Efficiently finding duplicates in a list

Question

I have the function below which searches an array for duplicate entries and then returns a list of the duplicates. I'd like to speed this piece of my code up, can anyone suggest a more efficient way?

Code:

def findDupe(array):
    dupelist = []
    for i in range(len(array)):
        for j in range(len(array)):
            comp1 = array[i]
            comp2 = array[j]
            if comp1 == comp2 and i!=j:
                if comp2 not in dupelist:
                    dupelist.append(comp2)
    return dupelist

user_4909087 · Answer

The idea here is to do a single sweep in linear time. You can use a counter to do this. The idea is to count each element and then return all those which were counted multiple times.

from collections import Counter

def get_duplicates(array):
    c = Counter(array)
    return [k for k in c if c[k] > 1]

The approach above is linear in complexity, but makes two passes over the input - once, to count (this is abstracted away by the Counter constructor) and the other is to return non-unique values in the list comp. You can optimise this a bit using a yield statement, and return duplicates as you find them.

def get_duplicates(array):
    c = Counter()
    seen = set()
    for i in array: 
        c[i] += 1
        if c[i] > 1 and i not in seen:
            seen.add(i)
            yield i

This results in a compulsory if check and extra space in the form of a set, but you reduce two passes to one.

user_8258338 · Answer

What is the type of elements in your lists?

Storing elements in a Set, as explained above, gives you average complexity Θ(n) but requires the elements to be hashable (a Set in Python is implemented with hash table, see https://wiki.python.org/moin/TimeComplexity)
If you have a comparison function, you can sort the list in worst case Θ(nlog(n)) then compare each element of the list to the next.
If you have a comparison function, you can also implement a set with a (balanced) BST and do the same as 1 to achieve complexity Θ(nlog(n)) in average (in worst case).

Efficiently finding duplicates in a list

2 Answers