Minimum Jumps to Reach Home

• Define a recursive function, say dfs(position, jumps, isLastJumpBackward, pathVisited).
• pathVisited is a set of states (position, isLastJumpBackward) visited in the current recursive path to detect cycles.
• The main function initializes a global minimum jumps variable, minJumps, to infinity and calls dfs(0, 0, false, new HashSet<>()).
• Inside dfs:
  • If the current number of jumps is already greater than or equal to minJumps, prune the search by returning.
  • If position == x, a path is found. Update minJumps = min(minJumps, jumps) and return.
  • Check for invalid states: if position is negative, forbidden, too far, or the state (position, isLastJumpBackward) is already in pathVisited, return to avoid cycles and invalid moves.
  • Add the current state to pathVisited to mark it as part of the current path.
  • Recursively call for the next possible moves:
    • Forward jump: dfs(position + a, jumps + 1, false, pathVisited).
    • Backward jump: If isLastJumpBackward is false, dfs(position - b, jumps + 1, true, pathVisited).
  • Backtrack by removing the current state from pathVisited before returning.
• After the initial call returns, if minJumps is still infinity, it means x is unreachable. Otherwise, minJumps holds the answer.

The brute-force Depth First Search (DFS) approach tries to find a solution by exploring every possible path from the starting position. A recursive function is defined to navigate through the state space, where a state is defined by the bug's current position and whether the last jump was backward.

To prevent infinite loops (e.g., 0 -> a -> 0 -> a ...), we must keep track of the states visited along the current path. A major drawback is that this method will explore paths exhaustively. For instance, it might find a very long path to the solution first and will only find the shortest path after exploring many other (potentially shorter) paths. This leads to a very high time complexity, making it unsuitable for the given constraints.

1import java.util.HashSet;2import java.util.Set;3 4class Solution {5    int minJumps = Integer.MAX_VALUE;6    Set<Integer> forbiddenSet;7    int a, b, x;8    int limit;9 10    public int minimumJumps(int[] forbidden, int a, int b, int x) {11        this.forbiddenSet = new HashSet<>();12        int maxForbidden = 0;13        for (int f : forbidden) {14            this.forbiddenSet.add(f);15            maxForbidden = Math.max(maxForbidden, f);16        }17        this.a = a;18        this.b = b;19        this.x = x;20        // A loose upper bound to prune search space21        this.limit = Math.max(x, maxForbidden) + a + b;22 23        // pathVisited stores "pos,isBackward" strings24        dfs(0, 0, false, new HashSet<String>());25 26        return minJumps == Integer.MAX_VALUE ? -1 : minJumps;27    }28 29    private void dfs(int pos, int jumps, boolean isLastBackward, Set<String> pathVisited) {30        if (pos == x) {31            minJumps = Math.min(minJumps, jumps);32            return;33        }34 35        // Pruning conditions36        if (pos < 0 || pos > limit || forbiddenSet.contains(pos) || jumps >= minJumps) {37            return;38        }39 40        String state = pos + "," + isLastBackward;41        if (pathVisited.contains(state)) {42            return; // Cycle detected in current path43        }44 45        pathVisited.add(state);46 47        // Forward jump48        dfs(pos + a, jumps + 1, false, pathVisited);49 50        // Backward jump51        if (!isLastBackward) {52            dfs(pos - b, jumps + 1, true, pathVisited);53        }54 55        pathVisited.remove(state); // Backtrack56    }57}

• Initialize a Set with all forbidden positions for O(1) lookups.
• Determine a safe upper bound for the search space. A value like max(x, max(forbidden)) + a + b is sufficient to ensure we don't miss any potential solutions.
• Initialize a queue for BFS and add the starting state (position=0, lastJumpWasBackward=false).
• Use a 2D boolean array visited[position][wasBackward] to keep track of visited states to avoid redundant computations and cycles.
• Start the BFS loop, proceeding level by level. Each level corresponds to one additional jump.
• In each level, dequeue all states added in the previous level.
• For each dequeued state (pos, lastWasBackward):
  • If pos == x, the shortest path is found. Return the current jump count.
  • Explore the forward jump: Calculate nextPos = pos + a. If nextPos is within the bounds, not forbidden, and the state (nextPos, false) has not been visited, add it to the queue and mark it as visited.
  • Explore the backward jump: If lastWasBackward is false, calculate nextPos = pos - b. If nextPos is non-negative, not forbidden, and the state (nextPos, true) has not been visited, add it to the queue and mark it as visited.
• If the queue becomes empty and x has not been reached, it's impossible to get home. Return -1.

The key insight is to represent the problem as a shortest path problem on a state graph. A state is defined by a pair: (position, was_last_jump_backward). The was_last_jump_backward flag is crucial because it determines whether a backward jump is allowed from the current position.

We start a BFS from the initial state (0, false). A queue stores the states to be visited. To avoid infinite loops and redundant work, we use a visited data structure (e.g., a 2D boolean array) to keep track of states we've already processed. The BFS proceeds in levels, where each level corresponds to one jump. When we first encounter the target position x, we are guaranteed to have found a path with the minimum number of jumps because BFS explores all paths of length k before exploring any path of length k+1.

An important consideration is the search space boundary. The bug can jump beyond x. We need to define a reasonable upper limit for positions to explore. A safe and effective limit is max(x, max(forbidden)) + a + b, which is well within the limits imposed by the problem constraints.

1import java.util.*;2 3class Solution {4    public int minimumJumps(int[] forbidden, int a, int b, int x) {5        if (x == 0) {6            return 0;7        }8 9        Set<Integer> forbiddenSet = new HashSet<>();10        int maxVal = x;11        for (int f : forbidden) {12            forbiddenSet.add(f);13            maxVal = Math.max(maxVal, f);14        }15 16        // A safe upper bound for position to explore.17        int limit = maxVal + a + b;18 19        Queue<int[]> queue = new LinkedList<>();20        // State: {position, isLastJumpBackward} (0 for false, 1 for true)21        queue.offer(new int[]{0, 0});22 23        // visited[position][isLastJumpBackward]24        boolean[][] visited = new boolean[limit + 1][2];25        visited[0][0] = true;26 27        int jumps = 0;28        while (!queue.isEmpty()) {29            int levelSize = queue.size();30            for (int i = 0; i < levelSize; i++) {31                int[] current = queue.poll();32                int pos = current[0];33                int lastWasBackward = current[1];34 35                if (pos == x) {36                    return jumps;37                }38 39                // Try forward jump40                int forwardPos = pos + a;41                if (forwardPos <= limit && !visited[forwardPos][0] && !forbiddenSet.contains(forwardPos)) {42                    visited[forwardPos][0] = true;43                    queue.offer(new int[]{forwardPos, 0});44                }45 46                // Try backward jump (only if last jump was not backward)47                if (lastWasBackward == 0) {48                    int backwardPos = pos - b;49                    if (backwardPos >= 0 && !visited[backwardPos][1] && !forbiddenSet.contains(backwardPos)) {50                        visited[backwardPos][1] = true;51                        queue.offer(new int[]{backwardPos, 1});52                    }53                }54            }55            jumps++;56        }57 58        return -1;59    }60}