Number of Ways to Rearrange Sticks With K Sticks Visible

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Dynamic Programming

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

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on `n` (number of sticks) and `k` (number of visible sticks)? What is the maximum possible value for each?
Can `n` or `k` be zero? Can `k` ever be greater than `n`?
Are we only dealing with integer numbers of sticks, and should the result be an integer as well?
The problem asks for the number of ways to rearrange sticks, but what should I return if the result exceeds the maximum integer value? Should I use modulo?
If no arrangement results in exactly `k` visible sticks, what value should I return?

Brute Force Solution

Approach

The core idea is to try every possible arrangement of sticks. We then count how many arrangements have exactly the required number of visible sticks. This is very inefficient, but guaranteed to find the answer.

Here's how the algorithm would work step-by-step:

First, consider all possible orderings of the sticks.
For each ordering, go through the sticks from left to right.
Keep track of the tallest stick seen so far.
If a stick is taller than the tallest stick seen so far, it's visible.
Count the number of visible sticks in each arrangement.
If the number of visible sticks matches the required number, increase the count of valid arrangements.
After checking all possible arrangements, the final count is the answer.

Code Implementation

import itertools

def number_of_ways_brute_force(number_of_sticks, number_of_visible_sticks):
    number_of_arrangements = 0
    sticks = list(range(1, number_of_sticks + 1))

    # Iterate through all possible permutations
    for permutation in itertools.permutations(sticks):
        visible_sticks_count = 0
        tallest_stick_seen_so_far = 0

        # Iterate through each stick in the current arrangement
        for stick in permutation:
            # If current stick is taller, it is visible
            if stick > tallest_stick_seen_so_far:

                visible_sticks_count += 1
                tallest_stick_seen_so_far = stick

        # Check if this arrangement matches the number of visible sticks
        if visible_sticks_count == number_of_visible_sticks:

            number_of_arrangements += 1

    return number_of_arrangements

Big(O) Analysis

Time Complexity

O(n!) – The algorithm considers all possible orderings of n sticks, which is n! (n factorial). For each of these n! permutations, the algorithm iterates through the n sticks to count the visible sticks. Therefore, the time complexity is dominated by generating and processing all permutations, resulting in O(n! * n). However, the n! term overwhelms the linear factor of n, simplifying the time complexity to O(n!).

Space Complexity

O(N!) – The plain English explanation outlines generating all possible permutations of the N sticks. Generating all permutations inherently requires storing those permutations, at least temporarily during the recursive exploration. Although not explicitly stated, a common implementation of generating permutations will use a call stack to keep track of the current permutation being built, and that call stack can grow up to a depth of N. Furthermore, generating all N! permutations implies that at some point, we are likely to store a permutation which takes O(N) space. Therefore the recursive generation contributes a space complexity tied to the total number of possible arrangements, which is factorial N (N!).

Optimal Solution

Approach

The core idea is to build the solution incrementally using dynamic programming. We consider adding one stick at a time and see how it affects the number of visible sticks, avoiding recalculating redundant information.

Here's how the algorithm would work step-by-step:

Imagine we're placing sticks one by one from smallest to largest.
Consider the last stick we place, the tallest one. It can either be visible or hidden.
If the tallest stick is visible, it must be placed at the beginning of the arrangement. Then, we need to arrange the remaining sticks to have the required number of visible sticks minus one.
If the tallest stick is hidden, we can place it in any position except at the beginning. The number of visible sticks remains the same as if it weren't present.
We can build a table where rows represent the number of sticks placed, and columns represent the number of visible sticks. Each cell will store the number of arrangements for that situation.
We calculate each cell based on the two possibilities: tallest stick visible or tallest stick hidden, using the values from previously filled cells in the table.
The final answer will be the value in the cell corresponding to placing all the sticks and having the required number of visible sticks.

Code Implementation

def rearrange_sticks(number_of_sticks, number_of_visible_sticks):
    modulo = 10**9 + 7
    # Table to store the number of ways for each subproblem.
    dp_table = [[0] * (number_of_visible_sticks + 1) for _ in range(number_of_sticks + 1)]
    dp_table[0][0] = 1

    for current_stick_count in range(1, number_of_sticks + 1):
        for current_visible_count in range(1, number_of_visible_sticks + 1):
            # If the tallest stick is visible, it must be at the beginning.
            dp_table[current_stick_count][current_visible_count] = dp_table[current_stick_count - 1][current_visible_count - 1]

            # If the tallest stick is hidden, it can be in any of the other positions.
            if current_stick_count > 1:
                dp_table[current_stick_count][current_visible_count] += (dp_table[current_stick_count - 1][current_visible_count] * (current_stick_count - 1)) % modulo

            dp_table[current_stick_count][current_visible_count] %= modulo

    # The result is stored in the bottom-right cell.
    return dp_table[number_of_sticks][number_of_visible_sticks]

Big(O) Analysis

Time Complexity

O(n*k) – The dynamic programming solution builds a table of size n x k, where n is the number of sticks and k is the number of visible sticks. The algorithm iterates through each cell of the table to calculate the number of arrangements. The outer loop goes from 1 to n, and the inner loop goes from 1 to k. Thus, the time complexity is proportional to the product of n and k because calculating each cell takes constant time. This results in an overall time complexity of O(n*k).

Space Complexity

O(N*K) – The algorithm uses a dynamic programming table to store intermediate results. This table has rows representing the number of sticks placed (up to N) and columns representing the number of visible sticks (up to K). Therefore, the table requires N*K cells to store the number of arrangements, which is the dominant factor in the space complexity. The space used by the table approximates to N*K. Thus the auxiliary space complexity is O(N*K).

Edge Cases

Case	How to Handle
n = 0, k > 0	Return 0 as there are no sticks to arrange to have any visible.
n > 0, k = 0	Return 0 as we must have at least one stick visible.
n = 1, k = 1	Return 1 as there is only one arrangement with one visible stick.
n > 0, k > n	Return 0 as it's impossible to have more visible sticks than total sticks.
n and k are moderately large (e.g., n = 50, k = 25)	Dynamic programming handles this without stack overflow or excessive recursion; memoization is key.
Integer overflow in intermediate calculations for larger n and k	Use modulo operator throughout calculation to prevent exceeding the integer limit.
n and k are both very large and close to each other (e.g., n = 1000, k = 999)	Ensure the dynamic programming table is large enough to accommodate large values of n and k without causing an out-of-bounds error.
k = 1 and n > 0	The visible stick must be the tallest; return (n-1)! ways to arrange the others.

n = 0, k > 0

How to Handle:

Return 0 as there are no sticks to arrange to have any visible.

n > 0, k = 0

How to Handle:

Return 0 as we must have at least one stick visible.

n = 1, k = 1

How to Handle:

Return 1 as there is only one arrangement with one visible stick.

n > 0, k > n

How to Handle:

Return 0 as it's impossible to have more visible sticks than total sticks.

n and k are moderately large (e.g., n = 50, k = 25)

How to Handle:

Dynamic programming handles this without stack overflow or excessive recursion; memoization is key.

Integer overflow in intermediate calculations for larger n and k

How to Handle:

Use modulo operator throughout calculation to prevent exceeding the integer limit.

n and k are both very large and close to each other (e.g., n = 1000, k = 999)

How to Handle:

Ensure the dynamic programming table is large enough to accommodate large values of n and k without causing an out-of-bounds error.

k = 1 and n > 0

How to Handle:

The visible stick must be the tallest; return (n-1)! ways to arrange the others.