Problem Description

The clumsy factorial is a variation of the regular factorial where we apply operations in a specific rotating pattern.

In a regular factorial, we multiply all positive integers from 1 to n. For example, factorial(10) = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1.

In a clumsy factorial, we start with n and work our way down to 1, but instead of only using multiplication, we rotate through four operations in this exact order:

1. Multiply *
2. Divide / (using floor division)
3. Add +
4. Subtract -

For example, clumsy(10) = 10 * 9 / 8 + 7 - 6 * 5 / 4 + 3 - 2 * 1.

The operations follow standard arithmetic rules:

• Multiplication and division are performed before addition and subtraction
• Multiplication and division are evaluated from left to right
• Division uses floor division (rounds down to the nearest integer)

Given an integer n, you need to calculate and return the clumsy factorial of n.

The key insight is that the expression needs to be evaluated respecting operator precedence. In the example above:

• First calculate: 10 * 9 = 90, then 90 / 8 = 11
• Then calculate: 6 * 5 = 30, then 30 / 4 = 7
• Then calculate: 2 * 1 = 2
• Finally combine: 11 + 7 - 7 + 3 - 2 = 12

Intuition

When we look at the clumsy factorial expression like 10 * 9 / 8 + 7 - 6 * 5 / 4 + 3 - 2 * 1, we need to respect operator precedence. This means multiplication and division must be evaluated first before addition and subtraction.

The key observation is that we can group the operations into chunks based on operator precedence:

• 10 * 9 / 8 forms one chunk (evaluates to 11)
• + 7 is added
• - 6 * 5 / 4 forms another chunk (evaluates to -7)
• + 3 is added
• - 2 * 1 forms the last chunk (evaluates to -2)

This naturally suggests using a stack to handle the calculation. Why? Because:

1. When we encounter multiplication or division, we need to immediately evaluate it with the previous number (since these have higher precedence)
2. When we encounter addition or subtraction, we can defer the calculation until the end

The pattern becomes clear:

• For the first number n, push it onto the stack
• When we see *, pop the top, multiply with the current number, and push back
• When we see /, pop the top, divide by the current number, and push back
• When we see +, just push the current number (it will be added later)
• When we see -, push the negative of the current number (so we can sum everything at the end)

By tracking which operation we're on using a counter k that cycles through 0, 1, 2, 3 (representing *, /, +, -), we can simulate the entire process. At the end, the sum of all elements in the stack gives us our answer because we've already handled the multiplication/division immediately, and the remaining stack contains terms to be added or subtracted.

Learn more about Stack and Math patterns.

Solution Approach

We implement the solution using a stack and simulation approach.

Initialization:

• Create a stack stk and push the initial value n into it
• Initialize a counter k = 0 to track which operation we're currently on in the rotation

Main Process: We iterate through numbers from n-1 down to 1, and for each number x, we perform different operations based on the current value of k:

1. When k = 0 (Multiplication):

   • Pop the top element from the stack
   • Multiply it by x
   • Push the result back onto the stack
   • This handles expressions like 10 * 9

2. When k = 1 (Division):

   • Pop the top element from the stack
   • Divide it by x using floor division (int(stk.pop() / x))
   • Push the result back onto the stack
   • This completes chunks like 90 / 8 = 11

3. When k = 2 (Addition):

   • Simply push x onto the stack
   • We don't need to perform the addition immediately since it has lower precedence

4. When k = 3 (Subtraction):

   • Push -x onto the stack (negative value)
   • By storing the negative, we can sum all values at the end

Update Counter: After each operation, update k = (k + 1) % 4 to rotate through the four operations cyclically.

Final Result: Return sum(stk) - the sum of all elements in the stack. This works because:

• Multiplication and division operations have already been evaluated and their results are in the stack
• Addition terms are stored as positive values
• Subtraction terms are stored as negative values
• Summing them all gives the correct final result

Example Walkthrough for n = 10:

• Start: stk = [10], k = 0
• x = 9: multiply → stk = [90], k = 1
• x = 8: divide → stk = [11], k = 2
• x = 7: add → stk = [11, 7], k = 3
• x = 6: subtract → stk = [11, 7, -6], k = 0
• x = 5: multiply → stk = [11, 7, -30], k = 1
• x = 4: divide → stk = [11, 7, -7], k = 2
• x = 3: add → stk = [11, 7, -7, 3], k = 3
• x = 2: subtract → stk = [11, 7, -7, 3, -2], k = 0
• x = 1: multiply → stk = [11, 7, -7, 3, -2], k = 1
• Final: sum([11, 7, -7, 3, -2]) = 12

Example Walkthrough

Let's walk through calculating clumsy(5) = 5 * 4 / 3 + 2 - 1.

