Solution Explanation for LeetCode 1006: Clumsy Factorial

This problem requires calculating a "clumsy factorial," a modified version of the standard factorial where the operations alternate between multiplication, division, addition, and subtraction in a specific pattern. The key is to correctly handle the order of operations (multiplication and division before addition and subtraction) and the integer division.

Approach: Stack-based Simulation

The most efficient approach uses a stack to simulate the calculation process. Here's a breakdown:

1. Initialization:

   • Start with a stack containing only the input number n.
   • Initialize a variable k to 0. k will track the current operation (0: multiply, 1: divide, 2: add, 3: subtract).

2. Iteration:

   • Iterate from n-1 down to 1. For each number x:
     • Based on the value of k, perform the corresponding operation:
       • k=0 (multiply): Pop the top element from the stack, multiply it by x, and push the result back onto the stack.
       • k=1 (divide): Pop the top element, perform integer division (/), and push the result.
       • k=2 (add): Push x onto the stack.
       • k=3 (subtract): Push -x onto the stack.
     • Update k to (k + 1) % 4 to cycle through the operations.

3. Final Summation:

   • After the loop, the stack will contain a series of numbers. Sum the elements in the stack to get the final result.

Time and Space Complexity Analysis

• Time Complexity: O(n). The loop iterates n-1 times, and each operation within the loop takes constant time. The final summation also takes linear time with respect to the size of the stack (which is at most n).
• Space Complexity: O(n). In the worst case, the stack can hold up to n elements.

Code Implementation (Python)

class Solution:
    def clumsy(self, n: int) -> int:
        stk = [n]  # Initialize stack with n
        k = 0      # Operation counter (0: *, 1: /, 2: +, 3: -)
        for x in range(n - 1, 0, -1):
            if k == 0:
                stk.append(stk.pop() * x)
            elif k == 1:
                stk.append(stk.pop() // x) # Integer division
            elif k == 2:
                stk.append(x)
            else:
                stk.append(-x)
            k = (k + 1) % 4
        return sum(stk)

The code in other languages (Java, C++, Go, TypeScript) follows the same logic, using their respective stack implementations and syntax. The core algorithm remains consistent across all languages. Note the use of integer division (// in Python, / in Java, C++, Go, and | 0 in Typescript) to correctly handle the floor division requirement.