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* In place of calculating the factorial several times we can run a loop k times to calculate the combination for example: 5 C 3 = 5! / (3! * (5-3)! ) = (5*4*3*2*1)/[(3*2*1)*(2*1)] =(5*4*3)/(3*2*1) so running a loop k times will reduce the time complexity to O(k) * [pre-commit.ci] auto fixes from pre-commit.com hooks for more information, see https://pre-commit.ci * Update maths/combinations.py * [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> Co-authored-by: Tianyi Zheng <tianyizheng02@gmail.com>
61 lines
1.4 KiB
Python
61 lines
1.4 KiB
Python
"""
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https://en.wikipedia.org/wiki/Combination
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"""
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def combinations(n: int, k: int) -> int:
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"""
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Returns the number of different combinations of k length which can
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be made from n values, where n >= k.
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Examples:
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>>> combinations(10,5)
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252
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>>> combinations(6,3)
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20
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>>> combinations(20,5)
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15504
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>>> combinations(52, 5)
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2598960
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>>> combinations(0, 0)
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1
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>>> combinations(-4, -5)
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...
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Traceback (most recent call last):
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ValueError: Please enter positive integers for n and k where n >= k
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"""
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# If either of the conditions are true, the function is being asked
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# to calculate a factorial of a negative number, which is not possible
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if n < k or k < 0:
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raise ValueError("Please enter positive integers for n and k where n >= k")
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res = 1
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for i in range(k):
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res *= n - i
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res //= i + 1
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return res
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if __name__ == "__main__":
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print(
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"The number of five-card hands possible from a standard",
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f"fifty-two card deck is: {combinations(52, 5)}\n",
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)
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print(
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"If a class of 40 students must be arranged into groups of",
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f"4 for group projects, there are {combinations(40, 4)} ways",
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"to arrange them.\n",
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)
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print(
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"If 10 teams are competing in a Formula One race, there",
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f"are {combinations(10, 3)} ways that first, second and",
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"third place can be awarded.",
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)
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