1654. Minimum Jumps to Reach Home
Problem Description
In this problem, we have to find the minimum number of jumps a bug needs to take to get from position 0 to its home at a specified position x
on the x-axis. The bug has specific rules for jumping; it can jump forward by a
positions, or backward by b
positions, but cannot jump backward twice in a row. Additionally, there are forbidden positions the bug cannot land on, provided in an array forbidden
. The goal is to calculate the fewest jumps required for the bug to reach position x
. If the bug cannot reach its home following the jump rules and avoiding forbidden positions, we should return -1.
Flowchart Walkthrough
Let's follow the algorithm selection using the Flowchart. Here's a step-by-step analysis for LeetCode 1654. Minimum Jumps to Reach Home:
Is it a graph?
- Yes: The problem can be modeled as a graph where each position x is a node and each jump (either forward or backward) to other positions represents edges.
Is it a tree?
- No: The problem isn't about hierarchies or branching structures without cycles typical of trees. We're essentially dealing with various possible jump paths from a starting point.
Is the problem related to directed acyclic graphs (DAGs)?
- No: We're not specifically dealing with dependencies or topological ordering, which are characteristics of problems related to DAGs.
Is the problem related to shortest paths?
- Yes: The goal is to find the minimum number of jumps to reach a specific home location, similar to finding the shortest path in a graph.
Is the graph weighted?
- No: Each possible jump (forwards and backwards) can be considered equally weighted, as each counts as a single step.
Conclusion: The flowchart suggests using BFS for this shortest path problem in an unweighted graph, where BFS is efficient for tracking levels or layers of nodes, exploring all possible moves uniformly.
Intuition
The problem can be approached using the Breadth-First Search (BFS) algorithm. The idea behind using BFS here is that it can help us traverse the possible positions level by level, starting from 0, with each level representing a jump, until we reach the desired position x
. This approach ensures that we get the minimum number of jumps because we visit nodes in order of their distance from the starting point.
To implement BFS, we typically use a queue to keep track of the positions we need to visit and the direction of the last jump (forward or backward). We also use a set to keep track of the forbidden positions, and another set to record visited positions along with the direction (this is important to avoid repeat visits to the same spot with the same last jump direction).
The thought process includes the following considerations:
- We treat each position and its last jump direction as a state
(position, direction)
, wheredirection
can be 1 for a forward jump and 0 for a backward jump. - The queue is initialized with the starting state
(0, 1)
, meaning the bug starts at position 0 and can jump in either direction. - A loop is used to process all states in the queue, where for each state:
- If the current position is
x
, we can return the current number of jumps (the level number in BFS terms). - We generate possible next states from the current state, one for jumping forward and, if it's allowed, one for jumping backward.
- For each next state, if the new position is non-negative, not forbidden, not already visited, and less than a certain threshold (
6000
in this case, set to prevent infinite searching where the bug keeps jumping forward and backward), we add it to the queue for the next level.
- If the current position is
- If none of the states in the queue lead us to the position
x
, and there are no more states to consider, we return -1, as it's impossible to reach the bug's home.
Through BFS, we guarantee we're exploring the shortest path first, and once we reach the home, we immediately have the minimum number of jumps that it took to get there.
Learn more about Breadth-First Search and Dynamic Programming patterns.
Solution Approach
The solution uses the Breadth-First Search algorithm (BFS) to explore positions the bug can jump to. The algorithm uses the following data structures and concepts:
- Queue: A
deque
is used to implement the queue functionality needed for BFS. It stores tuples in the form(position, direction)
, where 'position' is the current position of the bug and 'direction' is a binary flag indicating the last jump (1 for forward and 0 for backward). - Set: Two sets are used:
- One set
s
to store theforbidden
positions which the bug should not jump to. - Another set
vis
(visited) to store positions along with the last jump's direction the bug has already been to in order to prevent revisiting the same states.
- One set
- Loop: The code iterates over each level of the queue until it is empty or the home position
x
is found. This represents a single jump made by the bug.
Hereās a breakdown of how the code works:
-
Initialize a set
s
with forbidden positions to efficiently check if a given position is forbidden inO(1)
time. -
Start the BFS from position 0 with a direction of 1 (forward), indicating that the bug can jump in any direction from the start.
-
Use a
while
loop to process all positions in the queue:- For each position in the queue at the current level (
ans
), there can be two possible actions: jump forward or jump backward (if not already taken as the last action). - Create a
nxt
list to hold possible next states. - Add a state for moving forward (position +
a
). - If the last jump was not backward (checked by
if k & 1:
), also add a state for moving backward (position -b
).
