Minimum Sideway Jumps

Medium

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Dynamic Programming

There is a 3 lane road of length n that consists of n + 1 points labeled from 0 to n. A frog starts at point 0 in the second lane and wants to jump to point n. However, there could be obstacles along the way.

You are given an array obstacles of length n + 1 where each obstacles[i] (ranging from 0 to 3) describes an obstacle on the lane obstacles[i] at point i. If obstacles[i] == 0, there are no obstacles at point i. There will be at most one obstacle in the 3 lanes at each point.

For example, if obstacles[2] == 1, then there is an obstacle on lane 1 at point 2.

The frog can only travel from point i to point i + 1 on the same lane if there is not an obstacle on the lane at point i + 1. To avoid obstacles, the frog can also perform a side jump to jump to another lane (even if they are not adjacent) at the same point if there is no obstacle on the new lane.

For example, the frog can jump from lane 3 at point 3 to lane 1 at point 3.

Return the minimum number of side jumps the frog needs to reach any lane at point n starting from lane 2 at point 0.

Note: There will be no obstacles on points 0 and n.

Example 1:

Input: obstacles = [0,1,2,3,0]
Output: 2 
Explanation: The optimal solution is shown by the arrows above. There are 2 side jumps (red arrows).
Note that the frog can jump over obstacles only when making side jumps (as shown at point 2).

Example 2:

Input: obstacles = [0,1,1,3,3,0]
Output: 0
Explanation: There are no obstacles on lane 2. No side jumps are required.

Example 3:

Input: obstacles = [0,2,1,0,3,0]
Output: 2
Explanation: The optimal solution is shown by the arrows above. There are 2 side jumps.

Constraints:

obstacles.length == n + 1
1 <= n <= 5 * 10⁵
0 <= obstacles[i] <= 3
obstacles[0] == obstacles[n] == 0

Try coding it in LeetCode

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:

What are the constraints on the size of the `obstacles` array (number of points on the path)?
What are the possible values within the `obstacles` array? Are they limited to just 0, 1, 2, and 3, representing no obstacle or an obstacle on the corresponding lane?
If there's no possible path to the end, what value should I return?
Is it guaranteed that the starting and ending points will always be clear (no obstacle at indices 0 and n-1, respectively, on lane 2)?
Could the starting or ending point be blocked (i.e., obstacles[0][2] != 0 or obstacles[n-1][2] != 0)?

Brute Force Solution

Approach

Imagine you are a frog trying to cross a road with lanes, hopping forward and potentially sideways to avoid obstacles. The brute force method involves trying every single possible path the frog could take across the road. We will check each one to see which uses the fewest sideways jumps.

Here's how the algorithm would work step-by-step:

Start the frog at the beginning of the road in the starting lane.
From the frog's current position, explore all possible next moves: either hop forward in the same lane if it's clear, or jump sideways to another lane if that lane is clear at the next position.
Repeat this process for every possible path the frog could take until it either reaches the end of the road or gets stuck because there's no valid move forward.
Keep track of the number of sideways jumps for each path that successfully gets the frog to the end of the road.
Compare the number of sideways jumps for all successful paths and pick the path with the fewest jumps. That's our answer!

Code Implementation

def minimum_sideway_jumps_brute_force(obstacles):
    number_of_lanes = 3
    start_lane = 2
    start_position = 0

    min_jumps = float('inf')

    def explore_path(current_lane, current_position, jumps_so_far):
        nonlocal min_jumps

        # If we reached the end, update min_jumps
        if current_position == len(obstacles) - 1:
            min_jumps = min(min_jumps, jumps_so_far)
            return

        # Check if we hit an obstacle in the current lane
        if obstacles[current_position] == current_lane:
            return

        # Try moving forward in the same lane
        next_position = current_position + 1
        if next_position < len(obstacles) and obstacles[next_position] != current_lane:
            explore_path(current_lane, next_position, jumps_so_far)

        # Try jumping to other lanes
        for next_lane in range(1, number_of_lanes + 1):
            if next_lane != current_lane:

                # Prevent jumping into an obstacle
                if obstacles[current_position] != next_lane and next_position < len(obstacles) and obstacles[next_position] != next_lane:
                    explore_path(next_lane, next_position, jumps_so_far + 1)

    # Start the exploration from the initial position
    explore_path(start_lane, start_position, 0)

    # Return the minimum jumps or -1 if no path exists
    return min_jumps if min_jumps != float('inf') else -1

Big(O) Analysis

Time Complexity

O(3^n) – The brute force approach explores every possible path. At each position, the frog can either move forward (if no obstacle) or jump to one of the two other lanes (if no obstacle). In the worst-case scenario, at almost every position, the frog has three choices (forward and two sideway jumps). Therefore, the number of possible paths grows exponentially with the length of the road, n. This leads to a time complexity of approximately O(3^n), representing the branching factor of 3 at each step.

