Introduction

In this blog post, we will explore how to solve the LeetCode problem "1006" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

The factorial of a positive integer n is the product of all positive integers less than or equal to n. For example, factorial(10) = 10 9 8 7 6 5 4 3 2 1. We make a clumsy factorial using the integers in decreasing order by swapping out the multiply operations for a fixed rotation of operations with multiply '', divide '/', add '+', and subtract '-' in this order. For example, clumsy(10) = 10 9 / 8 + 7 - 6 5 / 4 + 3 - 2 1. However, these operations are still applied using the usual order of operations of arithmetic. We do all multiplication and division steps before any addition or subtraction steps, and multiplication and division steps are processed left to right. Additionally, the division that we use is floor division such that 10 9 / 8 = 90 / 8 = 11. Given an integer n, return the clumsy factorial of n. Example 1: Input: n = 4 Output: 7 Explanation: 7 = 4 3 / 2 + 1 Example 2: Input: n = 10 Output: 12 Explanation: 12 = 10 9 / 8 + 7 - 6 5 / 4 + 3 - 2 1 Constraints: 1 <= n <= 104

Explanation

Here's the breakdown of the solution:

• Decomposition: The clumsy factorial can be broken down into groups of four operations (*, /, +, -). The last group might have fewer than four numbers.
• Calculation within groups: Perform the multiplication and division within each group first, then the addition and subtraction.

• Accumulation: Accumulate the results of each group, considering the sign of the first number in each group after the initial one.

• Runtime Complexity: O(n)

• Storage Complexity: O(1)

Code

    def clumsy(n):
    if n == 1:
        return 1
    if n == 2:
        return 2
    if n == 3:
        return 6

    result = n * (n - 1) // (n - 2)
    result += (n - 3)
    n -= 4

    while n > 0:
        term = - (n * (n - 1) // (n - 2))
        if n > 3:
            term += (n - 3)
            n -= 4
        elif n == 3:
            term += (n-3)
            n -=3
        elif n==2:
            term += (n-2)
            n-=2
        elif n==1:
            term += (n-1)
            n-=1

        result += term
    return result