Python – All Possible unique K size combinations till N

Sometimes, while working with Python domain, we can have a problem in which we need to produce various combination of elements. This can be K sized unique combinations till N. This problem can have application in data domains and school programming. Let’s discuss certain ways in which this task can be performed.

Input : N = 2, K = 3
Output : [(0, 0, 0), (0, 0, 1), (0, 1, 1), (1, 1, 1)]

Input : N = 4, K = 2
Output : [(0, 0), (0, 1), (0, 2), (0, 3), (1, 1), (1, 2), (1, 3), (2, 2), (2, 3), (3, 3)]

Method #1 : Using product() + setdefault() + loop
The combination of above functions offers an approach to this problem. In this, we use product to perform all combinations and eliminate duplicates using setdefault() and loop by brute force approach.

# Python3 code to demonstrate working of
# All Possible unique K size combinations till N
# Using product() + setdefault() + loop
from itertools import product

# initializing N
N = 4

# Initialize K
K = 3

# All Possible unique K size combinations till N
# Using product() + setdefault() + loop
temp = list(product(range(N), repeat = K))
res = {}
for tup in temp:
    key = tuple(sorted(tup))
    res.setdefault(key, []).append(tup)
res = list(res.keys())

# printing result
print("The unique combinations : " + str(res))

Output :

The unique combinations : [(0, 0, 0), (0, 0, 1), (0, 0, 2), (0, 0, 3), (0, 1, 1), (0, 1, 2), (0, 1, 3), (0, 2, 2), (0, 2, 3), (0, 3, 3), (1, 1, 1), (1, 1, 2), (1, 1, 3), (1, 2, 2), (1, 2, 3), (1, 3, 3), (2, 2, 2), (2, 2, 3), (2, 3, 3), (3, 3, 3)]

Method #2 : Using combinations_with_replacement()
This offers an alternative method to solve this problem. This function performs internally, all that is required to get to solution.

# Python3 code to demonstrate working of
# All Possible unique K size combinations till N
# Using combinations_with_replacement()
from itertools import product, combinations_with_replacement

# initializing N
N = 4

# Initialize K
K = 3

# All Possible unique K size combinations till N
# Using combinations_with_replacement()
res = list(combinations_with_replacement(range(N), K))

# printing result
print("The unique combinations : " + str(res))

Output :

The unique combinations : [(0, 0, 0), (0, 0, 1), (0, 0, 2), (0, 0, 3), (0, 1, 1), (0, 1, 2), (0, 1, 3), (0, 2, 2), (0, 2, 3), (0, 3, 3), (1, 1, 1), (1, 1, 2), (1, 1, 3), (1, 2, 2), (1, 2, 3), (1, 3, 3), (2, 2, 2), (2, 2, 3), (2, 3, 3), (3, 3, 3)]