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Points That Intersect With Cars

Easy

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ArraysGreedy Algorithms

You are given a 0-indexed 2D integer array nums representing the coordinates of the cars parking on a number line. For any index i, nums[i] = [start_i, end_i] where start_i is the starting point of the i^th car and end_i is the ending point of the i^th car.

Return the number of integer points on the line that are covered with any part of a car.

Example 1:

Input: nums = [[3,6],[1,5],[4,7]]
Output: 7
Explanation: All the points from 1 to 7 intersect at least one car, therefore the answer would be 7.

Example 2:

Input: nums = [[1,3],[5,8]]
Output: 7
Explanation: Points intersecting at least one car are 1, 2, 3, 5, 6, 7, 8. There are a total of 7 points, therefore the answer would be 7.

Constraints:

1 <= nums.length <= 100
nums[i].length == 2
1 <= start_i <= end_i <= 100

Try coding it in LeetCode

Points That Intersect With Cars

Easy

Asked by:

14 views

Topics:

ArraysGreedy Algorithms

Return the number of integer points on the line that are covered with any part of a car.

Example 1:

Input: nums = [[3,6],[1,5],[4,7]]
Output: 7
Explanation: All the points from 1 to 7 intersect at least one car, therefore the answer would be 7.

Example 2:

Input: nums = [[1,3],[5,8]]
Output: 7
Explanation: Points intersecting at least one car are 1, 2, 3, 5, 6, 7, 8. There are a total of 7 points, therefore the answer would be 7.

Constraints:

1 <= nums.length <= 100
nums[i].length == 2
1 <= start_i <= end_i <= 100

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 data types and range of values for the car start, end coordinates and point coordinates? Can they be floats or just integers?
Can a point have the same coordinate as the start or end coordinate of a car? If so, is that considered an intersection?
How should I handle edge cases such as an empty list of points or an empty list of cars? Should I return an empty list or throw an exception?
If a point intersects with multiple cars, should I include it multiple times in the output, or only once?
What is the expected output format? Should I return a list of the point coordinates that intersect, or a list of indices corresponding to those points in the input list?

Brute Force Solution

Approach

We need to find which points are covered by any of the cars. The brute force method checks each point against every car to see if it's covered.

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

Take the first point.
Look at the first car. Is the point within the start and end position of the car?
If it is, mark that the point is covered. If it isn't, move to the next car.
Repeat the previous step for all the cars. If the point is covered by at least one car, we know it's a valid intersection.
Move to the next point and repeat the entire process, checking it against all the cars.
Do this for every point. In the end, you'll know exactly which points are intersected by at least one car.

Code Implementation

def find_intersecting_points_brute_force(points, cars):
    intersecting_points = []

    for point in points:
        is_point_covered = False

        # Iterate through each car to see if the current point is covered
        for car_start, car_end in cars:

            # Check if the point falls within the start and end positions of the current car.
            if car_start <= point <= car_end:
                is_point_covered = True
                break

        # Add the point to the result if it's covered by at least one car
        if is_point_covered:
            intersecting_points.append(point)

    return intersecting_points

Big(O) Analysis

Time Complexity

O(n*m) – Let n be the number of points and m be the number of cars. The algorithm iterates through each of the n points. For each point, it iterates through all m cars to check if the point lies within the car's range. Therefore, the total number of operations is proportional to the product of the number of points and the number of cars, resulting in a time complexity of O(n*m).

Space Complexity

O(1) – The provided brute force solution iterates through points and cars, checking each point against every car. It only requires a boolean variable to mark if a point is covered, which is updated in place during the inner loop. No additional data structures that scale with the input size are used; therefore, the auxiliary space remains constant regardless of the number of points or cars. The algorithm's space complexity is O(1).

Optimal Solution

Approach

The most efficient way to solve this problem is to imagine a timeline. We track when each car's path 'starts' and 'ends', allowing us to quickly determine how many cars a given point overlaps with. This avoids checking each car individually for every point.

