In place of calculating the factorial several times we can run a loop k times to calculate the combination #10051
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… 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)
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… k times to calculate the combination (TheAlgorithms#10051) * 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>
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Describe your change:
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)
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