Number of Ways to Rearrange Sticks With K Sticks Visible

// Accepted solution for LeetCode #1866: Number of Ways to Rearrange Sticks With K Sticks Visible
class Solution {
    public int rearrangeSticks(int n, int k) {
        final int mod = (int) 1e9 + 7;
        int[][] f = new int[n + 1][k + 1];
        f[0][0] = 1;
        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= k; ++j) {
                f[i][j] = (int) ((f[i - 1][j - 1] + f[i - 1][j] * (long) (i - 1)) % mod);
            }
        }
        return f[n][k];
    }
}

// Accepted solution for LeetCode #1866: Number of Ways to Rearrange Sticks With K Sticks Visible
func rearrangeSticks(n int, k int) int {
	const mod = 1e9 + 7
	f := make([][]int, n+1)
	for i := range f {
		f[i] = make([]int, k+1)
	}
	f[0][0] = 1
	for i := 1; i <= n; i++ {
		for j := 1; j <= k; j++ {
			f[i][j] = (f[i-1][j-1] + (i-1)*f[i-1][j]) % mod
		}
	}
	return f[n][k]
}

# Accepted solution for LeetCode #1866: Number of Ways to Rearrange Sticks With K Sticks Visible
class Solution:
    def rearrangeSticks(self, n: int, k: int) -> int:
        mod = 10**9 + 7
        f = [[0] * (k + 1) for _ in range(n + 1)]
        f[0][0] = 1
        for i in range(1, n + 1):
            for j in range(1, k + 1):
                f[i][j] = (f[i - 1][j - 1] + f[i - 1][j] * (i - 1)) % mod
        return f[n][k]

// Accepted solution for LeetCode #1866: Number of Ways to Rearrange Sticks With K Sticks Visible
/**
 * [1866] Number of Ways to Rearrange Sticks With K Sticks Visible
 *
 * There are n uniquely-sized sticks whose lengths are integers from 1 to n. You want to arrange the sticks such that exactly k sticks are visible from the left. A stick is visible from the left if there are no longer sticks to the left of it.
 *
 * 	For example, if the sticks are arranged [<u>1</u>,<u>3</u>,2,<u>5</u>,4], then the sticks with lengths 1, 3, and 5 are visible from the left.
 *
 * Given n and k, return the number of such arrangements. Since the answer may be large, return it modulo 10^9 + 7.
 *  
 * Example 1:
 *
 * Input: n = 3, k = 2
 * Output: 3
 * Explanation: [<u>1</u>,<u>3</u>,2], [<u>2</u>,<u>3</u>,1], and [<u>2</u>,1,<u>3</u>] are the only arrangements such that exactly 2 sticks are visible.
 * The visible sticks are underlined.
 *
 * Example 2:
 *
 * Input: n = 5, k = 5
 * Output: 1
 * Explanation: [<u>1</u>,<u>2</u>,<u>3</u>,<u>4</u>,<u>5</u>] is the only arrangement such that all 5 sticks are visible.
 * The visible sticks are underlined.
 *
 * Example 3:
 *
 * Input: n = 20, k = 11
 * Output: 647427950
 * Explanation: There are 647427950 (mod 10^9 + 7) ways to rearrange the sticks such that exactly 11 sticks are visible.
 *
 *  
 * Constraints:
 *
 * 	1 <= n <= 1000
 * 	1 <= k <= n
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/number-of-ways-to-rearrange-sticks-with-k-sticks-visible/
// discuss: https://leetcode.com/problems/number-of-ways-to-rearrange-sticks-with-k-sticks-visible/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn rearrange_sticks(n: i32, k: i32) -> i32 {
        0
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    #[ignore]
    fn test_1866_example_1() {
        let n = 3;
        let k = 2;

        let result = 3;

        assert_eq!(Solution::rearrange_sticks(n, k), result);
    }

    #[test]
    #[ignore]
    fn test_1866_example_2() {
        let n = 5;
        let k = 5;

        let result = 1;

        assert_eq!(Solution::rearrange_sticks(n, k), result);
    }

    #[test]
    #[ignore]
    fn test_1866_example_3() {
        let n = 20;
        let k = 11;

        let result = 647427950;

        assert_eq!(Solution::rearrange_sticks(n, k), result);
    }
}

// Accepted solution for LeetCode #1866: Number of Ways to Rearrange Sticks With K Sticks Visible
function rearrangeSticks(n: number, k: number): number {
    const mod = 10 ** 9 + 7;
    const f: number[][] = Array.from({ length: n + 1 }, () =>
        Array.from({ length: k + 1 }, () => 0),
    );
    f[0][0] = 1;
    for (let i = 1; i <= n; ++i) {
        for (let j = 1; j <= k; ++j) {
            f[i][j] = (f[i - 1][j - 1] + (i - 1) * f[i - 1][j]) % mod;
        }
    }
    return f[n][k];
}