remove duplicated elements with a special rule

Question

The middle element in the sequence of the duplicated elements can be retained only. For example, the list like [1, 2, 1, 3, 1] is processed and its output should be [2, 1, 3] because there are three ‘1’ in the list, and the red one can be retained as it’s the middle one, and the first and third ones are removed. If the number of the duplicated elements is even, the right one over the middle is retained. For example, the output of [ 1, 2, 1, 3, 1, 1] is [2, 3, 1] because there are four ones, and the third (orred) one is the one just over the middle. There are more examples as follows to help you to understand the rule. The red elements in the lists are the elements which should be retained.

[2, 1, 2, 3, 1] -> [2, 3, 1]

[3, 2, 1] -> [3, 2, 1]

[1, 2, 3, 3, 2, 1] -> [3, 2, 1]

[3, 2, 1, 1, 2, 3, 2, 1, 3, 2] -> [1, 3, 2]

Photo of the question

I was trying to implement this but I got the following output. Here is my implementation.

def remove_duplicates(numbers):
    # Step 1: Initialize dictionary to track count and index
    count_dict = {}

    # Step 2: Count occurrences and store index
    for index, num in enumerate(numbers):
        if num in count_dict:
            count_dict[num].append(index)
        else:
            count_dict[num] = [index]

    # Step 3: Initialize list for final unique elements
    unique_elements = []

    # Step 4: Determine unique elements based on the rule
    for num, indices in count_dict.items():
        count = len(indices)
        if count == 1:
            unique_elements.append(num)
        else:
            middle_index = indices[count // 2 + count % 2 - 1]
            unique_elements.append(numbers[middle_index])

    # Step 5: Return the list of unique elements
    return unique_elements

Output:

# expected 
[1, 3, 2]

# got
[3, 2, 1]

Answer 1

Group indexes by value using collections.defaultdict. Pull the middle index of each group. Sort the resulting indexes and use them to retrieve the corresponding values from the original list.

from collections import defaultdict

def remove_duplicates(arr):
    d = defaultdict(list)
    for i, k in enumerate(arr):
        d[k].append(i)
    return [arr[i] for i in sorted(v[len(v) // 2] for v in d.values())]

Answer 2

I'd simply create a Counter of elements. Then iterate over the list, decrease the counter and if the count == 0, add the item to output:

from collections import Counter

test_cases = [
    ([1, 2, 1, 3, 1, 1], [2, 3, 1]),
    ([2, 1, 2, 3, 1], [2, 3, 1]),
    ([3, 2, 1], [3, 2, 1]),
    ([1, 2, 3, 3, 2, 1], [3, 2, 1]),
    ([3, 2, 1, 1, 2, 3, 2, 1, 3, 2], [1, 3, 2]),
]

def fn(l):
    c = Counter(l)
    out = []
    for v in l:
        c[v] -= 1
        if c[v] == 0:
            out.append(v)
    return out


for l, ans in test_cases:
    assert fn(l) == ans