- For each position in the queue at the current level (
-
For each of the next states:
- Check if the new position is non-negative, not forbidden, and not yet visited. There is also a threshold of 6000 to limit the spaceāpositions beyond this are not considered.
- If all conditions are met, add the new state
(j, k)
to the queue to explore in the next level and mark it as visited in thevis
set.
-
Increment the counter
ans
after each level is processed, which represents the number of jumps made. -
If at any point the current position
i
is the home positionx
, return the current jump counterans
as the minimum number of jumps needed.
If the queue is exhausted and the home position x
has not been reached, return -1
, indicating that it is impossible for the bug to reach its home.
This BFS approach ensures that the shortest path to position x
is found first, accounting for forbidden positions and jump rules, thus providing the correct minimum number of jumps required.
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Let's consider a small example to illustrate the solution approach with the following parameters:
- Position
x
(home) is at 10. - The bug jumps forward by
a = 6
and backward byb = 4
. - Forbidden positions are given by the array
forbidden = [14]
.
Initial Setup
- Create a set
s
with the forbidden positions:s = {14}
. - Initialize the starting state in the queue
deque
as(0, 1)
, meaning the bug is at position 0 and the last jump was forward (arbitrarily since the bug is at start), and initializeans = 0
which will keep count of the jumps.
BFS Traversal
- The queue initially contains one state:
(0, 1)
. This is the initial position with a forward jump.
First Jump
- Dequeue the first state
(0, 1)
. From this state, the bug can either jump to position0 + 6 = 6
or position0 - 4
, but it will not jump backward since that would result in a negative position. - Check if position 6 is forbidden or not (
6 not in s
) and not visited (not invis
). Since it's not forbidden and not visited, enqueue(6, 1)
into the queue and add it to the setvis
. - Increment
ans
to 1.
Continue BFS
- Dequeue
(6, 1)
. Now, the bug can jump to position12
by moving forward or2
by moving backward. Position12
is not forbidden but is beyond the target10
, so we skip it as it's not efficient. Position2
is not forbidden and not visited, therefore, enqueue(2, 0)
into the queue and add it tovis
. - Increment
ans
to 2.
Check for Home Position
- Dequeue
(2, 0)
. The bug can jump to8
(moving forward) or2 - 4 = -2
(moving backward), but backward is not allowed since it was the last jump direction. - The jump to
8
does not violate any constraints (forbidden, visited, or threshold). Add(8, 1)
to the queue andvis
. - Increment
ans
to 3.
Home Found
- Dequeue
(8, 1)
. The bug can now jump to14
(forward) or4
(backward). We skip14
as it's forbidden, but4
is not. - Now the bug can jump from
4
to10
with a forward jump. We check if10
is forbidden or visited. It's neither, and10
is our target home position. - Return the current jump counter
ans
, which is 4. This implies that the minimum number of jumps needed for the bug to reach its home at position10
is 4.
If Home Not Found
- If the bug had not found the home at any point, and all possible jumps had either been processed or discarded due to constraints, we would simply exit the loop and return
-1
.
The above steps demonstrate through a simple example how a BFS approach is used to find the minimum number of jumps required for the bug to reach home while avoiding forbidden positions and not jumping backward twice in a row.
