Last Moment Before All Ants Fall Out of a Plank

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We have a wooden plank of the length n units. Some ants are walking on the plank, each ant moves with a speed of 1 unit per second. Some of the ants move to the left, the other move to the right.

When two ants moving in two different directions meet at some point, they change their directions and continue moving again. Assume changing directions does not take any additional time.

When an ant reaches one end of the plank at a time t, it falls out of the plank immediately.

Given an integer n and two integer arrays left and right, the positions of the ants moving to the left and the right, return the moment when the last ant(s) fall out of the plank.

Example 1:

Input: n = 4, left = [4,3], right = [0,1]
Output: 4
Explanation: In the image above:
-The ant at index 0 is named A and going to the right.
-The ant at index 1 is named B and going to the right.
-The ant at index 3 is named C and going to the left.
-The ant at index 4 is named D and going to the left.
The last moment when an ant was on the plank is t = 4 seconds. After that, it falls immediately out of the plank. (i.e., We can say that at t = 4.0000000001, there are no ants on the plank).

Example 2:

Input: n = 7, left = [], right = [0,1,2,3,4,5,6,7]
Output: 7
Explanation: All ants are going to the right, the ant at index 0 needs 7 seconds to fall.

Example 3:

Input: n = 7, left = [0,1,2,3,4,5,6,7], right = []
Output: 7
Explanation: All ants are going to the left, the ant at index 7 needs 7 seconds to fall.

Constraints:

1 <= n <= 10⁴
0 <= left.length <= n + 1
0 <= left[i] <= n
0 <= right.length <= n + 1
0 <= right[i] <= n
1 <= left.length + right.length <= n + 1
All values of left and right are unique, and each value can appear only in one of the two arrays.

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 possible values for positions in the input arrays `left` and `right`, including the possibility of negative values or zero?
Can the same position be occupied by ants moving in opposite directions, or by multiple ants moving in the same direction?
If the plank is of length `n`, is it possible for an ant to start at position `0` or `n`?
What should be returned if the input arrays `left` and `right` are empty?
Is `n` guaranteed to be a non-negative integer?

Brute Force Solution

Approach

The ants are walking on the plank and bumping into each other. The brute force approach is to simulate the movement of each ant at every single moment in time until all ants fall off the plank.

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

Imagine the plank as a line and each ant as a tiny object moving along it.
At each tiny step of time, move each ant according to its direction (left or right).
If two ants bump into each other, reverse their directions.
Check after each tiny time step if any ants have fallen off either end of the plank.
Continue this process, step by step, until all ants have fallen off.
The time when the last ant falls off is the answer.

Code Implementation

def last_moment_before_all_ants_fall_out_of_a_plank(plank_length, left_positions, right_positions):
    ants = []
    for position in left_positions:
        ants.append([position, -1])  # -1 means moving left
    for position in right_positions:
        ants.append([position, 1])  # 1 means moving right

    time = 0
    while ants:
        # Move each ant based on its direction
        for ant_index in range(len(ants)):
            ants[ant_index][0] += ants[ant_index][1]

        # Handle collisions by reversing directions
        for first_ant_index in range(len(ants)): 
            for second_ant_index in range(first_ant_index + 1, len(ants)): 
                if ants[first_ant_index][0] == ants[second_ant_index][0]:
                    ants[first_ant_index][1] *= -1
                    ants[second_ant_index][1] *= -1

        # Remove ants that have fallen off the plank
        fallen_ants = []
        for ant_index in range(len(ants)): 
            if ants[ant_index][0] < 0 or ants[ant_index][0] > plank_length:
                fallen_ants.append(ant_index)

        #It's crucial to remove in reverse to avoid index errors
        fallen_ants.sort(reverse=True)
        for ant_index in fallen_ants:
            del ants[ant_index]

        time += 1

    return time - 1

Big(O) Analysis

Time Complexity

O(n*L) – The brute force approach simulates the movement of each ant at every time step. In the worst-case scenario, an ant might need to travel the entire length L of the plank to fall off. Since there are n ants, and for each time step, we need to update the position of each ant and check for collisions (which takes O(n) time in itself to resolve all collisions at the same location), the total time complexity becomes O(n*L). Therefore, the algorithm's runtime is directly proportional to both the number of ants and the length of the plank.

