到家的最少跳跃次数

/* Problem Statement: 
   A flea starts at position 0 on a number line and needs to reach its home at position x.
   It can jump forward by 'a' units and backward by 'b' units, but cannot jump to forbidden positions.
   It also cannot make two consecutive backward jumps.
   The goal is to find the minimum number of jumps to reach position x, or return -1 if it's not possible.
*/
import java.util.*;
import java.util.stream.*;

public class FleaJumpStream {
    public int minJumps(int[] forbidden, int a, int b, int x) {
        Set<Integer> forbiddenSet = Arrays.stream(forbidden).boxed().collect(Collectors.toSet());
        Queue<int[]> queue = new LinkedList<>();
        Set<String> visited = new HashSet<>();
        queue.offer(new int[]{0, 0, 0}); // {current position, jumps count, last jump was backward (1 for true)}
        visited.add("0_0");

        int upperBound = 6000;
        while (!queue.isEmpty()) {
            int[] curr = queue.poll();
            int pos = curr[0], jumps = curr[1], lastJump = curr[2];

            if (pos == x) return jumps;

            int forwardPos = pos + a;
            int backwardPos = pos - b;

            if (forwardPos < upperBound && !forbiddenSet.contains(forwardPos) && visited.add(forwardPos + "_0")) {
                queue.offer(new int[]{forwardPos, jumps + 1, 0});
            }

            if (lastJump == 0 && backwardPos >= 0 && !forbiddenSet.contains(backwardPos) && visited.add(backwardPos + "_1")) {
                queue.offer(new int[]{backwardPos, jumps + 1, 1});
            }
        }
        return -1;
    }
}