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LeetCode 1306: Jump Game III Solution

Master LeetCode problem 1306 (Jump Game III), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

Problem Explanation

Explanation:

To solve this problem, we can use a breadth-first search (BFS) algorithm starting from the start index. We will keep track of the indices we have visited and explore all possible jumps from each index until we reach an index with value 0 or until there are no more valid jumps left.

Algorithm:

  1. Initialize a queue with the start index and a set to keep track of visited indices.
  2. While the queue is not empty, pop an index from the queue and check if its value is 0. If so, return true.
  3. Otherwise, explore all possible jumps from the current index and add the new indices to the queue if they are within bounds and have not been visited before.
  4. If no index with value 0 is found after exploring all possible paths, return false.

Time Complexity: O(N) where N is the number of elements in the array. Space Complexity: O(N) where N is the number of elements in the array.

:

Solution Code

class Solution {
    public boolean canReach(int[] arr, int start) {
        Queue<Integer> queue = new LinkedList<>();
        Set<Integer> visited = new HashSet<>();
        
        queue.offer(start);
        
        while (!queue.isEmpty()) {
            int currentIndex = queue.poll();
            
            if (arr[currentIndex] == 0) {
                return true;
            }
            
            visited.add(currentIndex);
            
            int nextIndex1 = currentIndex + arr[currentIndex];
            int nextIndex2 = currentIndex - arr[currentIndex];
            
            if (nextIndex1 >= 0 && nextIndex1 < arr.length && !visited.contains(nextIndex1)) {
                queue.offer(nextIndex1);
            }
            
            if (nextIndex2 >= 0 && nextIndex2 < arr.length && !visited.contains(nextIndex2)) {
                queue.offer(nextIndex2);
            }
        }
        
        return false;
    }
}

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