Ant on the Boundary

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Arrays

An ant is on a boundary. It sometimes goes left and sometimes right.

You are given an array of non-zero integers nums. The ant starts reading nums from the first element of it to its end. At each step, it moves according to the value of the current element:

If nums[i] < 0, it moves left by -nums[i] units.
If nums[i] > 0, it moves right by nums[i] units.

Return the number of times the ant returns to the boundary.

Notes:

There is an infinite space on both sides of the boundary.
We check whether the ant is on the boundary only after it has moved |nums[i]| units. In other words, if the ant crosses the boundary during its movement, it does not count.

Example 1:

Input: nums = [2,3,-5]
Output: 1
Explanation: After the first step, the ant is 2 steps to the right of the boundary.
After the second step, the ant is 5 steps to the right of the boundary.
After the third step, the ant is on the boundary.
So the answer is 1.

Example 2:

Input: nums = [3,2,-3,-4]
Output: 0
Explanation: After the first step, the ant is 3 steps to the right of the boundary.
After the second step, the ant is 5 steps to the right of the boundary.
After the third step, the ant is 2 steps to the right of the boundary.
After the fourth step, the ant is 2 steps to the left of the boundary.
The ant never returned to the boundary, so the answer is 0.

Constraints:

1 <= nums.length <= 100
-10 <= nums[i] <= 10
nums[i] != 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 possible values for the ant's moves? Can they be negative, zero, or floating-point numbers?
What is the size of the boundary? Is it a fixed length or does it depend on the ant's moves?
What should I return if the ant never crosses the boundary?
If the ant crosses the boundary multiple times, should I return the number of times it crosses or is there a specific crossing I should identify?
Is the boundary defined as inclusive or exclusive? For example, if the boundary is [0, 10], does a position of 0 or 10 count as crossing the boundary?

Brute Force Solution

Approach

The ant starts at the origin (0, 0). We want to count how many times the ant crosses any of the four axes. The brute force solution simulates every single move the ant makes, step-by-step, and checks if that move results in crossing an axis.

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

Start the ant at the origin point.
Read the first instruction for the ant's movement.
Move the ant according to that instruction.
Check if the ant has crossed any of the axes during that move. This means, did its x or y coordinate change signs from positive to negative or negative to positive, or become exactly zero? If the ant lies exactly on the boundary, increment a counter variable.
Repeat steps 2-4 for every instruction in the sequence of moves.
Once all moves are processed, the counter variable stores the total number of times the ant crossed an axis. This is the result.

Code Implementation

def ant_on_the_boundary_brute_force(moves):
    ant_x_coordinate = 0
    ant_y_coordinate = 0
    crossings_count = 0

    for move in moves:
        previous_x_coordinate = ant_x_coordinate
        previous_y_coordinate = ant_y_coordinate

        if move == 'L':
            ant_x_coordinate -= 1
        elif move == 'R':
            ant_x_coordinate += 1
        elif move == 'U':
            ant_y_coordinate += 1
        elif move == 'D':
            ant_y_coordinate -= 1

        # Check x-axis crossing
        if (previous_y_coordinate > 0 and ant_y_coordinate < 0) or \
           (previous_y_coordinate < 0 and ant_y_coordinate > 0):

            crossings_count += 1

        # Check y-axis crossing
        if (previous_x_coordinate > 0 and ant_x_coordinate < 0) or \
           (previous_x_coordinate < 0 and ant_x_coordinate > 0):

            crossings_count += 1

        # Increment if exactly on x-axis
        if ant_y_coordinate == 0 and previous_y_coordinate != 0:

            crossings_count += 1

        # Increment if exactly on y-axis
        if ant_x_coordinate == 0 and previous_x_coordinate != 0:

            crossings_count += 1

    return crossings_count

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the sequence of moves, which has a size of n. For each move, it performs a constant amount of work: updating the ant's position and checking if it has crossed an axis. Since the work per move is constant and independent of n, the overall time complexity is directly proportional to the number of moves, n. Therefore, the time complexity is O(n).

Space Complexity

O(1) – The provided solution maintains the ant's current x and y coordinates, along with a counter variable to track axis crossings. These variables consume a fixed amount of memory, irrespective of the number of movement instructions (N). Since no auxiliary data structures that scale with the input size are used, the space complexity is constant.

Optimal Solution

Approach

The problem asks to figure out when an ant, taking random steps, will cross a boundary. Instead of simulating the ant's movements many times, we can focus on the final outcome of the ant's position being outside the allowed region and compute the probability of that outcome directly. We will need to consider how the steps affect the distance from the origin.

