Clumsy Factorial
Medium
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
Solution
Clarifying Questions
When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:
- What is the range of values for the input integer 'n'? Is there a maximum or minimum value I should be aware of?
- Should I handle the case where n is zero or negative, and if so, what should I return?
- Given that integer division truncates toward zero, can you provide an example of how that would affect the result with negative numbers? For instance, what is -5 / 2?
- Can you confirm the order of operations is strictly multiplication, division, addition, and subtraction, repeating this sequence until we reach 1?
- Are there any specific error conditions I should handle, or are we guaranteed that 'n' will always be a valid positive integer?
Brute Force Solution
Approach
The clumsy factorial is calculated with a specific pattern of multiplication, division, addition, and subtraction. The brute force approach simply performs the calculation exactly as the pattern dictates, step by step, following the rules to the letter. It calculates each operation one at a time in the proper order.
Here's how the algorithm would work step-by-step:
- Start with the first number.
- Then, multiply it by the second number.
- Next, divide the result by the third number.
- After that, add the fourth number to the result.
- Now, repeat these operations, starting again with multiplication. Always use the current result in the next operation.
- Continue until all the numbers have been used in the calculation.
- The final result is the clumsy factorial.
Code Implementation
def clumsy_factorial_brute_force(number): if number <= 0:
return 0 result = number
operation_counter = 0
for i in range(number - 1, 0, -1):
# Determine operation based on counter
if operation_counter % 4 == 0:
result *= i
elif operation_counter % 4 == 1:
# Integer division per prompt
result //= i
elif operation_counter % 4 == 2:
result += i
# Increment after addition because order matters
operation_counter += 1
else:
result -= i
operation_counter += 1
return resultBig(O) Analysis
Optimal Solution
Approach
The clumsy factorial problem can be solved efficiently by simulating the operations directly, rather than calculating the full factorial first. We process the numbers in groups of four, applying multiplication, division, addition, and subtraction in sequence, and then handle any remaining numbers at the end.
Here's how the algorithm would work step-by-step:
- Divide the input number into groups of four.
- Within each group of four, perform the multiplication first, then integer division, then addition, and finally subtraction. Make sure to handle division by zero appropriately.
- Sum the result from each group, remembering to properly account for the sign. Most of the time the group result will be added, however, the very last group result should be subtracted.
- If there are any remaining numbers (less than four) after processing the groups, handle them separately. Start by multiplying the remaining numbers, and then adding or subtracting them from the intermediate result based on their position.
Code Implementation
def clumsy(number) -> int:
if number == 0:
return 0
if number == 1:
return 1
if number == 2:
return 2
if number == 3:
return 6
total = 0
index = number
while index >= 4:
# Process the first four operations
group_result = index * (index - 1) // (index - 2) + (index - 3)
if index == number:
total += group_result
else:
total -= group_result
index -= 4
# Handle any remaining numbers
if index == 3:
total -= index * (index - 1) // (index - 2)
elif index == 2:
total -= index * (index - 1)
elif index == 1:
# Subtract last remaining index
total -= index
return totalBig(O) Analysis
Edge Cases
| Case | How to Handle |
|---|---|
| n = 0 | Return 0, as there are no operations to perform. |
| n = 1 | Return 1, as it is the base case and the only number to consider. |
| n = 2 | Return 2*1 = 2, handling the multiplication operation. |
| n = 3 | Return 3*2/1 = 6, handling multiplication and division operations. |
| n = 4 | Return 4*3/2+1 = 6+1 = 7, showing all four operations. |
| Large n causing integer overflow during multiplication | Consider using a larger data type like long or BigInteger depending on the language to prevent integer overflow. |
| Division by zero if an operation results in zero | This case is impossible since we are only dividing by the numbers from n to 1 in decreasing order. |
| Negative n is not a valid input | The problem specifies a positive integer n, so input validation should reject negative values. |
