Problem Formulation
Given an integer number n.
Get all divisors c of the number n so that c * i = n for another integer i. The desired output format is a list of integers (divisors).
Here are a couple of examples:
n = 10 # Output: [1, 2, 5, 10] n = 13 # Output: [1, 13] n = 24 # Output: [1, 2, 3, 4, 6, 8, 12]
Method 1: Naive Approach
Integer i is a divisor of n if n modulo i is zero.
We use this observation in the function divisors(). We create an initially empty list result and check for every integer number i between 0 and n/2 whether this number is a divisor of n. If it is, we append it to the list.
The following Python code accomplishes this:
def divisors(n):
result = []
for i in range(1, n//2 + 1):
if n % i == 0:
result.append(i)
result.append(n)
return result
print(divisors(24))
# [1, 2, 3, 4, 6, 8, 12, 24]This approach is not very efficient because we traverse every single number from 0 to n/2. If the number n becomes large such as n=1000000, we need to check every number i=0, i=1, i=2, i=3, ..., i=500000.
Runtime complexity: The runtime complexity of calculating the divisors of number n is O(n) using this approach assuming the modulo operation can be performed in one step.
Can we do better? Yes!
Method 2: Reducing the Number of Loop Iterations
We use two observations to reduce the number of loop iterations of the “naive algorithm”.
Observation 1: If number i is a divisor of n, number j = n/i must be an integer and a divisor of n as well because i * n/i = n. This means that each time we find a divisor i, we can also add the divisor n/i to the list of divisors.
Observation 2: For a pair of n-divisors (i, j), one of them must be smaller than or equal to the square root of n. The reason is simple: if both were larger than the square root, the multiplication i * j would be larger than n for sure because root(n) * root(n) == n. Thus, we can traverse the potential divisors from i=0 to i=root(n) and be sure to have found all divisors. This saves us all iterations from i=root(n) to i=n//2.
Here’s the simple tweak with significant performance benefits:
def divisors(n):
result = set()
for i in range(1, int(n**0.5)+1):
if n % i == 0:
result.add(i)
result.add(n//i)
return list(result)
print(divisors(24))
# [1, 2, 3, 4, 6, 8, 12, 24]
This code iterates only from 0 to the square root of the number n. If we find a divisor i, we also add n//i which is the other factor and a divisor of n as well.
Runtime complexity: The runtime complexity of calculating the divisors of number n is O(n^0.5) using this approach assuming the modulo operation is counted as one step.
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Where to Go From Here?
Enough theory. Let’s get some practice!
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