LeetCode 1306. Jump Game III

Question

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

Source: https://leetcode.com/problems/jump-game-iii/

Solution

For jump games, if one path reaches a visited point, we can terminate that path (pruning). Because it can be replaced by a previous path.

// BFS, Time Complexity O(n)
public boolean canReach(int[] arr, int start) {
    int len = arr.length;
    boolean[] visited = new boolean[len]; // initialized with false
    Queue<Integer> queue = new LinkedList<>(); // store indexes
    queue.add(start);

    while (!queue.isEmpty()) {
        int currIndex = queue.poll();
        // if current path includes a visited index, it can be replaced
        if (visited[currIndex]) {
            continue;
        }
        if (arr[currIndex] == 0) {
            return true;
        }
        visited[currIndex] = true;
        if (currIndex + arr[currIndex] < len) {
            queue.add(currIndex + arr[currIndex]);
        }
        if (currIndex - arr[currIndex] >= 0) {
            queue.add(currIndex - arr[currIndex]);
        }
    }
    return false;
}

private boolean dfs(int[] arr, boolean[] visited, int start, int len) {
    if (start < 0 || start >= len || visited[start]) {
        return false;
    }
    if (arr[start] == 0) {
        return true;
    }
    visited[start] = true;
    return dfs(arr, visited, start - arr[start], len) ||
            dfs(arr, visited, start + arr[start], len);
}

// DFS, Time Complexity O(n)
public boolean canReach2(int[] arr, int start) {
    int len = arr.length;
    boolean[] visited = new boolean[len]; // initialized with false
    return dfs(arr, visited, start, len);
}