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

First, think about each car's journey as starting at one point in time and ending at another. We only care about these 'start' and 'end' times.
Make a list of all the 'start' times and another list of all the 'end' times, keeping track of which car each time belongs to.
Combine and sort these lists into a single timeline of events, showing when cars start and end their journeys, like appointments on a calendar.
For each point we want to check, imagine drawing a vertical line on this timeline. Count how many car 'start' times are to the left of our line (the cars have already started their journey).
Then, count how many car 'end' times are also to the left of our line (the cars have already finished their journey).
The number of cars that our point intersects with is the difference between those two counts: cars that have started but haven't ended yet.

Code Implementation

def points_that_intersect_with_cars(car_start_times, car_end_times, points_to_check):
    start_times = []
    end_times = []

    for car_index in range(len(car_start_times)):
        start_times.append((car_start_times[car_index], 1))
        end_times.append((car_end_times[car_index], -1))

    # Combine start and end times into a single timeline.
    timeline = sorted(start_times + end_times)

    results = []

    for point in points_to_check:
        cars_present = 0
        # Iterate through the timeline to calculate intersections.
        for time, type in timeline:
            if time <= point:
                cars_present += type

        results.append(cars_present)

    return results

Big(O) Analysis

Time Complexity

O(n log n + m log n) – Let n be the number of cars and m be the number of points. Creating the start and end times lists takes O(n) time. Sorting the combined list of 2n events takes O(n log n) time. For each of the m points, we perform two binary searches (one for start times and one for end times) on the sorted list of 2n events, each taking O(log n) time. This results in O(m log n) time for processing all the points. The total time complexity is therefore O(n log n + m log n).

Space Complexity

O(N) – The algorithm creates two lists, one for start times and one for end times, each containing information for every car. If we denote the number of cars as N, then each of these lists will have a size of N. Combining and sorting these into a single timeline (events list) results in a list of size 2N. Therefore, the auxiliary space required is proportional to the number of cars, resulting in O(N) space complexity.

Edge Cases

Case	How to Handle
Cars array is null or empty	Return an empty list since there are no cars to intersect with.
Points array is null or empty	Return an empty list since there are no points to check for intersection.
Cars array has zero length segments	Filter out cars with start and end point same to avoid division by zero errors.
All points are outside of all car segments	Return an empty list as no points intersect with any cars.
Points array contains duplicate points	The solution should correctly identify if these points intersect regardless of duplication.
Integer overflow in calculations, especially for very large coordinates	Use long data type for intermediate calculations to avoid overflow issues.
Large input arrays exceeding memory limits	Consider processing points in batches or using a more memory-efficient data structure if necessary.
Points are on the car segment's boundaries (start or end point)	Define clearly whether points on the boundaries are considered intersecting based on the problem requirements and implement the condition accordingly.

Cars array is null or empty

How to Handle:

Return an empty list since there are no cars to intersect with.

Points array is null or empty

How to Handle:

Return an empty list since there are no points to check for intersection.

Cars array has zero length segments

How to Handle:

Filter out cars with start and end point same to avoid division by zero errors.

All points are outside of all car segments

How to Handle:

Return an empty list as no points intersect with any cars.

Points array contains duplicate points

How to Handle:

The solution should correctly identify if these points intersect regardless of duplication.

Integer overflow in calculations, especially for very large coordinates

How to Handle:

Use long data type for intermediate calculations to avoid overflow issues.

Large input arrays exceeding memory limits

How to Handle:

Consider processing points in batches or using a more memory-efficient data structure if necessary.

Points are on the car segment's boundaries (start or end point)

How to Handle:

Define clearly whether points on the boundaries are considered intersecting based on the problem requirements and implement the condition accordingly.

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 data types and range of values for the car start, end coordinates and point coordinates? Can they be floats or just integers?
Can a point have the same coordinate as the start or end coordinate of a car? If so, is that considered an intersection?
How should I handle edge cases such as an empty list of points or an empty list of cars? Should I return an empty list or throw an exception?
If a point intersects with multiple cars, should I include it multiple times in the output, or only once?
What is the expected output format? Should I return a list of the point coordinates that intersect, or a list of indices corresponding to those points in the input list?

