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Minimum Number of Lines to Cover Points

Medium

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Topics:

ArraysGreedy Algorithms

You are given an array points where points[i] = [x_i, y_i] represents points on the X-Y plane.

Return the minimum number of lines required to cover all the points.

Example 1:

Input: points = [[0,1],[2,3],[4,5],[4,3]]
Output: 2
Explanation:
The minimum number of lines to cover the points is 2.
One possible solution is to draw a line connecting the points [0, 1] and [2, 3] and another line connecting the points [4, 3] and [4, 5].

Example 2:

Input: points = [[0,2],[2,2],[3,1],[2,4],[4,2],[5,4]]
Output: 3
Explanation:
The minimum number of lines to cover the points is 3.
One possible solution is to draw a line connecting the points [0, 2] and [5, 4], another line connecting the points [2, 2] and [3, 1], and another line connecting the points [2, 4] and [4, 2].

Constraints:

1 <= points.length <= 100
points[i].length == 2
-100 <= x_i, y_i <= 100
All the points are unique.

Try coding it in LeetCode

Minimum Number of Lines to Cover Points

Medium

Asked by:

5 views

Topics:

ArraysGreedy Algorithms

You are given an array points where points[i] = [x_i, y_i] represents points on the X-Y plane.

Return the minimum number of lines required to cover all the points.

Example 1:

Input: points = [[0,1],[2,3],[4,5],[4,3]]
Output: 2
Explanation:
The minimum number of lines to cover the points is 2.
One possible solution is to draw a line connecting the points [0, 1] and [2, 3] and another line connecting the points [4, 3] and [4, 5].

Example 2:

Input: points = [[0,2],[2,2],[3,1],[2,4],[4,2],[5,4]]
Output: 3
Explanation:
The minimum number of lines to cover the points is 3.
One possible solution is to draw a line connecting the points [0, 2] and [5, 4], another line connecting the points [2, 2] and [3, 1], and another line connecting the points [2, 4] and [4, 2].

Constraints:

1 <= points.length <= 100
points[i].length == 2
-100 <= x_i, y_i <= 100
All the points are unique.

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 number of points in the input array? Could it be empty, or very large?
What is the range of values for the x and y coordinates of each point? Can they be negative, zero, or floating-point numbers?
If all points are collinear (lie on a single line), should I return 1? What if there are no points, should I return 0?
Are there duplicate points in the input array? If so, how should they be handled?
Is there an expected type for the coordinates in the array, and is the input array always valid (not null)?
Are we looking for the minimum number of infinite lines that pass through ALL points, or lines segments?

Brute Force Solution

Approach

The brute force approach for finding the minimum number of lines to cover points involves trying all possible combinations of lines. We consider every possible pair of points to define a line and check if other points fall on that line. We repeat this process, aiming to cover all points with the fewest lines possible by exhaustively testing every combination.

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

First, consider using just one line. Try all possible pairs of points to define that single line.
Check if every other point lies on that line. If so, you're done - one line covers all points.
If one line isn't enough, try using two lines. Pick any two points to define the first line.
Then, for each remaining pair of uncovered points, create a second line.
Check if all points are now covered by either the first or the second line. If so, record this solution.
Keep trying different combinations of points for the first and second lines until you've explored all possibilities.
If two lines aren't enough, repeat the process for three lines, four lines, and so on, until all points are covered.
Finally, compare all the recorded solutions (the number of lines needed for each successful combination) and choose the solution that uses the fewest lines.

Code Implementation

def minimum_lines_to_cover_points(points):
    number_of_points = len(points)

    if number_of_points <= 0:
        return 0

    for number_of_lines in range(1, number_of_points + 1):
        for combinations_of_lines in find_combinations_of_lines(points, number_of_lines):
            all_points_covered = True
            for point in points:
                covered = False
                for line in combinations_of_lines:
                    if is_point_on_line(point, line):
                        covered = True
                        break
                if not covered:
                    all_points_covered = False
                    break

            if all_points_covered:
                return number_of_lines

    return number_of_points

def find_combinations_of_lines(points, number_of_lines):
    if number_of_lines == 0:
        return [[]]

    if not points:
        return []

    combinations = []

    # Create the first line
    for first_point_index in range(len(points)):
        for second_point_index in range(first_point_index + 1, len(points)):
            first_line = (points[first_point_index], points[second_point_index])
            remaining_points = points[:first_point_index] + points[first_point_index+1:second_point_index] + points[second_point_index+1:]
            
            # Recursively find the combinations for the remaining lines
            for remaining_combinations in find_combinations_of_lines(points, number_of_lines - 1):
                combinations.append([first_line] + remaining_combinations)
    return combinations

def is_point_on_line(point, line):
    point_one, point_two = line

    if point_one[0] == point_two[0]:
        # Vertical Line
        return point[0] == point_one[0]
    if point_one[1] == point_two[1]:
        # Horizontal Line
        return point[1] == point_one[1]

    slope = (point_two[1] - point_one[1]) / (point_two[0] - point_one[0])
    intercept = point_one[1] - slope * point_one[0]