Space Complexity

O(3^N) – The brute force approach explores all possible paths. At each position, the frog can either move forward in the same lane or jump to one of the two other lanes. Thus, in the worst-case scenario, where there are no obstacles and the frog can always move in any direction, the algorithm explores roughly 3 possible paths at each position. This branching factor of 3 continues for each of the N positions along the road, leading to a maximum call stack depth proportional to N, and an exponential number of possible paths. This exploration is best represented by a decision tree where each node has a maximum of three children. The space needed for the call stack is proportional to the number of possible paths, which grows exponentially with N.

Optimal Solution

Approach

The best way to solve this is to always focus on the easiest way forward, remembering the minimum jumps at each point. Instead of exploring every possible path, we'll remember the best path to each point and use that to make smart choices going forward.

Here's how the algorithm would work step-by-step:

Start on lane 2, which is where the frog begins.
Keep track of the fewest jumps needed to reach each lane at each position.
At each position, if the current lane is clear, move forward using the fewest jumps already calculated.
If the current lane has an obstacle, consider jumping to a different lane.
When jumping, only jump to lanes that are clear at the current position.
Update the minimum number of jumps required to reach the new lane at that position. If the jump is better than a previous jump to that point, use the new one.
Continue moving forward, always choosing the path with the fewest jumps.
The final answer is the minimum number of jumps needed to reach the end position in any lane.

Code Implementation

def minimum_sideway_jumps(obstacles):
    number_of_positions = len(obstacles) - 1
    minimum_jumps = [[float('inf')] * 3 for _ in range(number_of_positions + 1)]
    minimum_jumps[0][1] = 0

    for position in range(1, number_of_positions + 1):
        for lane in range(3):
            # If there's an obstacle, we can't reach this lane at this position.
            if obstacles[position] == lane + 1:
                minimum_jumps[position][lane] = float('inf')
                continue

            # Moving forward from the previous position in the same lane.
            minimum_jumps[position][lane] = minimum_jumps[position - 1][lane]

            # Consider jumping from other lanes at the same position.
            for other_lane in range(3):
                if obstacles[position] == other_lane + 1:
                    continue

                # Need a jump if we are coming from a different lane
                if lane != other_lane:
                    minimum_jumps[position][lane] = min(
                        minimum_jumps[position][lane],
                        minimum_jumps[position][other_lane] + 1
                    )

    # Find the minimum jumps to reach the end in any lane.
    return min(minimum_jumps[number_of_positions])

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through each position in the obstacle array of length n. At each position, it potentially considers jumping to the other two lanes. Jumping decisions and updates to the minimum jumps for each lane at that position take constant time. Therefore, the overall time complexity is directly proportional to the number of positions, resulting in O(n).

Space Complexity

O(N) – The solution involves keeping track of the fewest jumps needed to reach each lane at each position. This implies the use of a data structure, most likely a 2D array or a similar structure with dimensions related to the number of lanes (fixed at 3) and the number of positions (length of the obstacles array, N). Therefore, the auxiliary space required grows linearly with the input size N. The minimum jumps array will be of size 3 x N which simplifies to O(N).

Edge Cases

Case	How to Handle
Null or empty obstacle array	Return 0 as no obstacles exist and no jumps are needed.
Obstacle array with only one element	Return 0, as the frog starts at position 0 and the path ends at n-1, so no jumps are needed when n = 1.
Obstacle in the starting lane (lane 2 at position 0)	Modify the starting lane according to the obstacle and begin DP from there, or return -1 if starting lane is fully blocked.
Obstacles in all three lanes at the same position	The frog cannot proceed if all lanes are blocked at any position, so return -1.
Large number of positions (n is very large)	Use dynamic programming or a similar optimized algorithm to avoid exceeding time limits.
Obstacles clustered in one specific lane	The algorithm should efficiently handle traversing multiple clear lanes to avoid the blocked lane.
Obstacles only on the start or end lane	The frog only needs to jump lanes to avoid the start or end obstacles.
Integer overflow potential when calculating distances or costs within DP	Ensure all distance/cost calculations are done with appropriately sized integers, or use a safe math library.

Null or empty obstacle array

How to Handle:

Return 0 as no obstacles exist and no jumps are needed.

Obstacle array with only one element

How to Handle:

Return 0, as the frog starts at position 0 and the path ends at n-1, so no jumps are needed when n = 1.

Obstacle in the starting lane (lane 2 at position 0)

How to Handle:

Modify the starting lane according to the obstacle and begin DP from there, or return -1 if starting lane is fully blocked.

Obstacles in all three lanes at the same position

How to Handle:

The frog cannot proceed if all lanes are blocked at any position, so return -1.

Large number of positions (n is very large)

How to Handle:

Use dynamic programming or a similar optimized algorithm to avoid exceeding time limits.

Obstacles clustered in one specific lane

How to Handle:

The algorithm should efficiently handle traversing multiple clear lanes to avoid the blocked lane.

Obstacles only on the start or end lane

How to Handle:

The frog only needs to jump lanes to avoid the start or end obstacles.

Integer overflow potential when calculating distances or costs within DP

How to Handle:

Ensure all distance/cost calculations are done with appropriately sized integers, or use a safe math library.