 0 && !visited[backward_jump]) q.push(backward_jump);
        if (forward_jump < n && !visited[forward_jump]) q.push(forward_jump);
    }
    return false;
}