Solution Implementation
1from collections import deque
2
3class Solution:
4 def minimumJumps(self, forbidden: List[int], forward_step: int, backward_step: int, target: int) -> int:
5 # Initialize a set of forbidden positions
6 forbidden_set = set(forbidden)
7
8 # Queue for breadth-first search with position and whether the last movement was forward (1) or backward (0)
9 queue = deque([(0, 1)])
10
11 # Visited set to keep track of positions and their last movement direction
12 visited = {(0, 1)}
13
14 # Step counter starting from 0
15 steps = 0
16
17 # Start breadth-first search
18 while queue:
19 # Process nodes level by level
20 for _ in range(len(queue)):
21 position, last_movement_forward = queue.popleft()
22
23 # If the current position is the target, return the number of steps
24 if position == target:
25 return steps
26
27 # List of possible next positions
28 next_positions = [(position + forward_step, 1)] # always can go forward
29
30 # Can only go backward if the last movement was forward
31 if last_movement_forward:
32 next_positions.append((position - backward_step, 0))
33
34 # Iterate over possible next positions
35 for next_position, next_movement_forward in next_positions:
36 # Check if within bounds, not forbidden, and not already visited
37 if (0 <= next_position < 6000 and
38 next_position not in forbidden_set and
39 (next_position, next_movement_forward) not in visited):
40
41 # Add to queue and mark as visited
42 queue.append((next_position, next_movement_forward))
43 visited.add((next_position, next_movement_forward))
44
45 # Increment steps after finishing processing the current level
46 steps += 1
47
48 # Return -1 if it's not possible to reach the target
49 return -1
50
1class Solution {
2 public int minimumJumps(int[] forbidden, int forwardJump, int backwardJump, int target) {
3 // Create a hash set to store forbidden positions for fast access
4 Set<Integer> forbiddenSet = new HashSet<>();
5 for (int position : forbidden) {
6 forbiddenSet.add(position);
7 }
8
9 // Declare a deque to perform BFS
10 Deque<int[]> queue = new ArrayDeque<>();
11 queue.offer(new int[] {0, 1}); // Starting at position 0; 1 indicates next jump can be either forward or backward
12
13 // Define upper bound of positions to avoid infinite searching
14 final int MAX_POSITION = 6000;
15
16 // Create a visited matrix to track visited positions; two states for forward [][1] or backward [][0] jump
17 boolean[][] visited = new boolean[MAX_POSITION][2];
18 visited[0][1] = true; // We start at position 0 and can move either forward or backward
19
20 // BFS to find minimum jumps to reach the target
21 for (int steps = 0; !queue.isEmpty(); ++steps) {
22 // Process nodes that are at the same level
23 for (int size = queue.size(); size > 0; --size) {
24 int[] currentPosition = queue.poll();
25 int position = currentPosition[0];
26 int canJumpBackward = currentPosition[1];
27
28 // If current position is the target, return the number of steps taken
29 if (position == target) {
30 return steps;
31 }
32
33 // Store next positions from current position
34 List<int[]> nextPositions = new ArrayList<>();
35 nextPositions.add(new int[] {position + forwardJump, 1}); // Always can jump forward
36
37 // Check if we can jump backward from the current position
38 if (canJumpBackward == 1) {
39 nextPositions.add(new int[] {position - backwardJump, 0}); // After jumping back, can't jump back again right away
40 }
41
42 // Explore next positions
43 for (int[] nextPos : nextPositions) {
44 int newPosition = nextPos[0];
45 int newCanJumpBackward = nextPos[1];
46
47 // Validate new position (not forbidden, within bounds, and not visited)
48 if (newPosition >= 0 &&
49 newPosition < MAX_POSITION &&
50 !forbiddenSet.contains(newPosition) &&
51 !visited[newPosition][newCanJumpBackward]) {
52
53 // Add valid position to the queue and mark it as visited
54 queue.offer(new int[] {newPosition, newCanJumpBackward});
55 visited[newPosition][newCanJumpBackward] = true;
56 }
57 }
58 }
59 }
60
61 // If we exhaust the queue without reaching the target, return -1 indicating it's impossible
62 return -1;
63 }
64}
65
1#include <vector>
2#include <unordered_set>
3#include <queue>
4#include <cstring>
5
6class Solution {
7public:
8 int minimumJumps(std::vector<int>& forbidden, int forwardJump, int backwardJump, int target) {
9 // Transform the forbidden list into a set for fast lookup.
10 std::unordered_set<int> forbiddenSet(forbidden.begin(), forbidden.end());
11
12 // Queue to hold the current position and jump direction (1 for forward, 0 for backward).
13 std::queue<std::pair<int, int>> queue;
14 queue.emplace(0, 1); // Start with position 0 and direction forward.
15
16 // Define the maximum possible position to consider, to avoid infinite searching.
17 const int MAX_POSITION = 6000;
18
19 // Visited positions array to keep track of places already visited and direction used.
20 bool visited[MAX_POSITION][2];
21
22 // Initialize the visited array to false.
23 std::memset(visited, false, sizeof(visited));
24 visited[0][1] = true; // Mark the start (0, forward) as visited.
25
26 // Variable to track the number of jumps.
27 int jumpCount = 0;
28
29 // BFS algorithm to find the minimum number of jumps to reach the target.
30 while (!queue.empty()) {
31 for (int levelSize = queue.size(); levelSize > 0; --levelSize) {
32 auto [currentPosition, canJumpBackward] = queue.front();
33 queue.pop();
34
35 // If the target position is reached, return the number of jumps.
