Formatted question description: https://leetcode.ca/all/1866.html

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

Level

Hard

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 10^9 + 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

Solution

Use dynamic programming. Create a 2D array dp with n + 1 rows and n + 1 columns, where dp[i][j] represents the number of ways to rearrange i sticks with j sticks visible, where i >= j >= 1.

The base case of dp is that dp[i][i] = 1 for all 1 <= i <= n. For the rest values, dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j] * (i - 1).

Finally, return dp[n][k].

  • class Solution {
        public int rearrangeSticks(int n, int k) {
            final int MODULO = 1000000007;
            int[][] dp = new int[n + 1][n + 1];
            for (int i = 1; i <= n; i++)
                dp[i][i] = 1;
            for (int i = 2; i <= n; i++) {
                dp[i][1] = (int) ((long) dp[i - 1][1] * (i - 1) % MODULO);
                for (int j = 2; j <= i; j++)
                    dp[i][j] = (int) ((dp[i - 1][j - 1] + (long) dp[i - 1][j] * (i - 1)) % MODULO);
            }
            return dp[n][k];
        }
    }
    
  • // OJ: https://leetcode.com/problems/number-of-ways-to-rearrange-sticks-with-k-sticks-visible/
    // Time: O(NK)
    // Space: O(Nk)
    long dp[1001][1001] = {}, fact[1001] = {}, mod = 1e9 + 7;
    class Solution {
        long dfs(int n, int k) {
            if (n < k) return 0;
            if (k == n) return 1;
            if (k == 1) return fact[n - 1];
            if (dp[n][k]) return dp[n][k];
            long ans = 0;
            ans = (ans + dfs(n - 1, k) * (n - 1) % mod) % mod;
            ans = (ans + dfs(n - 1, k - 1)) % mod;
            return dp[n][k] = ans;
        }
    public:
        int rearrangeSticks(int n, int k) {
            memset(dp, 0, sizeof(dp));
            fact[0] = 1;
            for (int i = 1; i <= n; ++i) {
                fact[i] = (fact[i - 1] * i) % mod; // fact(n) = n!
            }
            return dfs(n, k);
        }
    };
    
  • # 1866. Number of Ways to Rearrange Sticks With K Sticks Visible
    # https://leetcode.com/problems/number-of-ways-to-rearrange-sticks-with-k-sticks-visible
    
    class Solution:
        @functools.lru_cache(None)
        def rearrangeSticks(self, n, k, M = 10**9 + 7):
            if n == k: return 1
            if k == 0: return 0
            return (self.rearrangeSticks(n - 1, k - 1) + self.rearrangeSticks(n - 1, k) * (n - 1)) % M
            
    
    

All Problems

All Solutions