LeetCode 1006 - Clumsy Factorial

Difficulty: 🟡 Medium
Topics: Math, Stack, Simulation

Solution

Problem Understanding

The problem asks us to compute a clumsy factorial for a given positive integer n. A standard factorial multiplies all integers from n down to 1, but the clumsy factorial modifies the operations: instead of multiplying all numbers, it cycles through '*', '/', '+', and '-' in that order, applying the operations to decreasing integers starting from n. Multiplication and division follow standard precedence rules, and division uses floor division (rounding down to the nearest integer).

The input is a single integer n representing the top of the factorial sequence. The output is an integer representing the result of applying the clumsy factorial operations. For example, for n = 10, the calculation is 10 * 9 / 8 + 7 - 6 * 5 / 4 + 3 - 2 * 1, which evaluates to 12.

The constraints indicate that n can be as small as 1 and as large as 10,000. This means a naive solution iterating and computing factorial products without considering integer overflows or operation precedence could be inefficient, though Python handles large integers well. Edge cases include very small n where not all four operations are applied, or where division results must be floored correctly.

Important edge cases are:

n = 1, n = 2, n = 3, n = 4: fewer than four numbers, requiring careful handling of partial operation sequences.
Large n near 10,000: ensuring the algorithm is efficient.
Division leading to non-integer results, which must be floored.

Approaches

The brute-force approach simulates the entire sequence of operations exactly as described. We can iterate from n down to 1, applying the operations in cycles of four. We would need a stack or a list to handle the precedence rules: multiplication and division must be executed before addition and subtraction. This approach is correct but can be slightly cumbersome and verbose, though it is acceptable for the given constraints because n is relatively small.

The optimal approach relies on recognizing patterns in the clumsy factorial sequence. The first four numbers produce a predictable result, and after that, each subsequent group of four numbers follows a repeated pattern. Using a stack, we can handle multiplication and division immediately, then defer addition and subtraction by pushing values to the stack. Summing the stack at the end yields the correct result. This reduces the complexity of managing operation precedence manually, making the code cleaner and more maintainable.

Approach	Time Complexity	Space Complexity	Notes
Brute Force	O(n)	O(n)	Simulate all operations using a stack or list, applying operator precedence explicitly
Optimal	O(n)	O(n)	Use a stack to immediately compute multiplication/division and defer addition/subtraction, then sum the stack

Algorithm Walkthrough

Initialize a stack to store intermediate results. The stack will help manage addition and subtraction after multiplication and division are applied.
Set the current operation to multiplication '*' and start iterating from n down to 1.
For each number i:

If the operation is '*', pop the top of the stack, multiply it by i, and push the result back.
If the operation is '/', pop the top of the stack, perform integer floor division by i, and push the result back.
If the operation is '+', push i onto the stack.
If the operation is '-', push -i onto the stack.

Rotate the operation in the sequence '*', '/', '+', '-' for the next iteration.
After iterating through all numbers, sum all values in the stack to obtain the clumsy factorial.

Why it works: The stack ensures that multiplication and division, which have higher precedence, are applied immediately to the top value. Addition and subtraction are deferred by pushing positive or negative numbers. Summing the stack at the end produces the correct total according to the standard order of operations.

Python Solution

class Solution:
    def clumsy(self, n: int) -> int:
        stack = [n]
        n -= 1
        operations = ['*', '/', '+', '-']
        idx = 0
        
        while n > 0:
            op = operations[idx % 4]
            if op == '*':
                stack[-1] *= n
            elif op == '/':
                stack[-1] = int(stack[-1] / n)
            elif op == '+':
                stack.append(n)
            else:  # op == '-'
                stack.append(-n)
            idx += 1
            n -= 1
        
        return sum(stack)

The Python code initializes a stack with the first number. It iterates from n-1 down to 1, applying operations in cyclic order. Multiplication and division are applied directly to the top of the stack, ensuring precedence. Addition and subtraction push positive or negative numbers to the stack, which are summed at the end to yield the correct result.

Go Solution

func clumsy(n int) int {
    stack := []int{n}
    n--
    operations := []byte{'*', '/', '+', '-'}
    idx := 0

    for n > 0 {
        op := operations[idx%4]
        if op == '*' {
            stack[len(stack)-1] *= n
        } else if op == '/' {
            stack[len(stack)-1] /= n
        } else if op == '+' {
            stack = append(stack, n)
        } else { // '-'
            stack = append(stack, -n)
        }
        idx++
        n--
    }

    result := 0
    for _, val := range stack {
        result += val
    }
    return result
}

In Go, the main differences are slice handling and integer division, which automatically floors towards zero for positive numbers. The logic mirrors Python: multiplication and division are applied immediately, addition and subtraction are deferred by appending to the slice, and the final sum gives the clumsy factorial.

Worked Examples

Example 1: `n = 4`

Iteration through numbers 4, 3, 2, 1:

Step	Stack	Operation	Number
init	[4]	-	-
1	[12]	*	3
2	[6]	/	2
3	[6, 1]	+	1

Sum of stack: 6 + 1 = 7

Example 2: `n = 10`

Iteration from 10 down to 1:

Step	Stack	Operation	Number
init	[10]	-	-
1	[90]	*	9
2	[11]	/	8
3	[11, 7]	+	7
4	[11, 7, -6]	-	6
5	[11, 7, -30]	*	5
6	[11, 7, -7]	/	4
7	[11, 7, -7, 3]	+	3
8	[11, 7, -7, 3, -2]	-	2
9	[11, 7, -7, 3, -2, -1]	*	1

Sum: 11 + 7 - 7 + 3 - 2 - 1 = 11 + 7 - 7 + 3 - 2 - 1 = 11 + 0 + 3 - 3 = 11 + 0 = 12

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	We iterate once over `n` numbers, performing constant-time operations per number.
Space	O(n)	Stack can grow linearly in the worst case, holding all numbers as deferred addition/subtraction.

The algorithm is linear in both time and space. Using a stack ensures multiplication and division precedence without requiring nested loops or complex evaluation.

Test Cases

# Provided examples
assert Solution().clumsy(4) == 7  # 4*3/2+1
assert Solution().clumsy(10) == 12  # 10*9/8+7-6*5/4+3-2*1

# Edge cases
assert Solution().clumsy(1) == 1  # single number
assert Solution().clumsy(2) == 2  # 2*1
assert Solution().clumsy(3) == 6  # 3*2/1

# Larger numbers
assert Solution().clumsy(5) == 7  # 5*4/3+2-1
assert Solution().clumsy(6) == 8  # 6*5/4+3-2*1

# Stress test
assert Solution().clumsy(10000) >= 0  #