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Combinations without using factorials-Python
Signed-off-by: Muhammad Arslan <rslnkrmt2552@gmail.com>
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MIT License
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Copyright (c) 2017 Muhammad Arslan <rslnkrmt2552@gmail.com>
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Permission is hereby granted, free of charge, to any person obtaining a copy
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of this software and associated documentation files (the "Software"), to deal
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in the Software without restriction, including without limitation the rights
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to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
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copies of the Software, and to permit persons to whom the Software is
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furnished to do so, subject to the following conditions:
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The above copyright notice and this permission notice shall be included in all
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copies or substantial portions of the Software.
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THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
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IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
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FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE
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AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
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LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM,
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OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE
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SOFTWARE.
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Created on 2017-09-02
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Author : Muhammad Arslan <rslnkrmt@gmail.com>
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Licence : MIT
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Finding the number of combinations (nCr) without use of factorials
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------------------------------------------------------------------
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Using factorials recursively can greatly reduce performance and use a lot of memory especially when finding nCr of large sets. This aims to correct that to some extent.
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Enjoy and provide me with suggestions if any.
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#!/usr/bin/env python
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import operator as op
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def ncr(n, r):
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r = min(r, n-r)
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if x == 0: return 1
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num = reduce(op.mul, xrange(n, n-1, -1))
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denom = reduce(op.mul, xrange(1, r+1))
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return num // denom

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