How do you solve Clumsy Factorial on LeetCode?

Clumsy Factorial (LeetCode #1006) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding the order of operations is crucial.

What is the brute force solution for Clumsy Factorial?

The brute force approach for Clumsy Factorial is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach simulates the clumsy factorial by directly applying the operations in the specified order. We calculate the result step by step, maintaining the correct order of operations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Clumsy Factorial?

Clumsy Factorial uses the following concepts: Math, Stack, Simulation. The recommended patterns to study are: Stack, Simulation.

What are the interview tips for Clumsy Factorial?

Clarify the order of operations before coding. Consider edge cases, like small values of n. Explain your thought process clearly while coding.

What are common mistakes when solving Clumsy Factorial?

Forgetting to apply floor division correctly. Not managing the operation order properly.

#1006

Clumsy Factorial

Medium

MathStackSimulationStackSimulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a stack to manage the operations efficiently, allowing us to handle multiplication and division first, followed by addition and subtraction. This approach reduces unnecessary calculations and simplifies the process.

⚙️

Algorithm

3 steps

1Step 1: Initialize a stack and a variable to hold the current result.
2Step 2: Loop from n down to 1, applying operations based on the current operation index.
3Step 3: Push the result onto the stack for addition and subtraction, and calculate directly for multiplication and division.

solution.py16 lines

1def clumsy(n):
2    stack = []
3    current = n
4    op = 0
5    while current > 0:
6        if op == 0:  # Multiply
7            stack.append(current if not stack else stack.pop() * current)
8        elif op == 1:  # Divide
9            stack.append(stack.pop() // current)
10        elif op == 2:  # Add
11            stack.append(current)
12        elif op == 3:  # Subtract
13            stack.append(-current)
14        current -= 1
15        op = (op + 1) % 4
16    return sum(stack)

ℹ

Complexity note: The time complexity is O(n) since we process each number once, and the space complexity is O(n) due to the stack used to store intermediate results.

1Understanding the order of operations is crucial.
2Using a stack can simplify handling multiple operations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.