Problem Description

You have n points on a one-dimensional line, positioned at coordinates x = 0, 1, 2, ..., n-1. Your task is to draw exactly k non-overlapping line segments where:

1. Each segment must cover at least two points (a segment from point i to point j covers all points between and including i and j)

2. Segments cannot overlap - no two segments can share any interior points, though they can share endpoints

3. All endpoints must be at integral coordinates (at the given points)

4. You must draw exactly k segments - not more, not less

5. Not all points need to be covered - some points can remain uncovered by any segment

The problem asks you to count the total number of different ways to draw these k segments following all the rules above. Since the answer can be very large, return it modulo 10^9 + 7.

For example, if n = 4 (points at positions 0, 1, 2, 3) and k = 2, one valid way would be to draw segments [0,1] and [2,3]. Another valid way would be [0,2] and [2,3] (sharing endpoint at position 2).

The solution uses dynamic programming with two arrays:

• f[i][j]: number of ways to place j segments using first i points where the j-th segment does NOT end at point i-1
• g[i][j]: number of ways to place j segments using first i points where the j-th segment DOES end at point i-1

The recurrence relations handle the transitions between these states, considering whether to extend an existing segment or start a new one at each position.

Intuition

When thinking about this problem, we need to consider how line segments can be placed on a 1D line. The key insight is that at any given point, we need to make a decision: either we're continuing an existing segment or we're not part of any segment at that position.

Let's think about building our solution from left to right. At each point i, we have two states to track:

1. The last segment ends before point i (the point is "free")
2. The last segment ends exactly at point i (the point is an "endpoint")

This naturally leads us to define two DP states:

• f[i][j]: ways to place j segments using points 0 to i-1, where point i-1 is NOT the end of a segment
• g[i][j]: ways to place j segments using points 0 to i-1, where point i-1 IS the end of a segment

Why do we need both states? Because they transition differently:

1. If the previous point wasn't the end of a segment (f state), we can either:

   • Keep it free and move to the next point
   • Start a new segment from that point

2. If the previous point was the end of a segment (g state), we can:

   • Leave it as is and move to the next point
   • Extend the segment that just ended (making it longer)

The beauty of this approach is that it automatically handles non-overlapping constraints. When we're in state g (segment just ended), we can either start fresh or extend that specific segment - we never create overlaps.

For transitions:

• f[i][j] = f[i-1][j] + g[i-1][j]: Point i-1 is free if it was free before OR if a segment ended at i-2
• g[i][j] = g[i-1][j] + f[i-1][j-1] + g[i-1][j-1]: A segment ends at i-1 if we extend a previous segment OR start a new one

The final answer is f[n][k] + g[n][k] since the last point can be either free or the end of a segment.

Learn more about Math, Dynamic Programming and Combinatorics patterns.

Solution Approach

The implementation uses dynamic programming with two 2D arrays to track the different states:

Data Structures:

• f[i][j]: 2D array where f[i][j] represents the number of ways to place exactly j segments using the first i points, with point i-1 NOT being the endpoint of any segment
• g[i][j]: 2D array where g[i][j] represents the number of ways to place exactly j segments using the first i points, with point i-1 being the endpoint of a segment

Initialization:

• f[1][0] = 1: With only 1 point and 0 segments, there's exactly 1 way (the point remains uncovered)
• All other values start at 0

Transition Logic:

For each point i from 2 to n and for each number of segments j from 0 to k:

1. Calculate f[i][j] (point i-1 is not an endpoint):

   f[i][j] = (f[i-1][j] + g[i-1][j]) % mod

   This means point i-1 is free if:

   • It was already free in the previous state (f[i-1][j])
   • A segment ended at point i-2 (g[i-1][j])

2. Calculate g[i][j] (point i-1 is an endpoint):

   g[i][j] = g[i-1][j]  // Extend existing segment
   if j > 0:
       g[i][j] += f[i-1][j-1]  // Start new segment from free point
       g[i][j] += g[i-1][j-1]  // Start new segment after previous ended

   A segment can end at point i-1 if:

   • We extend a segment that was already ending at i-2 (g[i-1][j])
   • We start a new segment when we have j-1 segments already placed:
     • From a free point (f[i-1][j-1])
     • From a point where a segment just ended (g[i-1][j-1])

Final Answer:

return (f[n][k] + g[n][k]) % mod

The total number of ways is the sum of:

• Ways where the last point n-1 is not part of any segment
• Ways where the last point n-1 is the endpoint of a segment

Time Complexity: O(n * k) - We iterate through n points and for each point, we calculate values for up to k segments

Space Complexity: O(n * k) - We use two 2D arrays of size (n+1) × (k+1)

The modulo operation 10^9 + 7 is applied at each step to prevent integer overflow while maintaining the correctness of the result.

Example Walkthrough

Let's walk through a small example with n = 4 points (at positions 0, 1, 2, 3) and k = 2 segments.

