# Solving Leetcode Interviews in Seconds with AI: Number of Sets of K Non-Overlapping Line Segments


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1621" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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.    Constraints:  2 <= n <= 1000 1 <= k <= n-1  

	# Explanation
	Here's the solution:

*   **Dynamic Programming:** Utilize dynamic programming to store the number of ways to draw `j` segments using the first `i` points.
*   **State Definition:** `dp[i][j]` represents the number of ways to draw `j` segments using the first `i+1` points (points from 0 to i).
*   **Transitions:** Consider whether the `j`-th segment ends at point `i`. If it does, iterate through all possible starting points of the segment to compute the contribution. Otherwise, inherit the result from `dp[i-1][j]`.

*   **Runtime Complexity:** O(n<sup>3</sup>k), **Storage Complexity:** O(n\*k)

	
	# Code
	```python
	def solve():
    n, k = map(int, input().split())
    MOD = 10**9 + 7
    
    dp = [[0] * (k + 1) for _ in range(n)]
    
    # Base case: 0 segments, 1 way
    # for i in range(n):
    #    dp[i][0] = 1
    
    for i in range(n):
        dp[i][0] = 1  # Initialize the number of ways to draw 0 segments up to point i as 1 (doing nothing)

    for j in range(1, k + 1):
        for i in range(1, n):
            dp[i][j] = dp[i-1][j] # Inherit from not including ith point in a segment

            for l in range(i):
                if (i - l) >= 1: # Ensure the segment has at least two points
                    dp[i][j] = (dp[i][j] + (dp[l][j-1])) % MOD
                    
    print(dp[n-1][k])

solve()
	```