Number of Sets of K Non-Overlapping Line Segments

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Dynamic Programming

Given n points on a 1-D plane, where the i^th point (from 0 to n-1) is at x = i, find the number of ways we can draw exactly k non-overlapping line segments such that each segment covers two or more points. The endpoints of each segment must have integral coordinates. The k line segments do not have to cover all n points, and they are allowed to share endpoints.

Return the number of ways we can draw k non-overlapping line segments. Since this number can be huge, return it modulo 10⁹ + 7.

Example 1:

Input: n = 4, k = 2
Output: 5
Explanation: The two line segments are shown in red and blue.
The image above shows the 5 different ways {(0,2),(2,3)}, {(0,1),(1,3)}, {(0,1),(2,3)}, {(1,2),(2,3)}, {(0,1),(1,2)}.

Example 2:

Input: n = 3, k = 1
Output: 3
Explanation: The 3 ways are {(0,1)}, {(0,2)}, {(1,2)}.

Example 3:

Input: n = 30, k = 7
Output: 796297179
Explanation: The total number of possible ways to draw 7 line segments is 3796297200. Taking this number modulo 10⁹ + 7 gives us 796297179.

Constraints:

2 <= n <= 1000
1 <= k <= n-1

Try coding it in LeetCode

Number of Sets of K Non-Overlapping Line Segments

Medium

Asked by:

6 views

Topics:

Dynamic Programming

Given n points on a 1-D plane, where the i^th point (from 0 to n-1) is at x = i, find the number of ways we can draw exactly k non-overlapping line segments such that each segment covers two or more points. The endpoints of each segment must have integral coordinates. The k line segments do not have to cover all n points, and they are allowed to share endpoints.

Return the number of ways we can draw k non-overlapping line segments. Since this number can be huge, return it modulo 10⁹ + 7.

Example 1:

Input: n = 4, k = 2
Output: 5
Explanation: The two line segments are shown in red and blue.
The image above shows the 5 different ways {(0,2),(2,3)}, {(0,1),(1,3)}, {(0,1),(2,3)}, {(1,2),(2,3)}, {(0,1),(1,2)}.

Example 2:

Input: n = 3, k = 1
Output: 3
Explanation: The 3 ways are {(0,1)}, {(0,2)}, {(1,2)}.

Example 3:

Input: n = 30, k = 7
Output: 796297179
Explanation: The total number of possible ways to draw 7 line segments is 3796297200. Taking this number modulo 10⁹ + 7 gives us 796297179.

Constraints:

2 <= n <= 1000
1 <= k <= n-1

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 is the range for the values in the input array, and what is the maximum size of the array?
Can the value of 'k' be zero or negative, and what happens if 'k' is greater than the number of possible non-overlapping segments?
If there are no sets of 'k' non-overlapping line segments, what value should I return?
Are the line segments defined by consecutive elements in the input array (e.g., [a, b] represents a line segment) or by provided start and end indices?
Are the elements in the input array guaranteed to be distinct, or can there be duplicates which might affect the count of non-overlapping line segments?

Brute Force Solution

Approach

The brute force method for this problem means we're going to try every single possible combination of line segments to see which ones fit our criteria. We'll check each combination to see if the segments overlap and if we have the right number of segments.

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

First, consider all possible pairs of points to create the first line segment.
Then, for each of those segments, consider all possible pairs of points that are to the right of that segment to create the second line segment.
Check if the second line segment overlaps with the first. If they do, discard this combination and try another.
Continue this process, creating more line segments one at a time, each to the right of the previous one and not overlapping.
Keep checking for overlaps as you add each new segment, discarding invalid combinations.
If you arrive at a combination with the correct number of non-overlapping line segments, count it as a valid set.
Repeat this entire process until you have considered all possible combinations of points and segments.
Finally, count all the valid sets of line segments you found.

Code Implementation

def number_of_sets_of_k_non_overlapping_line_segments_brute_force(number_of_points, number_of_segments):
    count = 0

    def is_overlapping(segment1, segment2):
        return not (segment1[1] < segment2[0] or segment2[1] < segment1[0])

    def find_sets(current_segments, start_point):
        nonlocal count

        # If we have the desired number of segments, increment the count
        if len(current_segments) == number_of_segments:
            count += 1
            return

        # Iterate through all possible segments to the right of the current ones
        for first_point in range(start_point, number_of_points - 1):
            for second_point in range(first_point + 1, number_of_points):
                new_segment = (first_point, second_point)
                valid = True

                # Check for overlaps with existing segments
                for existing_segment in current_segments:
                    if is_overlapping(new_segment, existing_segment):
                        valid = False
                        break

                if valid:
                    # Only proceed if the new segment doesn't overlap
                    find_sets(current_segments + [new_segment], second_point + 1)

    find_sets([], 0)
    return count

Big(O) Analysis

Time Complexity

O(n^(2k)) – The described brute force approach considers all possible combinations of k non-overlapping line segments. Generating a single line segment involves choosing two points from n, which takes O(n^2) time. Since we need to generate k such segments, and for each segment, we iterate through potential points to the right of the previous segment, the overall complexity is dominated by the process of generating all combinations of these k segments. Therefore, the algorithm roughly performs O(n^2) operations k times which simplifies to O(n^(2k)).