Brute Force Solution

Approach

We need to find which points are covered by any of the cars. The brute force method checks each point against every car to see if it's covered.

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

Take the first point.
Look at the first car. Is the point within the start and end position of the car?
If it is, mark that the point is covered. If it isn't, move to the next car.
Repeat the previous step for all the cars. If the point is covered by at least one car, we know it's a valid intersection.
Move to the next point and repeat the entire process, checking it against all the cars.
Do this for every point. In the end, you'll know exactly which points are intersected by at least one car.

Code Implementation

def find_intersecting_points_brute_force(points, cars):
    intersecting_points = []

    for point in points:
        is_point_covered = False

        # Iterate through each car to see if the current point is covered
        for car_start, car_end in cars:

            # Check if the point falls within the start and end positions of the current car.
            if car_start <= point <= car_end:
                is_point_covered = True
                break

        # Add the point to the result if it's covered by at least one car
        if is_point_covered:
            intersecting_points.append(point)

    return intersecting_points

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

First, think about each car's journey as starting at one point in time and ending at another. We only care about these 'start' and 'end' times.
Make a list of all the 'start' times and another list of all the 'end' times, keeping track of which car each time belongs to.
Combine and sort these lists into a single timeline of events, showing when cars start and end their journeys, like appointments on a calendar.
For each point we want to check, imagine drawing a vertical line on this timeline. Count how many car 'start' times are to the left of our line (the cars have already started their journey).
Then, count how many car 'end' times are also to the left of our line (the cars have already finished their journey).
The number of cars that our point intersects with is the difference between those two counts: cars that have started but haven't ended yet.

Code Implementation

def points_that_intersect_with_cars(car_start_times, car_end_times, points_to_check):
    start_times = []
    end_times = []

    for car_index in range(len(car_start_times)):
        start_times.append((car_start_times[car_index], 1))
        end_times.append((car_end_times[car_index], -1))

    # Combine start and end times into a single timeline.
    timeline = sorted(start_times + end_times)

    results = []

    for point in points_to_check:
        cars_present = 0
        # Iterate through the timeline to calculate intersections.
        for time, type in timeline:
            if time <= point:
                cars_present += type

        results.append(cars_present)

    return results

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Cars array is null or empty	Return an empty list since there are no cars to intersect with.
Points array is null or empty	Return an empty list since there are no points to check for intersection.
Cars array has zero length segments	Filter out cars with start and end point same to avoid division by zero errors.
All points are outside of all car segments	Return an empty list as no points intersect with any cars.
Points array contains duplicate points	The solution should correctly identify if these points intersect regardless of duplication.
Integer overflow in calculations, especially for very large coordinates	Use long data type for intermediate calculations to avoid overflow issues.
Large input arrays exceeding memory limits	Consider processing points in batches or using a more memory-efficient data structure if necessary.
Points are on the car segment's boundaries (start or end point)	Define clearly whether points on the boundaries are considered intersecting based on the problem requirements and implement the condition accordingly.

Cars array is null or empty

How to Handle:

Return an empty list since there are no cars to intersect with.

Points array is null or empty

How to Handle:

Return an empty list since there are no points to check for intersection.

Cars array has zero length segments

How to Handle:

Filter out cars with start and end point same to avoid division by zero errors.

All points are outside of all car segments

How to Handle:

Return an empty list as no points intersect with any cars.

Points array contains duplicate points

How to Handle:

The solution should correctly identify if these points intersect regardless of duplication.

Integer overflow in calculations, especially for very large coordinates

How to Handle:

Use long data type for intermediate calculations to avoid overflow issues.

Large input arrays exceeding memory limits

How to Handle:

Consider processing points in batches or using a more memory-efficient data structure if necessary.

Points are on the car segment's boundaries (start or end point)

How to Handle:

Define clearly whether points on the boundaries are considered intersecting based on the problem requirements and implement the condition accordingly.