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Gets all permutations, and choose one of those. Good solution could be to avoid the list and use an iterator Problem 26 seems though :s. Signed-off-by: Julien Lengrand-Lambert <julien@lengrand.fr>
41 lines
1.4 KiB
Python
41 lines
1.4 KiB
Python
#!/usr/bin/env python
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"""
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##---
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# Julien Lengrand-Lambert
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#Created on : Thu Jan 19 10:12:06 CET 2012
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#
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# DESCRIPTION : Solves problem 24 of Project Euler
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A permutation is an ordered arrangement of objects. For example, 3124 is one
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possible permutation of the digits 1, 2, 3 and 4. If all of the permutations are
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listed numerically or alphabetically, we call it lexicographic order.
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The lexicographic permutations of 0, 1 and 2 are:
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012 021 102 120 201 210
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What is the millionth lexicographic permutation of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9?
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##---
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"""
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def lexi_perm(str, num):
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"""
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Returns the numst lexicographic permutations of all values in str
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"""
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# already sorted by all_permutations!
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return int(all_permutations(str)[num - 1])
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def all_permutations(seq):
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"""permutate a sequence and return a list of the permutations"""
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if not seq:
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return [seq] # is an empty sequence
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else:
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temp = []
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for k in range(len(seq)):
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part = seq[:k] + seq[k+1:]
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#print k, part # test
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for m in all_permutations(part):
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temp.append(seq[k:k+1] + m)
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#print m, seq[k:k+1], temp # test
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return temp
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if __name__ == '__main__' :
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print "Answer is : %d" %(lexi_perm("0123456789", 1000000))
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raw_input() # USed to keep Windows terminal open
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