Given n points on a 1-D plane, where the ith 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.


Example 4:

Input: n = 5, k = 3
Output: 7


Example 5:

Input: n = 3, k = 2
Output: 1

Constraints:

• 2 <= n <= 1000
• 1 <= k <= n-1

## Solution 1: Naive DP (TLE)

dp[n][k] := ans of problem(n, k)
dp[n] = n * (n – 1) / 2 # C(n,2)
dp[n][k] = 1 if k == n – 1
dp[n][k] = 0 if k >= n
dp[n][k] = sum((i – 1) * dp(n – i + 1, k – 1) 2 <= i < n

Time complexity: O(n^2*k)
Space complexity: O(n*k)

## Solution 2: DP w/ Prefix Sum

Time complexity: O(nk)
Space complexity: O(nk)

## Solution 3: DP / 3D State

Time complexity: O(nk)
Space complexity: O(nk)

## Solution 4: DP / Mathematical induction

Time complexity: O(nk)
Space complexity: O(nk)

## Solution 5: DP / Reduction

This problem can be reduced to: given n + k – 1 points, pick k segments (2*k points).
if two consecutive points were selected by two segments e.g. i for A and i+1 for B, then they share a point in the original space.
Answer C(n + k – 1, 2*k)

Time complexity: O((n+k)*2) Pascal’s triangle
Space complexity: O((n+k)*2)

## C++

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