LeetCode 1866. Number of Ways to Rearrange Sticks With K Sticks Visible Solution in Java, C++, Python & More | Explanation + Code

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] × (i - 1).

Therefore, the state transition equation is:

f[i][j] = f[i - 1][j - 1] + f[i - 1][j] × (i - 1)

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

The time complexity is O(n × k), and the space complexity is O(n × k). Where n and k are the two integers given in the problem.

PythonJavaC++GoTypeScript

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](code-box)

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]; } }(code-box)

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]; } };(code-box)

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] }(code-box)

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]; }(code-box)

Solution 2: Dynamic Programming (Space Optimization)

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 × k), and the space complexity is O(k). Here, n and k are the two integers given in the problem.

PythonJavaC++GoTypeScript

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](code-box)

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]; } }(code-box)

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]; } };(code-box)

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] }(code-box)

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]; }(code-box)