    # Check if point satisfies y = mx + c
    return abs(point[1] - (slope * point[0] + intercept)) < 1e-6

Big(O) Analysis

Time Complexity

O(n^n) – The brute force approach considers all possible combinations of lines formed by pairs of points to cover all n points. In the worst case, we might need up to n lines. For each additional line we consider, we have to iterate through all possible combinations of points to define that line. Since we are choosing pairs of points for each line, there are O(n^2) possibilities for each line. In the worst-case scenario, we might have to consider up to n lines, resulting in (n^2) * (n^2) * ... (n times) which approximates O(n^(2n)), which can be simplified to O(n^n). Therefore, trying all these combinations results in a time complexity of O(n^n).

Space Complexity

O(1) – The brute-force algorithm primarily utilizes a constant amount of extra space. It iterates through pairs of points and checks if other points lie on the defined line, but it doesn't store combinations of lines or intermediate covered point states beyond what is needed for the current line being evaluated. Therefore, the space required remains constant regardless of the number of input points (N), as it mainly involves comparing coordinates and counting. The only auxiliary space used is for a few variables to keep track of indices and counts, leading to constant space complexity.

Optimal Solution

Approach

The goal is to find the fewest lines needed to connect a bunch of points. The key idea is that any two points define a line, and the best strategy is to keep adding points to the same line as long as they fit on it. By doing this, we minimize the number of lines we have to draw.

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

Start with the first point and treat it as the start of a potential line.
Look at each of the following points, one at a time.
Check if the new point falls on the same line as the previous points. You can imagine drawing a line between the first and second point. Does the third point fall on that line?
If the point is on the same line, great! Add it to the current line and move on to the next point.
If the point is NOT on the same line, that means you need a new line. Increment the line count by one.
Start a new line with the point that didn't fit on the previous line.
Repeat steps 2-6 until you've considered all the points.
The total count of lines will be the minimum number needed to connect all the points.

Code Implementation

def minimum_lines_to_cover_points(points):
    if not points:
        return 0

    number_of_lines = 1
    current_line_start_point_x = points[0][0]
    current_line_start_point_y = points[0][1]

    # Iterate through the rest of the points
    for i in range(1, len(points)):
        if i == 1:
            second_point_x = points[i][0]
            second_point_y = points[i][1]
            continue

        current_point_x = points[i][0]
        current_point_y = points[i][1]

        # Check if the current point is on the same line
        if (second_point_x - current_line_start_point_x) * \
           (current_point_y - current_line_start_point_y) != \
           (second_point_y - current_line_start_point_y) * \
           (current_point_x - current_line_start_point_x):

            # Need a new line if the point is not on the same line
            number_of_lines += 1
            current_line_start_point_x = current_point_x
            current_line_start_point_y = current_point_y
            second_point_x = current_point_x
            second_point_y = current_point_y
            continue

    return number_of_lines

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through the input list of n points. For each point, it potentially checks all subsequent points to see if they lie on the same line. This line checking involves a calculation (determining if a point is collinear), which we consider to be O(1). The number of collinearity checks in the worst case approximates to (n-1) + (n-2) + ... + 1. This summation is equivalent to n*(n-1)/2. Therefore, the time complexity simplifies to O(n²).

Space Complexity

O(1) – The algorithm, as described, primarily uses a constant number of variables to keep track of the current line and iterate through the points. It doesn't create any auxiliary data structures like lists, hash maps, or arrays to store intermediate results or track visited points. Therefore, the space used remains constant irrespective of the number of input points, denoted as N. The space complexity is O(1).

Edge Cases

Case	How to Handle
Empty input array	Return 0 since no lines are needed to cover no points.
Null input array	Throw an IllegalArgumentException or return a predefined error code like -1 to indicate invalid input.
Input array with only one point	Return 1 as a single point requires only one (degenerate) line to cover it.
Input array with two points	Return 1 since two points define a line.
All points are identical	Return 1 because all identical points can be covered by a single point (degenerate line).
Large number of points (scalability)	Ensure the algorithm avoids unnecessary computations or quadratic complexity to maintain acceptable performance with large inputs.
Points forming vertical or horizontal lines	Handle division by zero errors when calculating slopes, typically by treating vertical lines as a special case.
Integer overflow during slope calculation	Use long data type or approximate floating point comparison for slope calculation to prevent integer overflow.

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 number of points in the input array? Could it be empty, or very large?
What is the range of values for the x and y coordinates of each point? Can they be negative, zero, or floating-point numbers?
If all points are collinear (lie on a single line), should I return 1? What if there are no points, should I return 0?
Are there duplicate points in the input array? If so, how should they be handled?
Is there an expected type for the coordinates in the array, and is the input array always valid (not null)?
Are we looking for the minimum number of infinite lines that pass through ALL points, or lines segments?