Setup:

• Points: [0, 1, 2, 3]
• We need exactly 2 non-overlapping segments
• Initialize f[1][0] = 1 (1 point, 0 segments = 1 way)

Building the DP tables step by step:

Step 1: i = 2 (considering points 0, 1)

• f[2][0] = f[1][0] + g[1][0] = 1 + 0 = 1 (no segments)
• f[2][1] = f[1][1] + g[1][1] = 0 + 0 = 0
• g[2][0] = g[1][0] = 0
• g[2][1] = g[1][1] + f[1][0] + g[1][0] = 0 + 1 + 0 = 1 (segment [0,1])

Step 2: i = 3 (considering points 0, 1, 2)

• f[3][0] = f[2][0] + g[2][0] = 1 + 0 = 1
• f[3][1] = f[2][1] + g[2][1] = 0 + 1 = 1 (segment [0,1], point 2 free)
• f[3][2] = f[2][2] + g[2][2] = 0 + 0 = 0
• g[3][0] = g[2][0] = 0
• g[3][1] = g[2][1] + f[2][0] + g[2][0] = 1 + 1 + 0 = 2
  • Extend [0,1] to [0,2]: 1 way
  • New segment [1,2] or [2,2]: 1 way
• g[3][2] = g[2][2] + f[2][1] + g[2][1] = 0 + 0 + 1 = 1
  • After [0,1], add [1,2]: 1 way

Step 3: i = 4 (considering all points 0, 1, 2, 3)

• f[4][2] = f[3][2] + g[3][2] = 0 + 1 = 1
• g[4][2] = g[3][2] + f[3][1] + g[3][1] = 1 + 1 + 2 = 4

Final Answer: f[4][2] + g[4][2] = 1 + 4 = 5

The 5 valid configurations are:

1. [0,1], [2,3] - two separate segments
2. [0,2], [2,3] - segments share endpoint at 2
3. [0,1], [1,3] - segments share endpoint at 1
4. [0,1], [1,2] - segments share endpoint at 1
5. [1,2], [2,3] - segments share endpoint at 2

This demonstrates how the DP tracks whether points are free (f) or endpoints (g), enabling us to count all valid non-overlapping segment placements.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n * k)

The algorithm uses nested loops where:

• The outer loop iterates from 2 to n + 1, which gives us n - 1 iterations
• The inner loop iterates from 0 to k, which gives us k + 1 iterations
• Inside the inner loop, all operations (additions and modulo operations) are O(1)

Therefore, the total time complexity is O((n - 1) * (k + 1)) which simplifies to O(n * k).

Space Complexity: O(n * k)

The algorithm creates two 2D arrays:

• Array f with dimensions (n + 1) × (k + 1)
• Array g with dimensions (n + 1) × (k + 1)

Each array requires O((n + 1) * (k + 1)) space, and since we have two such arrays, the total space complexity is O(2 * (n + 1) * (k + 1)), which simplifies to O(n * k).

Note: The space complexity could potentially be optimized to O(k) since each iteration only depends on the previous row, allowing us to use rolling arrays instead of storing all rows.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Confusing Points with Segments

A critical misunderstanding is thinking that n points can form n-1 segments at most. While this is true for consecutive single-unit segments, segments can span multiple points. For example, with 4 points (0,1,2,3), segment [0,3] covers all 4 points but counts as just 1 segment.

Solution: Remember that a segment from point i to point j covers all points in between, and the maximum number of non-overlapping segments depends on how you partition the points, not just the gaps between consecutive points.

2. Incorrect Base Case Initialization

Many implementations incorrectly initialize dp_closed[0][0] = 1 or dp_open[0][0] = 1, thinking that 0 points with 0 segments is a valid configuration.

Solution: The correct base case is dp_closed[1][0] = 1 because you need at least 1 point to have a valid configuration. With 1 point and 0 segments, there's exactly one way (leave the point uncovered).

3. Off-by-One Errors in Array Indexing

The problem uses 0-indexed points (0 to n-1) but the DP arrays often use 1-indexed counting for the number of points. Mixing these can lead to accessing wrong indices or missing edge cases.

Solution: Be consistent with your indexing scheme. In the solution, dp[i][j] represents using i points (points 0 through i-1), not point index i.

4. Forgetting Modulo Operations

Missing modulo operations at any step can cause integer overflow, especially for large values of n and k. This leads to incorrect results even if the logic is correct.

Solution: Apply modulo after every addition operation:

# Wrong
dp_closed[i][j] = dp_closed[i-1][j] + dp_open[i-1][j]
result = result % MOD  # Too late!

# Correct
dp_closed[i][j] = (dp_closed[i-1][j] + dp_open[i-1][j]) % MOD

5. Misunderstanding State Transitions

A common error is thinking that an "open" segment at position i must start at position i. Actually, an open segment at position i means it ends at or extends through position i-1.

Solution: Clearly define what each state represents:

• dp_closed[i][j]: Using points 0 to i-1, with j segments, where no segment ends at point i-1
• dp_open[i][j]: Using points 0 to i-1, with j segments, where a segment ends at point i-1

6. Not Handling the Minimum Segment Length

Forgetting that each segment must cover at least 2 points. A segment cannot be a single point.

Solution: Ensure that when starting a new segment, there's room for it to extend to at least one more point. This is implicitly handled in the DP transitions but should be kept in mind when understanding the problem.