A certain bug's home is on the x-axis at position x. Help them get there from position 0.
The bug jumps according to the following rules:
a positions forward (to the right).b positions backward (to the left).forbidden positions.The bug may jump forward beyond its home, but it cannot jump to positions numbered with negative integers.
Given an array of integers forbidden, where forbidden[i] means that the bug cannot jump to the position forbidden[i], and integers a, b, and x, return the minimum number of jumps needed for the bug to reach its home. If there is no possible sequence of jumps that lands the bug on position x, return -1.
Example 1:
Input: forbidden = [14,4,18,1,15], a = 3, b = 15, x = 9 Output: 3 Explanation: 3 jumps forward (0 -> 3 -> 6 -> 9) will get the bug home.
Example 2:
Input: forbidden = [8,3,16,6,12,20], a = 15, b = 13, x = 11 Output: -1
Example 3:
Input: forbidden = [1,6,2,14,5,17,4], a = 16, b = 9, x = 7 Output: 2 Explanation: One jump forward (0 -> 16) then one jump backward (16 -> 7) will get the bug home.
Constraints:
1 <= forbidden.length <= 10001 <= a, b, forbidden[i] <= 20000 <= x <= 2000forbidden are distinct.x is not forbidden.When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:
The brute force approach to finding the fewest jumps to reach home is like trying out every single possible path. We'll explore all jump combinations, going forward and backward, until we reach the home location.
Here's how the algorithm would work step-by-step:
def minimum_jumps_to_reach_home_brute_force(start_position, forward_jump, backward_jump, home_position, forbidden_positions):
queue = [(start_position, 0, False)] # (position, jumps, last_jump_backward)
visited = set()
visited.add((start_position, False))
min_jumps = float('inf')
while queue:
current_position, number_of_jumps, last_jump_backward = queue.pop(0)
if current_position == home_position:
min_jumps = min(min_jumps, number_of_jumps)
continue
# We shouldn't explore paths longer than the minimum already found
if number_of_jumps >= min_jumps:
continue
forward_position = current_position + forward_jump
if (forward_position <= 6000) and (forward_position not in forbidden_positions) and ((forward_position, False) not in visited):
queue.append((forward_position, number_of_jumps + 1, False))
visited.add((forward_position, False))
# We skip making a backward jump if we just made one
if not last_jump_backward:
backward_position = current_position - backward_jump
# Must explore if we can move backward
if (backward_position >= 0) and (backward_position not in forbidden_positions) and ((backward_position, True) not in visited):
queue.append((backward_position, number_of_jumps + 1, True))
visited.add((backward_position, True))
if min_jumps == float('inf'):
return -1
else:
return min_jumpsThis problem is about finding the fewest jumps to reach home, while avoiding obstacles. We use a method that explores the possible paths in order of the number of jumps, stopping as soon as we find the home location.
Here's how the algorithm would work step-by-step:
def minimum_jumps_to_reach_home(forbidden_positions, start_position,
forward_jump, backward_jump, home_position):
forbidden_set = set(forbidden_positions)
queue = [(start_position, 0, False)]
visited = set()
visited.add((start_position, False))
farthest_reach = max(home_position + forward_jump,
max(forbidden_positions) + forward_jump) + backward_jump
while queue:
current_position, jumps, went_backward = queue.pop(0)
if current_position == home_position:
return jumps
# Explore forward jump
next_position = current_position + forward_jump
if (next_position <= farthest_reach and next_position not in forbidden_set
and (next_position, False) not in visited):
queue.append((next_position, jumps + 1, False))
visited.add((next_position, False))
# Avoid jumping backward twice in a row
if not went_backward:
previous_position = current_position - backward_jump
# Backward jump constraints
if (previous_position >= 0 and previous_position not in forbidden_set
and (previous_position, True) not in visited):
# Mark True for went_backward to prevent consecutive backward jumps.
queue.append((previous_position, jumps + 1, True))
visited.add((previous_position, True))
# Return -1 if home is unreachable
return -1| Case | How to Handle |
|---|---|
| Forbidden positions overlapping with the starting position (0) | The algorithm needs to correctly handle the case where the starting position is in the forbidden array; if so, return -1. |
| Forward jump length equals the distance to the target | The algorithm should return 1 in this optimal case where only one forward jump is needed. |
| The target is unreachable due to the arrangement of forbidden positions. | The algorithm should terminate and return -1 if all possible paths lead to blocked positions or exceed the maximum bound. |
| Backwards jumps leading to negative positions. | The algorithm should avoid exploring negative positions as they are invalid, effectively pruning those paths. |
| Integer overflow during distance calculations. | Use appropriate data types (e.g., long) to prevent integer overflow when calculating distances and maximum reachable bound. |
| Infinite loop due to alternating forward and backward jumps near forbidden zones. | Implement a visited set that tracks both position and direction (forward/backward) to avoid cycles, effectively pruning the search space. |
| Large forward and backward jump lengths making BFS take too long. | We might need to calculate an upper bound on the positions we need to visit to reduce the space, since we can't go beyond x+b. |
| Target is much further away from start and forbidden positions are clustered near start. | The algorithm should handle large target distances without getting stuck exploring local dead ends near the start, and it should terminate if unreachable. |