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

<p>Return <em>the number of ways we can draw </em><code>k</code><em> non-overlapping line segments</em><em>.</em> Since this number can be huge, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1600-1699/1621.Number%20of%20Sets%20of%20K%20Non-Overlapping%20Line%20Segments/images/ex1.png" style="width: 179px; height: 222px;" />
<pre>
<strong>Input:</strong> n = 4, k = 2
<strong>Output:</strong> 5
<strong>Explanation:</strong> 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)}.
</pre>

<p><strong class="example">Example 2:</strong></p>

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

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> n = 30, k = 7
<strong>Output:</strong> 796297179
<strong>Explanation:</strong> The total number of possible ways to draw 7 line segments is 3796297200. Taking this number modulo 10<sup>9</sup> + 7 gives us 796297179.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= n &lt;= 1000</code></li>
	<li><code>1 &lt;= k &lt;= n-1</code></li>
</ul>
Math; Dynamic Programming; Combinatorics
class Solution:
    def numberOfSets(self, n: int, k: int) -> int:
        mod = 10**9 + 7
        f = [[0] * (k + 1) for _ in range(n + 1)]
        g = [[0] * (k + 1) for _ in range(n + 1)]
        f[1][0] = 1
        for i in range(2, n + 1):
            for j in range(k + 1):
                f[i][j] = (f[i - 1][j] + g[i - 1][j]) % mod
                g[i][j] = g[i - 1][j]
                if j:
                    g[i][j] += f[i - 1][j - 1]
                    g[i][j] %= mod
                    g[i][j] += g[i - 1][j - 1]
                    g[i][j] %= mod
        return (f[-1][-1] + g[-1][-1]) % mod
