Number Of Ways To Rearrange Sticks With K Sticks Visible - Leetcode Solution

Problem Statement: Given n sticks of lengths 1, 2, ..., n in a linear arrangement. Rearrange the sticks such that exactly k sticks are visible from the left. A stick is visible from the left if there are no taller sticks in front of it.

Example Input: n = 3, k = 2

Example Output: 3

Explanation: There are three sticks of lengths 1, 2, 3. Rearranging them, we can get the following arrangements where exactly 2 sticks are visible from the left: 1 3 2 2 1 3 3 1 2

Detailed Solution:

To solve this problem, we can use backtracking. We will generate all the possible arrangements of the sticks and check which of them have exactly k visible sticks.

Algorithm:

1. Create a boolean array 'used' of size n to keep track of which sticks have been used and which are still available for use.

2. Create a list 'arrangement' to store the current arrangement of sticks.

3. Create a helper function 'backtrack' that takes two arguments: 'num_visible', which is the number of visible sticks in the current arrangement, and 'position' which is the index of the next stick to be added to the arrangement.

4. In this function, check if 'num_visible' equals 'k'. If yes, add the current arrangement to a counter variable.

5. If 'position' >= n, return.

6. Loop through all the sticks from 1 to n, and check if the current stick is not already used.

7. If the current stick is not used, add it to the arrangement, mark it as used in the 'used' array, and call 'backtrack' recursively with 'num_visible+1' and the next 'position'.

8. After the recursive call returns, remove the current stick from the arrangement and mark it as unused in the 'used' array.

9. At the end of the function, return the counter variable.

Code:

Here's the python code for the Backtracking solution of the Number Of Ways To Rearrange Sticks With K Sticks Visible problem:

class Solution: def rearrangeSticks(self, n: int, k: int) -> int: def backtrack(num_visible, position): nonlocal counter if num_visible == k: counter += 1 return if position >= n: return for i in range(1, n+1): if not used[i]: arrangement.append(i) used[i] = True backtrack(num_visible + (i == max(arrangement)), position + 1) used[i] = False arrangement.pop()

    used = [False] * (n+1)
    arrangement = []
    counter = 0
    backtrack(0, 0)
    return counter

Time Complexity:

The time complexity of this solution is O(n!), as we are generating all n! possible arrangements of the sticks.

Space Complexity:

The space complexity of this solution is O(n), as we are using a boolean array of size n to keep track of which sticks have been used and which are still available for use, and a list to store the current arrangement.

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