Python – K length Combinations from given characters
Last Updated :
16 Aug, 2022
Given a string, generate K-length combinations in a one-liner.
Input : test_str = ‘gfg’, K = 3
Output : [‘ggg’, ‘ggf’, ‘ggg’, ‘gfg’, ‘gff’, ‘gfg’, ‘ggg’, ‘ggf’, ‘ggg’, ‘fgg’, ‘fgf’, ‘fgg’, ‘ffg’, ‘fff’, ‘ffg’, ‘fgg’, ‘fgf’, ‘fgg’, ‘ggg’, ‘ggf’, ‘ggg’, ‘gfg’, ‘gff’, ‘gfg’, ‘ggg’, ‘ggf’, ‘ggg’]
Explanation : All combinations of K length extracted.
Input : test_str = ‘G4G’, K = 2
Output : [‘GG’, ‘G4’, ‘GG’, ‘4G’, ’44’, ‘4G’, ‘GG’, ‘G4’, ‘GG’]
Explanation : All combinations of K length extracted.
Method #1 : Using itertools.product() + join() + map()
The task can be performed by product() using repeat param, but returns the result in a tuples list of individual characters. These all can be joined using join() and map().
Python3
from itertools import product
test_str = 'gf4g'
print ( "The original string is : " + str (test_str))
K = 2
res = list ( map (''.join, product(test_str, repeat = K)))
print ( "The generated Combinations : " + str (res))
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Output
The original string is : gf4g
The generated Combinations : ['gg', 'gf', 'g4', 'gg', 'fg', 'ff', 'f4', 'fg', '4g', '4f', '44', '4g', 'gg', 'gf', 'g4', 'gg']
Time Complexity: O(n)
Auxiliary Space: O(1)
Method #2 : Using itertools.product() + list comprehension
In this, list comprehension is used for the task of joining elements.
Python3
from itertools import product
test_str = 'gfg'
print ( "The original string is : " + str (test_str))
K = 2
res = [''.join(ele) for ele in product(test_str, repeat = K)]
print ( "The generated Combinations : " + str (res))
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Output
The original string is : gfg
The generated Combinations : ['gg', 'gf', 'gg', 'fg', 'ff', 'fg', 'gg', 'gf', 'gg']
Time Complexity: O(n)
Auxiliary Space: O(1)
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