Clumsy Factorial - Leetcode Solution

The Clumsy Factorial problem on LeetCode is a mathematical problem that involves computing the factorial of a given number using a certain formula.

The problem statement provides the following information:

Given an integer n, calculate the value of the factorial of that number using the following formula:

f(n) = n * (n - 1) / (n - 2) + (n - 3) - (n - 4) / (n - 5) + ... + 1

where:

• f(0) = 1
• n >= 1 is an integer

Note: In this formula, the division is done using integer division.

To solve this problem, we can use a simple approach that involves using a stack and iteratively computing the factorial of the given number.

Algorithm:

1. Initialize a stack with the first two terms from the formula: [n, n-1]

2. Iterate through the remaining terms in the formula from n-2 to 1, and perform the following operations:

   a. If the current iteration is even (i.e., i%4==2), push the current term onto the stack with a negative sign. b. If the current iteration is odd (i.e., i%4==3), compute the multiplication of all the elements in the stack, and add it to the next term in the formula. Push the resulting value onto the stack.

3. Finally, compute the multiplication of all the elements in the stack and return the result as the factorial of the given number.

Example:

Let's take an example to understand the algorithm better.

Given n=5, we need to calculate the factorial of 5 using the formula:

f(n) = n * (n - 1) / (n - 2) + (n - 3) - (n - 4) / (n - 5) + ... + 1

Using the algorithm, we can perform the following steps:

1. Initialize a stack with the first two terms from the formula: [5, 4]

2. Iterate through the remaining terms in the formula from n-2 to 1:

   i=3:

   • i%4==2, push -3 onto the stack, stack=[5, 4, -3]

   i=2:

   • i%4==0, do nothing

   i=1:

   • i%4==3, compute multiplication of all elements on the stack, (-3)45= -60, add it to the next term in the formula (i.e., 2), result = -58
   • push -58 onto the stack, stack=[-58]

3. Compute multiplication of all the elements in the stack: (-58) = -58, which is the factorial of 5.

Therefore, the result of the computation is -58.

Complexity Analysis:

The time complexity of the algorithm is O(n), where n is the given number.

The space complexity of the algorithm is also O(n), as we are using a stack to store the intermediate results. However, the size of the stack does not exceed n/2, so the space complexity can be considered as O(n/2) = O(n).

Overall, the algorithm is efficient and has a linear time complexity, making it suitable for large values of n.

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