💡 Problem Formulation: We are tasked with determining whether a given string can be formed by interleaving the characters of two other distinct strings without changing the relative order of characters in the original strings. For instance, if our input strings are “abc” and “def”, and the string to check is “dabecf”, we want our program to confirm that “dabecf” is a valid shuffle of “abc” and “def”.
Method 1: Iterative Comparison
This method entails creating an iterative function that compares the characters of the strings in sequence and determines if the target string is a valid shuffle. The specification involves walking through each string’s characters while maintaining their order, utilizing index pointers.
Here’s an example:
def is_valid_shuffle(str1, str2, result):
if len(result) != len(str1) + len(str2):
return False
i = j = k = 0
while k != len(result):
if i < len(str1) and str1[i] == result[k]:
i += 1
elif j < len(str2) and str2[j] == result[k]:
j += 1
else:
return False
k += 1
return True
print(is_valid_shuffle("abc", "def", "dabecf"))Output:
True
This code snippet initializes three pointers, each for the two input strings and the result string. It proceeds to iterate through the result string and checks if the next character matches the current character in either of the input strings, moving the respective pointer ahead. If at any point a character doesn’t match, it returns False. If it completes the iteration successfully, it returns True.
Method 2: Using Sorting
This technique involves concatenating the two source strings, sorting them, and then comparing the sorted result with a sorted target string. If they match, the function returns True; otherwise, False. This assumes that the relative order of characters in individual strings is not important.
Here’s an example:
def is_valid_shuffle_by_sort(str1, str2, result):
return ''.join(sorted(str1 + str2)) == ''.join(sorted(result))
print(is_valid_shuffle_by_sort("abc", "def", "dabecf"))Output:
True
The code defines a simple method that sorts both the combined input strings and the result string and compares them. As sorting rearranges characters, this method only works where the input strings contain characters that don’t repeat, or where relative character position is not crucial.
Method 3: Recursion
With recursion, we solve the problem by reducing it to smaller instances of itself. This method recursively checks if each character in the target string could come from one of the input strings, without violating the order.
Here’s an example:
def is_valid_shuffle_recursive(str1, str2, result):
if not str1 and not str2 and not result:
return True
if not result:
return False
if str1 and str1[0] == result[0]:
if is_valid_shuffle_recursive(str1[1:], str2, result[1:]):
return True
if str2 and str2[0] == result[0]:
return is_valid_shuffle_recursive(str1, str2[1:], result[1:])
return False
print(is_valid_shuffle_recursive("abc", "def", "dabecf"))Output:
True
The provided function calls itself with a smaller substring for both input strings and the remaining part of the result string. It tries characters from both strings and moves ahead if it finds a match, ensuring the order is maintained. If no matches are found or not all characters are used, it returns False.
Method 4: Dynamic Programming
Dynamic programming provides an efficient solution by storing solutions to subproblems. The method uses a 2D array to store the answers to whether a substring of the target string is a valid shuffle of substrings of the two input strings.
Here’s an example:
def is_valid_shuffle_dp(str1, str2, result):
if len(str1) + len(str2) != len(result):
return False
dp = [[False] * (len(str2) + 1) for _ in range(len(str1) + 1)]
for i in range(len(str1) + 1):
for j in range(len(str2) + 1):
if i == 0 and j == 0:
dp[i][j] = True
elif i == 0:
dp[i][j] = dp[i][j-1] and str2[j-1] == result[i+j-1]
elif j == 0:
dp[i][j] = dp[i-1][j] and str1[i-1] == result[i+j-1]
else:
dp[i][j] = (dp[i-1][j] and str1[i-1] == result[i+j-1]) or \
(dp[i][j-1] and str2[j-1] == result[i+j-1])
return dp[len(str1)][len(str2)]
print(is_valid_shuffle_dp("abc", "def", "dabecf"))Output:
True
The code initializes a 2D list ‘dp’ to keep track of valid shuffles for different lengths of the input strings. It iterates through the strings, filling in the table with boolean values indicating if it is possible to reach this state. In the end, dp[len(str1)][len(str2)] contains the final answer.
Bonus One-Liner Method 5: Using itertools
In this one-liner approach, Python’s itertools library is leveraged to create all possible shuffles of the input strings and check if the target string is among them. Note that this method is not efficient and only recommended for small strings due to its high computational cost.
Here’s an example:
import itertools
def is_valid_shuffle_itertools(str1, str2, result):
return result in map(''.join, itertools.chain(itertools.permutations(str1 + str2, len(str1 + str2))))
print(is_valid_shuffle_itertools("abc", "def", "dabecf"))Output:
False
This succinct function uses itertools to generate all permutations of the concatenated input strings of the same length as the result string and checks if the result is in this set. The output here is False since permutations do not preserve the internal order of the given strings.
Summary/Discussion
- Method 1: Iterative Comparison. Simple and intuitive. Efficient. Works correctly only if input strings may contain repeating characters and respecting their relative order is important.
- Method 2: Using Sorting. Very concise. Not suitable for strings with repeating characters or where the relative character position matters.
- Method 3: Recursion. Elegant and simple recursive solution. May be slow and inefficient for very long strings due to large recursion depth.
- Method 4: Dynamic Programming. Highly efficient for larger inputs. Complexity may be overkill for simple or small cases. More complex to understand and implement.
- Method 5: Using itertools. One-liner and slick. Impractical for large input strings due to computational expense.