Skip to content
Related Articles

Related Articles

Improve Article

Python – All Possible unique K size combinations till N

  • Last Updated : 08 Jun, 2020

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)]

 Attention geek! Strengthen your foundations with the Python Programming Foundation Course and learn the basics.  

To begin with, your interview preparations Enhance your Data Structures concepts with the Python DS Course. And to begin with your Machine Learning Journey, join the Machine Learning – Basic Level Course




My Personal Notes arrow_drop_up
Recommended Articles
Page :