5 Best Ways to Check Whether a Given String Can Be Generated by Concatenating Other Strings in Python

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πŸ’‘ Problem Formulation: Python developers often encounter scenarios where they need to verify if a target string can be constructed by concatenating a collection of other strings. For example, if we’re given a list of strings ["the", "a", "theater"] and we want to know whether we can concatenate these in some order to create the string "thetheater", we should have a method that confirms this is possible. This article explores five methods to perform such a check using Python.

Method 1: Using simple iteration and string replacement

This method involves iterating through each of the given smaller strings and replacing their occurrences in the target string with an empty string. If we successfully reduce the target string to an empty string, then it is possible to generate the target by concatenating given strings. The method is straight-forward and does not require additional libraries.

Here’s an example:

def can_generate(target, strings):
    for s in strings:
        target = target.replace(s, "")
    return target == ""

print(can_generate("thetheater", ["the", "a", "theater"]))

Output:

True

This code defines a function can_generate() that takes a target string and a list of strings. It iterates through the list, replacing instances of each string in the target with an empty string. If the target string is empty after all replacements, the function returns True.

Method 2: Using regular expressions

Regular expressions are a powerful tool for string manipulation. We can use them to check if the target string is composed exclusively of a combination of given strings. This is done by pattern matching using a regular expression built dynamically from the list of strings.

Here’s an example:

import re

def can_generate_regex(target, strings):
    pattern = '^' + '|'.join(sorted(strings, key=lambda s: len(s), reverse=True)) + '+$'
    return bool(re.match(pattern, target))

print(can_generate_regex("thetheater", ["the", "a", "theater"]))

Output:

True

Here, the can_generate_regex() function first creates a regular expression pattern that matches any of the strings provided. The strings are sorted in descending size to prioritize longer matches and encapsulated with ‘^’ and ‘+$’ to ensure the entire target string is matched. A successful match indicates the string can be generated by concatenation.

Method 3: Dynamic programming

Dynamic programming can be an efficient approach to solve this problem by breaking it down into smaller subproblems. At each point, we check if a substring of the target string can be formed using the given strings and build the solution incrementally.

Here’s an example:

def can_generate_dp(target, strings):
    dp = [False] * (len(target) + 1)
    dp[0] = True
    for i in range(1, len(target) + 1):
        for s in strings:
            if dp[i-len(s)] and target.startswith(s, i-len(s)):
                dp[i] = True
                break
    return dp[-1]

print(can_generate_dp("thetheater", ["the", "a", "theater"]))

Output:

True

The can_generate_dp() function uses dynamic programming by creating a boolean array dp to store whether the substring ending at each index can be created. It iterates through the target string and uses information from previous computations to determine if the current substring position can be generated.

Method 4: Recursive backtracking

Recursive backtracking is a brute-force approach where we attempt to build the target string from the given strings, backtracking whenever we reach a dead-end. This method is not the most efficient but can be simple to implement and understand.

Here’s an example:

def can_generate_recursively(target, strings):
    if not target:
        return True
    for s in strings:
        if target.startswith(s):
            if can_generate_recursively(target[len(s):], strings):
                return True
    return False

print(can_generate_recursively("thetheater", ["the", "a", "theater"]))

Output:

True

The can_generate_recursively() function checks whether a string starts with any of the provided strings, and if so, it recursively calls itself with the remainder of the target string. This process repeats until the target string is empty, at which point we have successfully constructed the target string.

Bonus One-Liner Method 5: Using itertools permutations

Python’s itertools library contains a permutations function that we can use to generate all possible concatenations of the given strings and then check if the target string is in this set. Note that this method is highly inefficient for large sets of strings.

Here’s an example:

from itertools import permutations

def can_generate_permutations(target, strings):
    return any(target == ''.join(p) for p in permutations(strings))

print(can_generate_permutations("thetheater", ["the", "a", "theater"]))

Output:

True

The can_generate_permutations() function leverages the permutations method from itertools to check whether any permutation of the provided strings results in the target string. If at least one permutation matches, it returns True.

Summary/Discussion

  • Method 1: Simple Iteration – Easy to implement and understand. Not efficient for all scenarios, especially if the target string contains repeated occurrences of the given strings.
  • Method 2: Regular Expressions – Very powerful and can be efficient, but might be complex for those not familiar with regex. Not suitable for extremely long strings or a large number of given strings due to potential regex engine limitations.
  • Method 3: Dynamic Programming – Typically more efficient than other methods, as it avoids redundant computations. However, it may be harder to implement correctly, and it has higher space complexity compared to simple iteration.
  • Method 4: Recursive Backtracking – Conceptually simple but has potentially exponential time complexity. Suitable for small datasets, where the number of given strings and their lengths are limited.
  • Bonus Method 5: Permutations with itertools – The most straightforward approach in terms of coding, but can be extremely inefficient, as it deals with the full set of permutations, which is factorial in size relative to the input.