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1866. Number of Ways to Rearrange Sticks With K Sticks Visible

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Description

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 [1,3,2,5,4], 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 109 + 7.

 

Example 1:

Input: n = 3, k = 2
Output: 3
Explanation: [1,3,2], [2,3,1], and [2,1,3] 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: [1,2,3,4,5] 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 109 + 7) ways to rearrange the sticks such that exactly 11 sticks are visible.

 

Constraints:

• 1 <= n <= 1000
• 1 <= k <= n

Solutions

Python3

class Solution:
    def rearrangeSticks(self, n: int, k: int) -> int:
        mod = 10**9 + 7
        f = [1] + [0] * k
        for i in range(1, n + 1):
            for j in range(k, 0, -1):
                f[j] = (f[j] * (i - 1) + f[j - 1]) % mod
            f[0] = 0
        return f[k]

class Solution:
    def rearrangeSticks(self, n: int, k: int) -> int:
        mod = 10**9 + 7
        f = [1] + [0] * k
        for i in range(1, n + 1):
            for j in range(k, 0, -1):
                f[j] = (f[j] * (i - 1) + f[j - 1]) % mod
            f[0] = 0
        return f[k]

Java

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];
    }
}

class Solution {
    public int rearrangeSticks(int n, int k) {
        final int mod = (int) 1e9 + 7;
        int[] f = new int[k + 1];
        f[0] = 1;
        for (int i = 1; i <= n; ++i) {
            for (int j = k; j > 0; --j) {
                f[j] = (int) ((f[j] * (i - 1L) + f[j - 1]) % mod);
            }
            f[0] = 0;
        }
        return f[k];
    }
}

C++

class Solution {
public:
    int rearrangeSticks(int n, int k) {
        const int mod = 1e9 + 7;
        int f[n + 1][k + 1];
        memset(f, 0, sizeof(f));
        f[0][0] = 1;
        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= k; ++j) {
                f[i][j] = (f[i - 1][j - 1] + (i - 1LL) * f[i - 1][j]) % mod;
            }
        }
        return f[n][k];
    }
};

class Solution {
public:
    int rearrangeSticks(int n, int k) {
        const int mod = 1e9 + 7;
        int f[k + 1];
        memset(f, 0, sizeof(f));
        f[0] = 1;
        for (int i = 1; i <= n; ++i) {
            for (int j = k; j; --j) {
                f[j] = (f[j - 1] + f[j] * (i - 1LL)) % mod;
            }
            f[0] = 0;
        }
        return f[k];
    }
};

Go

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]
}

func rearrangeSticks(n int, k int) int {
	const mod = 1e9 + 7
	f := make([]int, k+1)
	f[0] = 1
	for i := 1; i <= n; i++ {
		for j := k; j > 0; j-- {
			f[j] = (f[j-1] + f[j]*(i-1)) % mod
		}
		f[0] = 0
	}
	return f[k]
}

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