Open In App
Related Articles

Python | Check if given string can be formed by concatenating string elements of list

Improve
Improve
Improve
Like Article
Like
Save Article
Save
Report issue
Report

Given a string ‘str’ and a list of string elements, write a Python program to check whether given string can be formed by concatenating the string elements of list or not. 

Examples: 

Input : str = 'python'
        lst = ['co', 'de', 'py', 'ks', 'on']
Output : False

Input : str = 'geeks'
        lst = ['for', 'ge', 'abc', 'ks', 'e', 'xyz']
Output : True

Approach #1 : Using itertools.permutations

We can use all the permutations of the given list and then perform join operation on them. If any join result is equal to the given string, return true, otherwise false.  

Python3

# Python3 program to Check if given string can
# be formed by concatenating string elements
# of list
from itertools import permutations
 
def checkList(str, lst):
    for i in range(2, len(lst)+1):
        for perm in permutations(lst, i):
            if ''.join(perm)== str:
                return True
                 
    return False
     
# Driver code
str = 'geeks'
lst = ['for', 'ge', 'abc', 'ks', 'e', 'xyz']
print(checkList(str, lst))

                    

Output: 
True

 

Approach #2 : Python RegEx

Python3

# Python3 program to Check if given string can
# be formed by concatenating string elements
# of list
import re
 
def checkList(str, lst):
 
    r = re.compile("(?:" + "|".join(lst) + ")*$")
    if r.match(str) != None:
        return True
         
    return False
     
# Driver code
str = 'geeks'
lst = ['for', 'ge', 'abc', 'ks', 'e']
print(checkList(str, lst))

                    

Output: 
True

 

Approach#3: Using String.replace() and loop: We iterate over the list element by element and remove that element in string. After all Iteration if string become empty then it can be form by using list of string else it cannot. 

Python3

# Python3 program to Check if given string can
# be formed by concatenating string elements
# of list
 
# Function to check string can be formed using list or not
def checkList(str1, list1):
    # Iterating over list of string
    for i in list1:
        # Replacing the sub-string from string
        str1 = str1.replace(i, "")
        # Checking the emptiness of string
        if str1 == "":
            return True
             
    # Else returning False
    else: return False
     
# Driver code
str = 'geeks'
lst = ['for', 'ge', 'abc', 'e', 'ks', 'xyz']
print(checkList(str, lst))

                    

Output:

True

Approach#4: Using dynamic programming

this approach uses dynamic programming to check if the given string can be formed by concatenating string elements of the list. It initializes a boolean list and checks if any prefix of the string till that index is present in the given list and if the previous index is also True. If it is, it sets the current index to True. It returns the value at the last index of the list.

Algorithm

1. Initialize a boolean list of length n+1 where n is the length of the given string.
2. Set the 0th index to True since an empty string can always be formed.
3. For each index i, starting from 1, check if any prefix of the string till that index is present in the given list and if the previous index is also True.
4. If it is, set the current index to True.
5. If the last index is True, return True else return False.

Python3

def check_concatenation(str, lst):
    n = len(str)
    dp = [False]*(n+1)
    dp[0] = True
    for i in range(1, n+1):
        for j in range(i):
            if dp[j] and str[j:i] in lst:
                dp[i] = True
                break
    return dp[n]
str = 'geeks'
lst = ['for', 'ge', 'abc', 'e', 'ks', 'xyz']
print(check_concatenation(str, lst))

                    

Output
True

Time complexity: O(n^3), where n is the length of the given string.

Auxiliary Space: O(n).

Approach#5: Using the lambda function

In this approach, we will define a lambda function checkList(str, lst) to check if the string ‘str’ can be formed by concatenating the elements of the list ‘lst’.

Algorithm: 
1. For each length i of sub-lists from 2 to length of lst:
    a. Generate all possible permutations of length i using itertools.permutations.
    b. If the concatenated string of any permutation equals the given string ‘str’, return True.
2. Return False if the function has not yet returned True.
3. In the main block, set the string ‘str‘ and the list ‘lst‘.
4. Call the checkList function with ‘str‘ and ‘lst‘ as arguments and print the result.

Python3

from itertools import permutations
 
checkList = lambda s, lst: any(''.join(perm) == s for i in range(2, len(lst) + 1) for perm in permutations(lst, i))
 
# Driver code
s = 'geeks'
lst = ['for', 'ge', 'abc', 'ks', 'e', 'xyz']
print(checkList(s, lst))

                    

Output
True

Time Complexity: O(n^3)

Auxiliary Space: O(n)

METHOD 6:Using recursion

APPROACH:

The program checks if a given string can be formed by concatenating string elements of a list. It does this by recursively checking if each substring of the input string is present in the list. 

ALGORITHM:

1.Define a function is_possible that takes two arguments, string and lst.
2..If string is empty, return True.
3.Loop through lst and check if string starts with any of the words.
4.If string starts with a word, call is_possible with remaining string and lst.
5.If recursive call returns True, return True.
6.If none of the words match, return False.

Python3

def is_possible(string, lst):
    if string == "":
        return True
    for word in lst:
        if string.startswith(word):
            if is_possible(string[len(word):], lst):
                return True
    return False
 
string = 'geeks'
lst = ['for', 'ge', 'abc', 'ks', 'e', 'xyz']
print(is_possible(string, lst)) # True

                    

Output
True

Time Complexity: O(n^m) where n is the length of the string and m is the length of the list.
Auxiliary Space: O(m) where m is the length of the list.



Last Updated : 07 Apr, 2023
Like Article
Save Article
Previous
Next
Share your thoughts in the comments
Similar Reads