Initial Setup:

• Stack: [5]
• Counter k = 0 (tracks which operation: 0=multiply, 1=divide, 2=add, 3=subtract)
• We'll process numbers from 4 down to 1

Step-by-Step Execution:

Step 1: Process number 4 (k=0, multiply)

• Current stack: [5]
• Operation: Pop 5, compute 5 * 4 = 20, push back
• New stack: [20]
• Update: k = 1

Step 2: Process number 3 (k=1, divide)

• Current stack: [20]
• Operation: Pop 20, compute 20 / 3 = 6 (floor division), push back
• New stack: [6]
• Update: k = 2

Step 3: Process number 2 (k=2, add)

• Current stack: [6]
• Operation: Push 2 directly (we'll add it later)
• New stack: [6, 2]
• Update: k = 3

Step 4: Process number 1 (k=3, subtract)

• Current stack: [6, 2]
• Operation: Push -1 (negative, so we can sum everything at the end)
• New stack: [6, 2, -1]
• Update: k = 0

Final Calculation:

• Stack contains: [6, 2, -1]
• Result: sum([6, 2, -1]) = 7

Verification: Let's verify by evaluating the original expression with proper precedence:

• 5 * 4 / 3 + 2 - 1
• First: 5 * 4 = 20, then 20 / 3 = 6
• Then: 6 + 2 - 1 = 7 ✓

The stack approach correctly handles operator precedence by immediately evaluating multiplication and division (popping, computing, pushing back), while deferring addition and subtraction (just pushing values or their negatives) until the final sum.

Solution Implementation

Time and Space Complexity

The time complexity is O(n) because the code iterates through a range from n-1 down to 1, which results in exactly n-1 iterations. In each iteration, the operations performed (arithmetic operations, stack append/pop) are all O(1) operations. Therefore, the overall time complexity is O(n-1) which simplifies to O(n).

The space complexity is O(n) because the stack stk can potentially store up to n elements in the worst case. Initially, the stack contains one element n, and then for each iteration of the loop, we either modify the last element (for multiplication and division operations when k=0 or k=1) or add a new element (for addition and subtraction operations when k=2 or k=3). In the pattern of operations (multiply, divide, add, subtract), half of the operations add new elements to the stack, so the stack can grow to approximately n/2 + 1 elements, which is O(n) space complexity.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Division Operation

The Pitfall: Using Python's floor division operator // instead of truncation toward zero. This causes incorrect results for negative numbers.

# WRONG approach
stack.append(stack.pop() // current_num)

Why it's wrong:

• Python's // performs floor division, which rounds toward negative infinity
• For negative results: -7 // 2 = -4 (floors to -4)
• The problem requires truncation toward zero: -7 / 2 = -3.5 → -3

The Solution:

# CORRECT approach
stack.append(int(stack.pop() / current_num))

This uses regular division followed by int() conversion, which truncates toward zero correctly.

2. Not Handling Operator Precedence Properly

The Pitfall: Trying to evaluate the expression left-to-right without respecting that multiplication and division have higher precedence than addition and subtraction.

# WRONG approach - evaluating sequentially
result = n
for i in range(n-1, 0, -1):
    if operation == '*':
        result = result * i
    elif operation == '/':
        result = int(result / i)
    elif operation == '+':
        result = result + i  # Wrong! This evaluates immediately
    elif operation == '-':
        result = result - i  # Wrong! This evaluates immediately

Why it's wrong:

• This would calculate 10 * 9 / 8 + 7 as ((10 * 9) / 8) + 7 = 18
• But we need to continue with multiplication/division first: 10 * 9 / 8 + 7 - 6 * 5 / 4
• The 6 * 5 / 4 should be evaluated before adding to the previous result

The Solution: Use a stack to defer addition and subtraction operations:

• Multiplication and division modify the top of the stack immediately
• Addition pushes a new positive value to the stack
• Subtraction pushes a new negative value to the stack
• Final sum combines all values respecting precedence

3. Edge Case: Small Values of n

The Pitfall: Not considering that when n is very small (1, 2, or 3), the pattern doesn't complete a full cycle.

Examples:

• n = 1: Result is simply 1 (no operations)
• n = 2: Result is 2 * 1 = 2 (only multiplication)
• n = 3: Result is 3 * 2 / 1 = 6 (multiply then divide)

The Solution: The stack-based approach naturally handles these cases without special logic:

• The loop range(n-1, 0, -1) correctly handles when there are fewer operations
• The operation index cycling ensures the right operations are applied in order
• No special edge case handling needed!