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1621. Number of Sets of K Non-Overlapping Line Segments

Description

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

Solutions

  • class Solution {
    private static final int MOD = (int) 1e9 + 7;

    public int numberOfSets(int n, int k) {
        int[][] f = new int[n + 1][k + 1];
        int[][] g = new int[n + 1][k + 1];
        f[1][0] = 1;
        for (int i = 2; i <= n; ++i) {
            for (int j = 0; j <= k; ++j) {
                f[i][j] = (f[i - 1][j] + g[i - 1][j]) % MOD;
                g[i][j] = g[i - 1][j];
                if (j > 0) {
                    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[n][k] + g[n][k]) % MOD;
    }
}

  • class Solution {
public:
    int f[1010][1010];
    int g[1010][1010];
    const int mod = 1e9 + 7;

    int numberOfSets(int n, int k) {
        memset(f, 0, sizeof(f));
        memset(g, 0, sizeof(g));
        f[1][0] = 1;
        for (int i = 2; i <= n; ++i) {
            for (int j = 0; j <= k; ++j) {
                f[i][j] = (f[i - 1][j] + g[i - 1][j]) % mod;
                g[i][j] = g[i - 1][j];
                if (j > 0) {
                    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[n][k] + g[n][k]) % mod;
    }
};

  • 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


  • func numberOfSets(n int, k int) int {
	f := make([][]int, n+1)
	g := make([][]int, n+1)
	for i := range f {
		f[i] = make([]int, k+1)
		g[i] = make([]int, k+1)
	}
	f[1][0] = 1
	var mod int = 1e9 + 7
	for i := 2; i <= n; i++ {
		for j := 0; j <= k; j++ {
			f[i][j] = (f[i-1][j] + g[i-1][j]) % mod
			g[i][j] = g[i-1][j]
			if j > 0 {
				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[n][k] + g[n][k]) % mod
}

  • function numberOfSets(n: number, k: number): number {
    const f = Array.from({ length: n + 1 }, _ => new Array(k + 1).fill(0));
    const g = Array.from({ length: n + 1 }, _ => new Array(k + 1).fill(0));
    f[1][0] = 1;
    const mod = 10 ** 9 + 7;
    for (let i = 2; i <= n; ++i) {
        for (let j = 0; j <= k; ++j) {
            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[n][k] + g[n][k]) % mod;
}

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