K-diff pairs in an array

sum_pairs.py

from collections import Counter


def find_pairs(nums, k) -> int:
    # if k < 0, the result is 0
    if k < 0:
        return 0

    count = Counter(nums)
    pairs = set([])
    for num in count.keys():
        print(num)
        if k == 0:
            # if k = 0, then we need TWO occurrences of hte same number
            if count[num] > 1:
                pairs.add((num, num))
        else:
            otherNum = num + k
            if otherNum in count:
                pairs.add((num, otherNum) if num <= otherNum else (otherNum, num))

    return pairs


num = [3, 1, 2, 7, 5]
k = 2

print(find_pairs(num, k))