Taro Logo

Minimum Jumps to Reach Home

Medium
Asked by:
Profile picture
Profile picture
18 views
Topics:
GraphsDynamic Programming

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:

  • It can jump exactly a positions forward (to the right).
  • It can jump exactly b positions backward (to the left).
  • It cannot jump backward twice in a row.
  • It cannot jump to any 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 <= 1000
  • 1 <= a, b, forbidden[i] <= 2000
  • 0 <= x <= 2000
  • All the elements in forbidden are distinct.
  • Position x is not forbidden.

Solution


Clarifying Questions

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:

  1. What are the constraints on the values of forwardStep, backwardStep, and home? Are they positive integers?
  2. Are there any constraints on the values and the size of the forbidden array? Can the values in forbidden be larger than home?
  3. Is it possible to have forwardStep equal to or greater than home, or backwardStep equal to zero?
  4. If it's impossible to reach home, what should the function return? Should I return -1?
  5. After jumping backward, can I jump backward again immediately, or is there a restriction on consecutive backward jumps?

Brute Force Solution

Approach

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:

  1. Start at the beginning location.
  2. From there, imagine all the possible jump lengths forward and backward that you can take.
  3. For each of those jumps, consider all the possible jumps you can make from that new location, again both forward and backward.
  4. Keep doing this, exploring every single possible sequence of jumps.
  5. As you explore, keep track of how many jumps each sequence takes to reach the home location.
  6. If a sequence leads to a forbidden location, stop exploring that path.
  7. Once you've explored all possible jump sequences, find the one that reaches home with the fewest number of jumps.

Code Implementation

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_jumps

Big(O) Analysis

Time Complexity
O(2^(max(forbidden) + forward + backward)) – The brute force approach explores all possible jump sequences, branching out at each position with two choices: jump forward or jump backward. The depth of this exploration is determined by the maximum reachable position, which depends on the largest forbidden location, the forward jump length, and the backward jump length. In the worst case, the algorithm could explore a tree-like structure where each node has two children, leading to an exponential growth in the number of paths explored. Therefore, the time complexity is approximately O(2^d), where d is the depth of the exploration, which is related to max(forbidden) + forward + backward.
Space Complexity
O(X) – The brute force approach explores all possible jump combinations using recursion. The maximum depth of the recursion is bounded by the furthest reachable position, which depends on the forward jump length, backward jump length, the starting position, the home position and the forbidden positions. Let X be the furthest reachable position. The recursion stack can grow up to a depth of X in the worst case, leading to O(X) space complexity. Additionally, to avoid cycles, a set or similar data structure to keep track of visited locations could take up space up to the furthest reachable position X, contributing O(X) space. Thus, the overall space complexity is O(X).

Optimal Solution

Approach

This 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:

  1. Imagine starting at the starting location.
  2. Explore all locations you can reach in one jump, both forward and backward.
  3. If home is among these locations, you're done!
  4. Otherwise, consider all locations you can reach in two jumps from the starting location (by jumping from the locations you reached in one jump).
  5. Keep track of the locations you've already visited to avoid going in circles or exploring the same path multiple times.
  6. Also, be careful to avoid jumping onto forbidden locations or going too far away from the starting point.
  7. Continue this process, exploring locations reachable in three jumps, then four jumps, and so on, until you reach home. The number of jumps it took is the minimum.
  8. If you explore all possible jump combinations within a reasonable distance and still don't find home, then it's impossible to reach home.

Code Implementation

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

Big(O) Analysis

Time Complexity
O(x*f) – The breadth-first search explores possible jump combinations. Let 'x' be the maximum reachable position from the start, considering forward and backward jumps and the forbidden positions. We must track visited locations to avoid cycles. The number of forbidden locations is 'f'. The time complexity is proportional to the product of the reachable space 'x' and the number of forbidden locations 'f' since checking forbidden locations impacts runtime.
Space Complexity
O(max(home, max(forbidden) + a + b)) – The algorithm maintains a queue for BFS traversal, potentially storing all reachable locations. It also uses a set to track visited locations (both forward and backward jumps) to avoid cycles. The space required for the queue and the visited set depends on the range of possible positions which is determined by the home location, the maximum forbidden position, and the forward and backward jump lengths (a and b), due to the problem's constraints to avoid going too far from the start. Thus, the space complexity is O(max(home, max(forbidden) + a + b)), where 'home' is the target location, 'forbidden' is the array of forbidden locations, 'a' is the forward jump length, and 'b' is the backward jump length.

Edge Cases

Forbidden positions overlapping with the starting position (0)
How to Handle:
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
How to Handle:
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.
How to Handle:
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.
How to Handle:
The algorithm should avoid exploring negative positions as they are invalid, effectively pruning those paths.
Integer overflow during distance calculations.
How to Handle:
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.
How to Handle:
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.
How to Handle:
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.
How to Handle:
The algorithm should handle large target distances without getting stuck exploring local dead ends near the start, and it should terminate if unreachable.