Solution Explanation: Minimum Jumps to Reach Home

This problem asks for the minimum number of jumps a bug needs to reach its home at position x, starting from position 0. The bug can jump a units forward, b units backward (but not two backward jumps consecutively), and cannot land on any forbidden positions listed in forbidden.

The optimal approach is Breadth-First Search (BFS). BFS guarantees finding the shortest path in an unweighted graph (like this one where each jump is a single edge).

Algorithm:

1. State Representation: We represent the bug's state as a tuple (position, direction), where position is the bug's current location, and direction is 1 (forward) or 0 (backward). The direction helps enforce the "no consecutive backward jumps" rule.

2. Queue and Visited Set: A queue q stores states to explore, starting with (0, 1). A set vis tracks visited states to avoid cycles. The maximum position to consider is 6000, as explained below.

3. BFS Iteration: The main loop repeatedly dequeues states from q.

4. Jump Actions: For each state:

   • It checks if the current position is x. If so, it returns the number of steps taken (the BFS level).
   • It generates possible next states:
     • Always jump forward (position + a, 1).
     • If the last jump was forward (direction was 1), also jump backward (position - b, 0).

5. Validity Check: Each next state is checked for validity:

   • Position must be within the bounds [0, 6000) (explanation below).
   • Position must not be in forbidden.
   • State must not have been visited.

6. Queueing and Visiting: If a next state is valid, it's added to q and marked as visited.

7. Failure Condition: If q becomes empty before finding x, there's no solution, and -1 is returned.

Why the upper bound is 6000:

The maximum reachable position is constrained. If a >= b, the furthest the bug can go is x + b because further forward jumps won't be helpful in reaching x. If a < b, a more complex analysis is needed, but a value of 6000 is sufficient as an upper bound in this problem's constraints (max position and max jump lengths).

Time Complexity: O(M), where M is the upper bound (6000 in this case). In the worst case, we might visit all possible states within the bounds.

Space Complexity: O(M), due to the queue and visited set, which can store up to M states.

Code (Python):

from collections import deque
 
def minimumJumps(forbidden, a, b, x):
    s = set(forbidden)
    q = deque([(0, 1)])
    vis = {(0, 1)}
    ans = 0
    while q:
        for _ in range(len(q)):
            i, k = q.popleft()
            if i == x:
                return ans
            nxt = [(i + a, 1)]
            if k == 1:
                nxt.append((i - b, 0))
            for j, k in nxt:
                if 0 <= j < 6000 and j not in s and (j, k) not in vis:
                    q.append((j, k))
                    vis.add((j, k))
        ans += 1
    return -1
 

The code in other languages (Java, C++, Go, TypeScript) follows the same algorithm and logic, differing only in syntax and data structure implementations. They all achieve the same time and space complexity.