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];
    }
}

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