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Reduced Time Complexity to O(sqrt(n)) (#7429)
* Reduced Time Complexity to O(sqrt(n)) * Added testmod * [pre-commit.ci] auto fixes from pre-commit.com hooks for more information, see https://pre-commit.ci Co-authored-by: pre-commit-ci[bot] <66853113+pre-commit-ci[bot]@users.noreply.github.com>
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@ -1,3 +1,7 @@
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from doctest import testmod
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from math import sqrt
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def factors_of_a_number(num: int) -> list:
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"""
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>>> factors_of_a_number(1)
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@ -9,10 +13,22 @@ def factors_of_a_number(num: int) -> list:
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>>> factors_of_a_number(-24)
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[]
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"""
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return [i for i in range(1, num + 1) if num % i == 0]
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facs: list[int] = []
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if num < 1:
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return facs
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facs.append(1)
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if num == 1:
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return facs
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facs.append(num)
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for i in range(2, int(sqrt(num)) + 1):
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if num % i == 0: # If i is a factor of num
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facs.append(i)
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d = num // i # num//i is the other factor of num
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if d != i: # If d and i are distinct
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facs.append(d) # we have found another factor
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facs.sort()
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return facs
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if __name__ == "__main__":
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num = int(input("Enter a number to find its factors: "))
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factors = factors_of_a_number(num)
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print(f"{num} has {len(factors)} factors: {', '.join(str(f) for f in factors)}")
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testmod(name="factors_of_a_number", verbose=True)
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