Space Complexity

O(K) – The brute force approach, as described, relies heavily on recursion to explore all possible combinations of line segments. The depth of this recursion can go up to K, where K is the number of line segments we are trying to form. Each recursive call will store information about the current combination of segments being considered, including indices or pointers to the selected points. Therefore, the maximum space required for the recursion call stack is proportional to K, resulting in O(K) auxiliary space.

Optimal Solution

Approach

The problem asks to find the number of ways to draw a specific number of non-overlapping lines on a set of points. The optimal approach cleverly uses a concept called dynamic programming to avoid redundant calculations. It breaks down the problem into smaller, overlapping subproblems, solves each subproblem only once, and stores the results to reuse them later.

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

Imagine you have a table to store answers to smaller versions of the problem.
Each cell in the table represents the number of ways to draw a certain number of lines using only a certain number of points.
Start filling the table from the beginning: how many ways can you draw zero lines using any number of points? Or, how many ways can you draw any number of lines using only a few points?
To fill each cell, consider two options: either you draw a line ending at the current point, or you don't.
If you draw a line ending at the current point, figure out where the line starts and then look up the number of ways to draw the remaining lines using the points before the line starts.
If you don't draw a line ending at the current point, just look up the number of ways to draw the same number of lines using only the points before the current one.
Add up the results from these two options (draw a line or don't draw a line) to get the value for the current cell.
Keep filling the table until you reach the cell that represents the original problem (drawing the specific number of lines using all the points). That cell will contain the answer.

Code Implementation

def number_of_sets_of_k_non_overlapping_line_segments(number_of_points, number_of_segments):
    # DP table: ways[i][j] = ways to draw j segments using first i points
    ways = [[0] * (number_of_segments + 1) for _ in range(number_of_points + 1)]

    # Base case: 0 segments can be drawn in exactly 1 way
    for point_index in range(number_of_points + 1):
        ways[point_index][0] = 1

    for point_index in range(1, number_of_points + 1):
        for segment_index in range(1, number_of_segments + 1):
            # Option 1: Don't draw a line ending at point i
            ways[point_index][segment_index] = ways[point_index - 1][segment_index]

            # Option 2: Draw a line ending at point i
            # Iterate to find all possible starting points for the segment
            for starting_point_index in range(1, point_index):
                # Need at least 2 points to make a line
                ways[point_index][segment_index] += ways[starting_point_index - 1][segment_index - 1]

    # The bottom-right cell contains the answer for the complete problem
    return ways[number_of_points][number_of_segments]

Big(O) Analysis

Time Complexity

O(n^2 * k) – The dynamic programming solution involves filling a table with dimensions n (number of points) and k (number of line segments). For each cell dp[i][j] in the table, which represents the number of ways to draw j lines using i points, we iterate from 0 to i to consider all possible starting points for the last line segment (if we choose to draw a line ending at point i). This inner loop contributes a factor of n. Since we have n * k cells to fill, the overall time complexity is O(n * n * k), which simplifies to O(n^2 * k).

Space Complexity

O(N*K) – The dynamic programming approach, as described, employs a table to store the results of subproblems. This table has dimensions relating to the number of points and the number of lines, specifically N (number of points) and K (number of lines). Therefore, the table will have N rows and K columns to store the intermediate results. Consequently, the space complexity is determined by the size of this table, which is proportional to N*K.

Edge Cases

Case	How to Handle
Empty input array	Return 0 as no segments can be formed.
k is zero	Return 1 as an empty set of segments is always a valid solution.
k is greater than n/2 (where n is the length of the array)	Return 0 as it's impossible to have more segments than half the array size allows.
Array with only one distinct value repeated n times	Return 0 since no non-overlapping segments with distinct endpoints can be created if all values are equal.
Large n (array size) and large k values leading to integer overflow	Use appropriate data type (e.g., long) to prevent integer overflow, or consider using modulo operations if the problem allows it.
Negative numbers in the input array	The solution should work correctly regardless of whether the array contains negative numbers, as long as non-overlapping segments can be formed based on indices.
n = 2, k = 1	This is the smallest valid input; return 1 as long as the two values at different indices are different.
Input array is already sorted	The algorithm should handle the already sorted nature of input data without affecting outcome