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

Think about what determines whether the ant is outside the boundary: its final distance from the starting point must be greater than some limit.
Consider all possible paths the ant can take. Each path is a sequence of steps the ant takes.
Calculate the probability of each possible path. Since each step is random, each path has an equal chance of happening.
For each path, check if the ant ends up outside the boundary.
Add up the probabilities of all the paths where the ant ends up outside the boundary. This gives you the overall probability of the ant crossing the boundary.
Because all the possible steps have to total to one, the opposite outcome must be one minus this number.

Code Implementation

def ant_on_boundary(number_of_steps, step_size, boundary):
    total_possible_paths = 2 ** number_of_steps
    paths_outside_boundary = 0

    # Iterate through all possible paths the ant can take.
    for i in range(total_possible_paths):
        ant_position = 0
        
        # Simulate the ant's movement for each path.
        for j in range(number_of_steps):
            # Determine step direction based on the binary representation of i.
            if (i >> j) & 1:
                ant_position += step_size
            else:
                ant_position -= step_size

        # Check if the ant ended up outside the boundary.
        if abs(ant_position) > boundary:
            paths_outside_boundary += 1

    # Calculate the probability of the ant being outside the boundary.
    probability_outside = paths_outside_boundary / total_possible_paths

    # Return the probability of the ant staying within the boundary.
    return 1 - probability_outside

def main():
    number_of_steps = 10
    step_size = 1
    boundary = 5
    
    #Call the function with sample inputs and printing out the result.
    probability_within_boundary = ant_on_boundary(number_of_steps, step_size, boundary)
    print(f'{probability_within_boundary=}')

if __name__ == "__main__":
    main()

Big(O) Analysis

Time Complexity

O(2^n) – The algorithm considers all possible paths the ant can take. If the ant takes n steps, and each step can be in one of several directions (implicitly assumed to be a fixed number, let's say 4: up, down, left, right), then there are 4^n possible paths. For each of these paths, the algorithm checks if the ant ends up outside the boundary. Thus, the time complexity is proportional to the number of paths, which grows exponentially with the number of steps. Therefore, the dominant factor is the enumeration of all possible paths, leading to O(4^n), which can be generalized to O(2^n) since constants are ignored in Big O notation, where each step is a binary choice that builds out the path.

Space Complexity

O(N!) – The provided explanation suggests considering all possible paths the ant can take. This implies storing or processing these paths, which can grow factorially with the number of steps, N, where N is the number of steps the ant takes. Although not explicitly stated, implicitly calculating the probability of each path will require space to store/process each path. Therefore, the space complexity is related to the number of possible paths, leading to O(N!).

Edge Cases

Case	How to Handle
Empty steps array	Return 0 since the ant hasn't moved and is therefore on the boundary.
Array containing only zeros	Return 1 since the ant remains at the origin and is on the boundary at t=0.
Single step that returns the ant to the origin	The ant starts on the boundary, takes a step, and returns to the boundary; therefore, the answer is 2.
Large number of steps potentially leading to integer overflow in position calculation	Use a data type (e.g., long long in C++, long in Java) capable of storing the maximum possible position to prevent overflow.
Steps oscillate between two values that keep the ant oscillating around the origin	Track all visited positions and count only those that are on the boundary.
The ant never returns to the boundary	After processing all steps, return 1 if the starting position (0) is considered on the boundary, otherwise 0.
Maximum size array of steps leading to very large ant movement	Ensure the algorithm's space complexity doesn't scale linearly with the number of steps to avoid memory issues by only tracking boundary visits.
Negative or very large step values	The position variable needs to handle negative positions and avoid potential overflow with large step values, using long long or similar.

Empty steps array

How to Handle:

Return 0 since the ant hasn't moved and is therefore on the boundary.

Array containing only zeros

How to Handle:

Return 1 since the ant remains at the origin and is on the boundary at t=0.

Single step that returns the ant to the origin

How to Handle:

The ant starts on the boundary, takes a step, and returns to the boundary; therefore, the answer is 2.

Large number of steps potentially leading to integer overflow in position calculation

How to Handle:

Use a data type (e.g., long long in C++, long in Java) capable of storing the maximum possible position to prevent overflow.

Steps oscillate between two values that keep the ant oscillating around the origin

How to Handle:

Track all visited positions and count only those that are on the boundary.

The ant never returns to the boundary

How to Handle:

After processing all steps, return 1 if the starting position (0) is considered on the boundary, otherwise 0.

Maximum size array of steps leading to very large ant movement

How to Handle:

Ensure the algorithm's space complexity doesn't scale linearly with the number of steps to avoid memory issues by only tracking boundary visits.

Negative or very large step values

How to Handle:

The position variable needs to handle negative positions and avoid potential overflow with large step values, using long long or similar.