Python program to concatenate Strings around K

Last Updated : 05 Apr, 2023

Given List of Strings, join all the strings which occurs around string K.

Input : test_list = ["Gfg", "*", "is", "best", "*", "love", "gfg"], K = "*" 
Output : ['Gfg*is', 'best*love', 'gfg'] 
Explanation : All elements around * are joined.

Input : test_list = ["Gfg", "$", "is", "best", "$", "love", "gfg"], K = "$" 
Output : ['Gfg$is', 'best$love', 'gfg'] 
Explanation : All elements around $ are joined.

Method 1: Using a loop

This is a brute way in which this task can be performed. In this, we iterate through all the elements and check for K, if found we perform the required concatenation with preceding and successive elements.

Python3

# Python3 code to demonstrate working of
# Concatenate Strings on K String
# Using loop

# initializing list
test_list = ["Gfg",  "+", "is", "best", "+", "love", "gfg"]

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

# initializing K for Concatenate
K = "+"

res = []
idx = 0

while idx < len(test_list):
  
    ele = test_list[idx]

    # concatenation if next symbol is K
    if (idx < len(test_list) - 1) and test_list[idx + 1] == K:
        ele = ele + K + test_list[idx + 2]

        # increasing counter by 2
        idx += 2
    res.append(ele)
    idx += 1

# printing result
print("Strings after required concatenation : " + str(res))

Output

The original list is : ['Gfg', '+', 'is', 'best', '+', 'love', 'gfg']
Strings after required concatenation : ['Gfg+is', 'best+love', 'gfg']

Time Complexity: O(n)
Auxiliary Space: O(n)

Method 2 : Using join() + replace() + split()

The combination of the above functions can be used to solve this problem. In this, we perform joining of all elements and then remove space around target K. Being treated as a single string, splitting the required string produces joined values around K.

Python3

# initializing list
test_list = ["Gfg",  "+", "is", "best", "+", "love", "gfg"]

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

# initializing K for Concatenate
K = "+"

# performing split after removing space around K
# splits assuming Strings joined around K
res = ' '.join(test_list).replace(' ' + K + ' ', K).split()

# printing result
print("Strings after required concatenation : " + str(res))

Output

The original list is : ['Gfg', '+', 'is', 'best', '+', 'love', 'gfg']
Strings after required concatenation : ['Gfg+is', 'best+love', 'gfg']

Time Complexity: O(n)
Auxiliary Space: O(n)

Method 3: Using list comprehension

Step-by-step approach:

Initialize the list res to an empty list.
Use a list comprehension to iterate over each element ele in the test_list.
If the element ele is equal to K, then continue the iteration.
If the element ele is not equal to K, then check the next element in the test_list.
If the next element is also K, then concatenate the current element and the next element using K as a separator.
Otherwise, append the current element to the res list.
Return the res list.

Below is the implementation of the above approach:

Python3

# Python3 code to demonstrate working of
# Concatenate Strings on K String
# Using list comprehension

# initializing list
test_list = ["Gfg",  "+", "is", "best", "+", "love", "gfg"]

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

# initializing K for Concatenate
K = "+"

# using list comprehension
res = [test_list[i] + K + test_list[i+1] if i+1<len(test_list) and test_list[i+1]==K else test_list[i] for i in range(len(test_list)) if test_list[i]!=K]

# printing result
print("Strings after required concatenation : " + str(res))

Output

The original list is : ['Gfg', '+', 'is', 'best', '+', 'love', 'gfg']
Strings after required concatenation : ['Gfg++', 'is', 'best++', 'love', 'gfg']

Time complexity: O(n), as we are iterating over the given list once using the list comprehension.
Auxiliary space: O(n), as we are creating a new list res of the same size as the given list.

Method 4: Using the reduce() function from the functools module:

Step-by-step approach:

The reduce() function takes a function (lambda function in this case), a sequence, and an initial value (empty list in this case).
The lambda function takes two arguments x and y, which represent the accumulated value and the next element of the sequence, respectively.
If y is equal to K, the lambda function concatenates the last element of x with K and y and replaces the last element of x with the result.
Otherwise, the lambda function simply adds y to x.
Finally, the result is a list of concatenated strings.

Below is the implementation of the above approach:

Python3

# importing functools module
from functools import reduce

# initializing list
test_list = ["Gfg",  "+", "is", "best", "+", "love", "gfg"]

# initializing K for Concatenate
K = "+"

# using reduce() function
res = reduce(lambda x, y: x[:-1] + [x[-1] + K + y] if y == K else x + [y], test_list, [])

# printing result
print("Strings after required concatenation : " + str(res))

Output

Strings after required concatenation : ['Gfg++', 'is', 'best++', 'love', 'gfg']

Time Complexity: O(n), where n is the length of the input list test_list.
Auxiliary Space: O(n), where n is the length of the input list test_list.

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