Space Complexity

O(N) – The brute force approach simulates each ant's movement. At each time step, we need to store the position and direction of each ant. Therefore, we essentially maintain two arrays (or a similar data structure) of size N, where N is the number of ants, to represent this information at each time step. This results in auxiliary space proportional to the number of ants. Thus, the space complexity is O(N).

Optimal Solution

Approach

The key idea is that ants effectively pass through each other when they collide. Therefore, we can ignore the collisions and assume each ant simply keeps moving in its initial direction. The last ant to fall off will be the one that started furthest from the edge it was initially moving towards.

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

Imagine the ants can walk right through each other without changing direction.
For each ant, figure out the distance to the end of the plank it's moving toward. If an ant is moving left, it's the distance to the left end. If it's moving right, it's the distance to the right end.
Find the longest of all these distances. This longest distance is the time when the last ant falls off the plank.

Code Implementation

def last_moment(plank_length, left_positions, right_positions):
    max_time = 0

    # Calculate time for ants moving left.
    for ant_position in left_positions:
        max_time = max(max_time, ant_position)

    # Calculate time for ants moving right.
    # The last ant will fall from plank_length - position.
    for ant_position in right_positions:
        max_time = max(max_time, plank_length - ant_position)

    # We need to return the maximum time.
    return max_time

Big(O) Analysis

Time Complexity

O(n) – The solution iterates through the array of ant positions once, where n is the number of ants. For each ant, it performs a constant-time calculation to determine the time it takes to fall off the plank. The algorithm then finds the maximum of these times, which again takes O(n) time. Therefore, the overall time complexity is determined by the single loop through the ant positions, resulting in O(n) complexity.

Space Complexity

O(1) – The provided steps calculate the time until the last ant falls off the plank without using any auxiliary data structures that scale with the input size. The algorithm iterates through each ant, calculates a distance, and updates a maximum distance variable. These calculations only require a constant amount of extra space, independent of the number of ants (N). Therefore, the auxiliary space complexity is O(1).

Edge Cases

Case	How to Handle
Empty left and right arrays	Return 0 immediately as no ants exist to fall off the plank.
Plank length n is zero	Return 0 immediately, as the plank has no length for ants to traverse.
All ants move in the same direction	The last ant to fall will be the one furthest from the edge in its direction of movement.
Large plank length (n) leading to potential integer overflow in calculations.	Use long data types where applicable to prevent integer overflows during calculations.
Left array contains the value 0	The ant at position 0 falls immediately, so use Math.max(0, current_max) to ensure return value is valid
Right array contains the value n	The ant at position n falls immediately, so use Math.max(0, current_max) to ensure return value is valid
Input arrays are very large, impacting memory usage.	The solution's space complexity is O(1) regardless of input array size as we are only tracking the maximum time, so no special handling is needed.
Plank length (n) is very small (e.g., 1) and both arrays are empty	Return 0 if both left and right arrays are empty; otherwise, determine the answer from the non-empty array based on direction.

Empty left and right arrays

How to Handle:

Return 0 immediately as no ants exist to fall off the plank.

Plank length n is zero

How to Handle:

Return 0 immediately, as the plank has no length for ants to traverse.

All ants move in the same direction

How to Handle:

The last ant to fall will be the one furthest from the edge in its direction of movement.

Large plank length (n) leading to potential integer overflow in calculations.

How to Handle:

Use long data types where applicable to prevent integer overflows during calculations.

Left array contains the value 0

How to Handle:

The ant at position 0 falls immediately, so use Math.max(0, current_max) to ensure return value is valid

Right array contains the value n

How to Handle:

The ant at position n falls immediately, so use Math.max(0, current_max) to ensure return value is valid

Input arrays are very large, impacting memory usage.

How to Handle:

The solution's space complexity is O(1) regardless of input array size as we are only tracking the maximum time, so no special handling is needed.

Plank length (n) is very small (e.g., 1) and both arrays are empty

How to Handle:

Return 0 if both left and right arrays are empty; otherwise, determine the answer from the non-empty array based on direction.