Checking if two strings are permutations of each other in Python

Question

I\'m checking if two strings a and b are permutations of each other, and I\'m wondering what the ideal way to do this is in Python. From the Zen of

Answer 1

I think the first one is the "obvious" way. It is shorter, clearer, and likely to be faster in many cases because Python's built-in sort is highly optimized.

Answer 2

In Python 3.1/2.7 you can just use collections.Counter(a) == collections.Counter(b).

But sorted(a) == sorted(b) is still the most obvious IMHO. You are talking about permutations - changing order - so sorting is the obvious operation to erase that difference.

Answer 3

"but the first one is slower when (for example) the first char of a is nowhere in b".

This kind of degenerate-case performance analysis is not a good idea. It's a rat-hole of lost time thinking up all kinds of obscure special cases.

Only do the O-style "overall" analysis.

Overall, the sorts are O( n log( n ) ).

The a.count(char) for char in a solution is O( n ² ). Each count pass is a full examination of the string.

If some obscure special case happens to be faster -- or slower, that's possibly interesting. But it only matters when you know the frequency of your obscure special cases. When analyzing sort algorithms, it's important to note that a fair number of sorts involve data that's already in the proper order (either by luck or by a clever design), so sort performance on pre-sorted data matters.

In your obscure special case ("the first char of a is nowhere in b") is this frequent enough to matter? If it's just a special case you thought of, set it aside. If it's a fact about your data, then consider it.

Answer 4

from collections import defaultdict
def permutation(s1,s2):
    h = defaultdict(int)
    for ch in s1:
        h[ch]+=1
    for ch in s2:
        h[ch]-=1
    for key in h.keys():
        if h[key]!=0 or len(s1)!= len(s2):
            return False
        return True
print(permutation("tictac","tactic"))

Answer 5

Here is a way which is O(n), asymptotically better than the two ways you suggest.

import collections

def same_permutation(a, b):
    d = collections.defaultdict(int)
    for x in a:
        d[x] += 1
    for x in b:
        d[x] -= 1
    return not any(d.itervalues())

## same_permutation([1,2,3],[2,3,1])
#. True

## same_permutation([1,2,3],[2,3,1,1])
#. False

Answer 6

In Swift (or another languages implementation), you could look at the encoded values ( in this case Unicode) and see if they match.

Something like:

let string1EncodedValues = "Hello".unicodeScalars.map() {
//each encoded value
$0
//Now add the values
}.reduce(0){ total, value in
    total + value.value
}

let string2EncodedValues = "oellH".unicodeScalars.map() {
    $0
    }.reduce(0) { total, value in
    total + value.value
}

let equalStrings = string1EncodedValues == string2EncodedValues ? true : false

You will need to handle spaces and cases as needed.