36 if (currentPosition == target) {
37 return jumpCount;
38 }
39
40 // Calculate the next forward position.
41 int nextForwardPosition = currentPosition + forwardJump;
42
43 // Consider the next positions to move (forward and maybe backward).
44 std::vector<std::pair<int, int>> nextPositions = {{nextForwardPosition, 1}};
45 if (canJumpBackward) {
46 int nextBackwardPosition = currentPosition - backwardJump;
47 nextPositions.emplace_back(nextBackwardPosition, 0);
48 }
49
50 // Process each valid next position.
51 for (auto [nextPosition, nextDirection] : nextPositions) {
52 // Check if the next position is within bounds, not forbidden, and not visited.
53 if (nextPosition >= 0 && nextPosition < MAX_POSITION && !forbiddenSet.count(nextPosition) && !visited[nextPosition][nextDirection]) {
54 visited[nextPosition][nextDirection] = true; // Mark the next position as visited.
55 queue.emplace(nextPosition, nextDirection); // Add to queue for further exploration.
56 }
57 }
58 }
59 jumpCount++; // Increase the jump count for the next level.
60 }
61
62 // If we exited the loop, there is no way to reach the target.
63 return -1;
64 }
65};
66
1function minimumJumps(forbidden: number[], forwardStep: number, backwardStep: number, target: number): number {
2 // A set to hold forbidden positions for quick access.
3 const forbiddenPositions: Set<number> = new Set(forbidden);
4
5 // Initialize a queue to perform BFS with a tuple containing position and the direction flag (forward = 1, backward = 0).
6 const queue: [number, number][] = [[0, 1]];
7
8 // Define a boundary since the problem allows for a finite search space.
9 const maxPosition = 6000;
10
11 // Visited array to keep track of positions already checked.
12 // vis[position][direction] indicates whether the position has been visited in a particular direction.
13 const visited: boolean[][] = Array.from({ length: maxPosition }, () => [false, false]);
14
15 // Mark the start position [0,1] as visited.
16 visited[0][1] = true;
17
18 // Start the BFS.
19 for (let steps = 0; queue.length; ++steps) {
20 for (let size = queue.length; size > 0; --size) {
21 // Dequeue the next position and direction.
22 const [position, direction] = queue.shift()!;
23
24 // If target is reached, return the number of steps taken.
25 if (position === target) {
26 return steps;
27 }
28
29 // Calculate the next positions to visit.
30 const nextPositions: [number, number][] = [[position + forwardStep, 1]];
31
32 // If the last step was forward, we can go backward.
33 if (direction === 1) {
34 nextPositions.push([position - backwardStep, 0]);
35 }
36
37 // Check the next possible positions.
38 for (const [nextPosition, nextDirection] of nextPositions) {
39 // Check that the position is within bounds, not forbidden, and not visited yet in that direction.
40 if (nextPosition >= 0 && nextPosition < maxPosition &&
41 !forbiddenPositions.has(nextPosition) && !visited[nextPosition][nextDirection]) {
42 // Mark the position as visited in the current direction.
43 visited[nextPosition][nextDirection] = true;
44
45 // Enqueue the position for further exploration.
46 queue.push([nextPosition, nextDirection]);
47 }
48 }
49 }
50 }
51
52 // If the target is not reachable, return -1.
53 return -1;
54}
55
Time and Space Complexity
The time complexity of the code is O(N + a + b)
, where N
is the maximum position that can be reached considering the constraints of the problem. Since the algorithm uses a breadth-first search and each position is visited at most once, the time complexity is linear in terms of the number of positions checked. The factor of a + b
comes from the maximum range we might need to explore to ensure we can reach position x
or determine it's impossible. In the worst case, the algorithm explores up to position x + b
(since jumping b
back from beyond x
is the furthest we would need to go) and the positions backwards until reaching 0
. Given the constraint on i
(which is that 0 <= i < 6000
), N
effectively becomes 6000
, and thus the time complexity simplifies to O(6000)
which is constant, so effectively O(1)
.
The space complexity of the code is O(N)
, where N
again represents the maximum range of positions considered, which is determined to be 6000
based on the limitations placed within the code (0 <= j < 6000
). The space complexity arises from the storage of the vis
set containing tuples (position, direction)
which tracks the visited positions and the direction from which they were reached. The queue q
will also store at most O(N)
elements as it holds the positions to be explored. Therefore, space complexity also reduces to O(1)
since 6000
is the upper bound regardless of the input size.
Learn more about how to find time and space complexity quickly using problem constraints.
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