Python – Character Replacement Combination

Given a String and dictionary with characters mapped to replacement characters values list,  construct all possible strings after replacing present characters with mapped values.

Input : test_str = “geeks”, test_dict = {‘s’ : [‘1’, ‘5’], ‘k’ : [‘3’]} 

Output : [‘gee31’, ‘geek1’, ‘gee35’, ‘geek5’, ‘gee3s’, ‘geeks’] 

Explanation : All possible replacement of strings, e.g in ‘gee35’, k is replaced by ‘3’ and s is replaced by ‘5’. 

Input : test_str = “geeks”, test_dict = {‘s’ : [‘1’], ‘k’ : [‘3’]} 

Output : [‘gee31’, ‘geek1’, ‘gee3s’, ‘geeks’] 

Explanation : All possible replacement of strings, e.g in ‘gee31’, k is replaced by ‘3’ and s is replaced by ‘1’.

Method : Using zip() + list comprehension + replace() + product()

The combination of the above functions can be used to solve this problem. In this, we extract all the combination characters using product and pair one at a time using zip(), and replacement using replace().

Python3

# Python3 code to demonstrate working of
# Character Replacement Combination
# Using zip() + list comprehension + replace() + product()
from itertools import product

# initializing string
test_str = "geeks"

# printing original string
print("The original string is : " + str(test_str))

# initializing dictionary
test_dict = {'s': ['1', '2'], 'k': ['3']}

# adding original character to possible characters
for key in test_dict.keys():
    if key not in test_dict[key]:
        test_dict[key].append(key)

res = []

# constructing all possible combination of values using product
# mapping using zip()
for sub in [zip(test_dict.keys(), chr) for chr in product(*test_dict.values())]:
    temp = test_str
    for repls in sub:

        # replacing all elements at once using * operator
        temp = temp.replace(*repls)
    res.append(temp)

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

Output

The original string is : geeks
All combinations : ['gee31', 'geek1', 'gee32', 'geek2', 'gee3s', 'geeks']

Time Complexity: O(n³)

Auxiliary Space: O(n)

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