Esmail: > oh, I forgot to mention that each list may contain duplicates.
Comparing the sorted lists is a possible O(n ln n) solution:
a.sort()
b.sort()
a == b
Another solution is to use frequency dicts, O(n):
from itertools import defaultdict
d1 = defaultdict(int)
for el in a:
d1[el] += 1
d2 = defaultdict(int)
for el in b:
d2[el] += 1
d1 == d2
As the arrays (Python lists) become large the second solution may
become faster.
Bye,
bearophile
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