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Saturday, August 13, 2022

[FIXED] How to efficiently compare two unordered lists (not sets)?

 August 13, 2022     algorithm, comparison, list, python     No comments   

Issue

a = [1, 2, 3, 1, 2, 3]
b = [3, 2, 1, 3, 2, 1]

a & b should be considered equal, because they have exactly the same elements, only in different order.

The thing is, my actual lists will consist of objects (my class instances), not integers.


Solution

O(n): The Counter() method is best (if your objects are hashable):

def compare(s, t):
    return Counter(s) == Counter(t)

O(n log n): The sorted() method is next best (if your objects are orderable):

def compare(s, t):
    return sorted(s) == sorted(t)

O(n * n): If the objects are neither hashable, nor orderable, you can use equality:

def compare(s, t):
    t = list(t)   # make a mutable copy
    try:
        for elem in s:
            t.remove(elem)
    except ValueError:
        return False
    return not t


Answered By - Raymond Hettinger
Answer Checked By - Mary Flores (PHPFixing Volunteer)
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