LeetCode 1654: Minimum Jumps to Reach Home Solution

Master LeetCode problem 1654 (Minimum Jumps to Reach Home), 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.

1654. Minimum Jumps to Reach Home

Medium

Array Dynamic Programming Breadth-First Search

Problem Explanation

Explanation:

To solve this problem, we can use a breadth-first search (BFS) approach. We will start from position 0 and explore all possible valid moves the bug can make. We will keep track of the visited positions and the number of jumps required to reach each position. If the bug reaches the home position x, we return the number of jumps required. If it is not possible to reach x, we return -1.

Algorithm:

Initialize a queue for BFS and a set to track visited positions.
Perform BFS starting from position 0 with 0 jumps.
At each step, consider two possible moves: a jump forward and a jump backward.
Check if the new positions are valid and not in the forbidden set.
If the bug reaches position x, return the number of jumps.
If the queue is empty and x is not reached, return -1.

Time Complexity:

The time complexity of this algorithm is O(max(x, max(forbidden))).

Space Complexity:

The space complexity of this algorithm is O(max(x, max(forbidden))).

Solution Code

class Solution {
    public int minimumJumps(int[] forbidden, int a, int b, int x) {
        Set<Integer> visited = new HashSet<>();
        Set<Integer> forbiddenSet = new HashSet<>();
        for (int f : forbidden) {
            forbiddenSet.add(f);
        }

        Queue<int[]> queue = new LinkedList<>();
        queue.offer(new int[]{0, 0});
        visited.add(0);

        int maxPos = x + 2 * a + b;

        while (!queue.isEmpty()) {
            int[] current = queue.poll();
            int pos = current[0];
            int jumps = current[1];

            if (pos == x) {
                return jumps;
            }

            if (pos + a < maxPos && !forbiddenSet.contains(pos + a) && visited.add(pos + a)) {
                queue.offer(new int[]{pos + a, jumps + 1});
            }

            if (pos - b >= 0 && pos - b != a && !forbiddenSet.contains(pos - b) && visited.add(pos - b)) {
                queue.offer(new int[]{pos - b, jumps + 1});
            }
        }

        return -1;
    }
}

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