Brute Force Solution

Approach

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

First, consider using just one line. Try all possible pairs of points to define that single line.
Check if every other point lies on that line. If so, you're done - one line covers all points.
If one line isn't enough, try using two lines. Pick any two points to define the first line.
Then, for each remaining pair of uncovered points, create a second line.
Check if all points are now covered by either the first or the second line. If so, record this solution.
Keep trying different combinations of points for the first and second lines until you've explored all possibilities.
If two lines aren't enough, repeat the process for three lines, four lines, and so on, until all points are covered.
Finally, compare all the recorded solutions (the number of lines needed for each successful combination) and choose the solution that uses the fewest lines.

Code Implementation

def minimum_lines_to_cover_points(points):
    number_of_points = len(points)

    if number_of_points <= 0:
        return 0

    for number_of_lines in range(1, number_of_points + 1):
        for combinations_of_lines in find_combinations_of_lines(points, number_of_lines):
            all_points_covered = True
            for point in points:
                covered = False
                for line in combinations_of_lines:
                    if is_point_on_line(point, line):
                        covered = True
                        break
                if not covered:
                    all_points_covered = False
                    break

            if all_points_covered:
                return number_of_lines

    return number_of_points

def find_combinations_of_lines(points, number_of_lines):
    if number_of_lines == 0:
        return [[]]

    if not points:
        return []

    combinations = []

    # Create the first line
    for first_point_index in range(len(points)):
        for second_point_index in range(first_point_index + 1, len(points)):
            first_line = (points[first_point_index], points[second_point_index])
            remaining_points = points[:first_point_index] + points[first_point_index+1:second_point_index] + points[second_point_index+1:]
            
            # Recursively find the combinations for the remaining lines
            for remaining_combinations in find_combinations_of_lines(points, number_of_lines - 1):
                combinations.append([first_line] + remaining_combinations)
    return combinations

def is_point_on_line(point, line):
    point_one, point_two = line

    if point_one[0] == point_two[0]:
        # Vertical Line
        return point[0] == point_one[0]
    if point_one[1] == point_two[1]:
        # Horizontal Line
        return point[1] == point_one[1]

    slope = (point_two[1] - point_one[1]) / (point_two[0] - point_one[0])
    intercept = point_one[1] - slope * point_one[0]

    # Check if point satisfies y = mx + c
    return abs(point[1] - (slope * point[0] + intercept)) < 1e-6

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Start with the first point and treat it as the start of a potential line.
Look at each of the following points, one at a time.
Check if the new point falls on the same line as the previous points. You can imagine drawing a line between the first and second point. Does the third point fall on that line?
If the point is on the same line, great! Add it to the current line and move on to the next point.
If the point is NOT on the same line, that means you need a new line. Increment the line count by one.
Start a new line with the point that didn't fit on the previous line.
Repeat steps 2-6 until you've considered all the points.
The total count of lines will be the minimum number needed to connect all the points.

Code Implementation

def minimum_lines_to_cover_points(points):
    if not points:
        return 0

    number_of_lines = 1
    current_line_start_point_x = points[0][0]
    current_line_start_point_y = points[0][1]

    # Iterate through the rest of the points
    for i in range(1, len(points)):
        if i == 1:
            second_point_x = points[i][0]
            second_point_y = points[i][1]
            continue

        current_point_x = points[i][0]
        current_point_y = points[i][1]

        # Check if the current point is on the same line
        if (second_point_x - current_line_start_point_x) * \
           (current_point_y - current_line_start_point_y) != \
           (second_point_y - current_line_start_point_y) * \
           (current_point_x - current_line_start_point_x):

            # Need a new line if the point is not on the same line
            number_of_lines += 1
            current_line_start_point_x = current_point_x
            current_line_start_point_y = current_point_y
            second_point_x = current_point_x
            second_point_y = current_point_y
            continue

    return number_of_lines

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty input array	Return 0 since no lines are needed to cover no points.
Null input array	Throw an IllegalArgumentException or return a predefined error code like -1 to indicate invalid input.
Input array with only one point	Return 1 as a single point requires only one (degenerate) line to cover it.
Input array with two points	Return 1 since two points define a line.
All points are identical	Return 1 because all identical points can be covered by a single point (degenerate line).
Large number of points (scalability)	Ensure the algorithm avoids unnecessary computations or quadratic complexity to maintain acceptable performance with large inputs.
Points forming vertical or horizontal lines	Handle division by zero errors when calculating slopes, typically by treating vertical lines as a special case.
Integer overflow during slope calculation	Use long data type or approximate floating point comparison for slope calculation to prevent integer overflow.