Python – Filter Strings combination of K substrings

Given a Strings list, extract all the strings that are a combination of K substrings.

Input : test_list = [“geeks4u”, “allbest”, “abcdef”], substr_list = [“s4u”, “est”, “al”, “ge”, “ek”, “def”], K = 3 

Output : [‘geeks4u’] 

Explanation : geeks4u made up of 3 substr – ge, ek and s4u. 

Input : test_list = [“geeks4u”, “allbest”, “abcdef”], substr_list = [“s4u”, “est”, “al”, “ge”, “def”, ‘lb’], K = 3 

Output : [‘allbest’] 

Explanation : allbest made up of 3 substr – al, lb and est.

Method #1 : Using permutations() + map() + join() + set() + loop

In this, we perform this task by getting all possible permutation of K substrings from substring list, and then perform task of join using join and map(). The set() is used to avoid duplication. At last, match from Strings list is done using loop.

Python3

# Python3 code to demonstrate working of
# Filter Strings  combination of K substrings
# Using permutations() + map() + join() + set() + loop
from itertools import permutations

# initializing list
test_list = ["geeks4u", "allbest", "abcdef"]

# printing string
print("The original list : " + str(test_list))

# initializing substring list
substr_list = ["s4u", "est", "al", "ge", "ek", "def", "lb"]

# initializing K
K = 3

# getting all permutations
perms = list(set(map(''.join, permutations(substr_list, r = K))))

# using loop to check permutations with list
res = []
for ele in perms:
    if ele in test_list:
        res.append(ele)

# printing results
print("Strings after joins : " + str(res))

Output

The original list : ['geeks4u', 'allbest', 'abcdef']
Strings after joins : ['geeks4u', 'allbest']

Method #2 : Using intersection()

This uses all functions of the above method, the last task of matching permutation list and original list is done by intersection.

Python3

# Python3 code to demonstrate working of
# Filter Strings  combination of K substrings
# Using permutations() + map() + join() + set() + intersection()
from itertools import permutations

# initializing list
test_list = ["geeks4u", "allbest", "abcdef"]

# printing string
print("The original list : " + str(test_list))

# initializing substring list
substr_list = ["s4u", "est", "al", "ge", "ek", "def", "lb"]

# initializing K
K = 3

# getting all permutations
perms = set(map(''.join, permutations(substr_list, r = K)))

# using intersection() to solve this problem
res = list(set(test_list).intersection(perms))

# printing results
print("Strings after joins : " + str(res))

Output

The original list : ['geeks4u', 'allbest', 'abcdef']
Strings after joins : ['geeks4u', 'allbest']

The Time and Space Complexity for all the methods are the same:

Time Complexity: O(n)

Auxiliary Space: O(n)