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, 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(n³k), Storage Complexity: O(n*k)

Code

    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()