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Number of Ways to Rearrange Sticks With K Sticks Visible

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

Solution Explanation for Number of Ways to Rearrange Sticks With K Sticks Visible

This problem asks to find the number of ways to arrange n uniquely-sized sticks (lengths 1 to n) such that exactly k sticks are visible from the left. A stick is visible if no longer sticks are to its left.

The most efficient approach is using dynamic programming.

1. Dynamic Programming Approach (with space optimization):

  • State: dp[i][j] represents the number of arrangements of i sticks where j sticks are visible.

  • Base Case: dp[0][0] = 1 (no sticks, no visible sticks).

  • Recurrence Relation: To calculate dp[i][j], consider two cases for the i-th stick (the longest stick):

    • Case 1: The i-th stick is visible. If it's visible, it must be the leftmost stick. The remaining i-1 sticks must have j-1 visible sticks. Therefore, we add dp[i-1][j-1] to our count.

    • Case 2: The i-th stick is not visible. This means it's placed somewhere among the other i-1 sticks. There are i-1 possible positions for this stick. For each position, we still need to arrange the other i-1 sticks such that j sticks are visible. So we add (i-1) * dp[i-1][j].

  • Final Result: The final answer is dp[n][k].

2. Space Optimization:

We can optimize the space complexity by observing that dp[i][j] only depends on the previous row (i-1). Therefore, we can use a 1D array to store the current row, reducing space from O(n*k) to O(k). The iterative calculation updates the array in-place.

Time and Space Complexity:

  • Time Complexity: O(n*k) due to the nested loops in the dynamic programming solution. We iterate through all possible numbers of sticks and visible sticks.

  • Space Complexity: O(k) after space optimization. We only need a 1D array of size k to store the DP states.

Code Implementation (Python with space optimization):

def rearrangeSticks(n: int, k: int) -> int:
    mod = 10**9 + 7
    dp = [0] * (k + 1)  # 1D DP array
    dp[0] = 1  # Base case
 
    for i in range(1, n + 1):
        new_dp = [0] * (k + 1) # Temporary array to store new row
        for j in range(1, k + 1):
            new_dp[j] = (dp[j - 1] + (i - 1) * dp[j]) % mod
        dp = new_dp # Update dp to the new row
 
    return dp[k]

The code in other languages (Java, C++, Go, TypeScript) follows a similar structure and logic, employing the space-optimized 1D DP array for efficient calculation. The modulo operation (% mod) ensures that the result remains within the specified range.