shashwatgokhe/deepcoder-14b 🔢📝 → 📝

The function should take an integer as input and return the factorial as an integer. Make sure to handle the case where the input is 0, which should return 1.

Okay, I need to write a Python function to calculate the factorial of a non-negative integer. Hmm, let's think about what factorial means. Oh right, the factorial of a number n is the product of all positive integers up to n. So for example, 5! is 54321 which equals 120. And 0! is defined as 1, right?

So the function should take an integer as input and return the factorial. I should make sure to handle the case when the input is 0. Oh wait, what if the input is negative? Oh, the problem says it's a non-negative integer, so I don't need to handle negative numbers. That's good.

How do I approach this? Well, I can think of two methods: iterative and recursive. Let's consider both.

For an iterative approach, I can loop from 1 to n, multiplying each number to a running total. That's straightforward. For n=0, I just return 1. Let's outline that:

def factorial(n):
if n == 0:
return 1
result = 1
for i in range(1, n+1):
result *= i
return result

Wait, no. Wait, if n is 0, I return 1. If n is 1, the loop runs once, multiplying 1. That should work. Let me test some cases in my mind. n=0: return 1. n=1: 1. n=2: 2. n=3: 6. Seems right.

Alternatively, a recursive approach. Factorial can be defined recursively as n * factorial(n-1), with the base case being factorial(0) = 1. So:

def factorial(n):
if n == 0:
return 1
else:
return n * factorial(n-1)

But wait, recursion in Python has a limit. For very large n, this might cause a stack overflow. But since the problem says the input is an integer, and I don't know the constraints, but for the purposes of this function, maybe it's acceptable. However, the iterative approach is more efficient and avoids potential recursion depth issues, especially if n is large.

So perhaps the iterative method is better. Also, recursion can be less efficient due to the function call overhead.

Wait, but for small n, recursion is fine. But I'm not sure what the expected input size is. Maybe the problem expects an iterative solution.

So I'll proceed with the iterative approach.

Wait, but let me think about the code structure. Oh right, for n=0, return 1. For n=1, same as 1. So in the iterative function, the loop runs from 1 to n. So if n is 0, the loop doesn't run, and the result is 1. Wait no, the initial code I thought about had an if statement for n=0. Wait no, no, the code I wrote before, for the iterative approach, if n is 0, returns 1. Otherwise, initializes result to 1 and loops from 1 to n, multiplying each i. So for n=0, returns 1. For n=1, the loop runs once, i=1, result becomes 1*1=1. That's correct.

Wait, but wait, maybe I can simplify the code. Like, handle the case where n is 0 in the loop. Let me think. If I don't have the if statement, and just initialize result as 1, then loop from 1 to n. So for n=0, the loop doesn't run, and result remains 1. That could simplify the code.

So the function can be written as:

def factorial(n):
result = 1
for i in range(1, n+1):
result *= i
return result

Wait, but what if n is 0? Then range(1, 0+1) is range(1,1), which is empty. So the loop doesn't run, result is 1. So yes, this handles n=0 correctly. So I don't need the if statement anymore. That's a cleaner code.

So that's better. So the function becomes:

def factorial(n):
result = 1
for i in range(1, n+1):
result *= i
return result

Testing some cases:

n=0: returns 1. Correct.

n=1: loop runs once, i=1, result=1*1=1. Correct.

n=5: 1*1=1, then *2=2, *3=6, *4=24, *5=120. Correct.

What about n=2: 1*1=1, *2=2. Correct.

Great. So this should handle all cases.

Another thing to consider: what if n is a float, like 5.0? The function expects an integer. But the problem says the function should take an integer as input, so perhaps we don't need to handle that. So the function assumes that the input is an integer, and it's non-negative.

So I think this function should work.

Wait, but wait, in Python, if someone passes a float that's effectively an integer, like 5.0, the loop will treat i as integer, but the function will proceed. But the problem says the function should take an integer as input, so perhaps we can assume that the input is always an integer, and non-negative.

So, I think this code should work.

Alternatively, perhaps we can add a check for n being a non-negative integer, but the problem statement says the function should take an integer as input, so maybe that's not necessary.

So the final code is:

def factorial(n):
result = 1
for i in range(1, n + 1):
result *= i
return result

Wait, but wait, what if n is zero? Let me test it in code. If n is 0, the loop runs from 1 to 0+1=1, but range(1,1) is empty. So no iterations, result remains 1. Correct.

Yes, that's correct.

So I think that's the solution.

To solve this problem, we need to write a Python function that calculates the factorial of a non-negative integer. The factorial of a number ( n ) (denoted as ( n! )) is the product of all positive integers up to ( n ). For example, the factorial of 5 (5!) is 120, as it is calculated as ( 5 imes 4 imes 3 imes 2 imes 1 ). The factorial of 0 (0!) is defined as 1.

Approach

The approach to solve this problem involves using an iterative method to compute the factorial. This method is efficient and avoids potential issues with recursion depth limits for large values of ( n ).

1. Initialization: Start with a result variable initialized to 1. This variable will hold the cumulative product as we compute the factorial.
2. Loop through integers: Use a loop to iterate from 1 up to the given integer ( n ). For each iteration, multiply the current value of the result by the loop variable.
3. Return the result: After completing the loop, the result will hold the factorial of the given integer.

This approach ensures that we handle the special case where ( n ) is 0 correctly, as the loop will not execute and the result will remain 1, which is the correct value for 0!.

Solution Code

def factorial(n):
    result = 1
    for i in range(1, n + 1):
        result *= i
    return result

Explanation

• Initialization: The variable result is initialized to 1 to start the multiplication process.
• Loop Execution: The loop runs from 1 to ( n ) (inclusive). In each iteration, result is multiplied by the current loop variable i, effectively building the product of all integers from 1 to ( n ).
• Handling Zero: If ( n ) is 0, the loop does not execute, and result remains 1, which is the correct factorial value for 0.

This method efficiently computes the factorial using a simple iterative approach, ensuring correctness and avoiding potential issues with recursion.