README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Hard	https://github.com/doocs/leetcode/edit/main/solution/1800-1899/1866.Number%20of%20Ways%20to%20Rearrange%20Sticks%20With%20K%20Sticks%20Visible/README_EN.md	2333	Weekly Contest 241 Q4	Math Dynamic Programming Combinatorics

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

中文文档

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

Solution 1: Dynamic Programming

We define $f[i][j]$ to represent the number of permutations of length $i$ in which exactly $j$ sticks can be seen. Initially, $f[0][0]=1$ and the rest $f[i][j]=0$. The answer is $f[n][k]$.

Consider whether the last stick can be seen. If it can be seen, it must be the longest. Then there are $i-1$ sticks in front of it, and exactly $j-1$ sticks can be seen, which is $f[i-1][j-1]$. If the last stick cannot be seen, it can be any one except the longest stick. Then there are $i-1$ sticks in front of it, and exactly $j$ sticks can be seen, which is $f[i-1][j]\times (i-1)$.

Therefore, the state transition equation is:

$$f[i][j]=f[i-1][j-1]+f[i-1][j]\times (i-1)$$

The final answer is $f[n][k]$.

We notice that $f[i][j]$ is only related to $f[i-1][j-1]$ and $f[i-1][j]$, so we can use a one-dimensional array to optimize the space complexity.

The time complexity is $O(n\times k)$, and the space complexity is $O(k)$. Here, $n$ and $k$ are the two integers given in the problem.

Python3

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]

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];
    }
}

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];
    }
};

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]
}

TypeScript

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];
}

Solution 2

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]

Java

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[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, 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]
}

TypeScript

function rearrangeSticks(n: number, k: number): number {
    const mod = 10 ** 9 + 7;
    const f: number[] = Array(n + 1).fill(0);
    f[0] = 1;
    for (let i = 1; i <= n; ++i) {
        for (let j = k; j; --j) {
            f[j] = (f[j] * (i - 1) + f[j - 1]) % mod;
        }
        f[0] = 0;
    }
    return f[k];
}