💡 Problem Formulation: The objective is to reorder the characters in a string such that all spaces are moved to the beginning (front) of the string in a single pass. For instance, if the input is "move spaces to front", the desired output would be " movespacestofront". This means the non-space characters should remain in the same order as they were, with spaces gathered at the front of the string.
Method 1: Count and Rebuild Approach
This method involves traversing the string to count the number of spaces and then building a new string with the counted spaces at the beginning followed by the non-space characters. The function, move_spaces_to_front, performs these steps efficiently.
Here’s an example:
def move_spaces_to_front(s):
space_count = s.count(' ')
new_string = ' ' * space_count + s.replace(' ', '')
return new_string
print(move_spaces_to_front("move spaces to front"))Output:
movespacestofront
This code snippet uses the built-in count method to determine the number of spaces and then reconstructs the string with the appropriate number of spaces followed by all non-space characters.
Method 2: Two-Pointer Technique
The two-pointer technique uses two indices to traverse the string from end to start. It places non-space characters from the end of a new string, preserving their order, while counting spaces to add them later at the front.
Here’s an example:
def move_spaces_to_front(s):
result = [''] * len(s)
j = len(s) - 1
for char in reversed(s):
if char != ' ':
result[j] = char
j -= 1
result[:j+1] = ' ' * (len(s) - j - 1)
return ''.join(result)
print(move_spaces_to_front("move spaces to front"))Output:
movespacestofront
This approach efficiently organizes the string in reverse, placing characters in their final position from the end while keeping track of spaces to place them subsequently at the beginning.
Method 3: In-place Swap with Python List
Converting the string to a list allows for in-place swapping. The method starts from the end of the list moving each non-space character to the correct position and shifting spaces to the front.
Here’s an example:
def move_spaces_to_front(s):
str_list = list(s)
non_space_index = len(s) - 1
for i in reversed(range(len(s))):
if str_list[i] != ' ':
str_list[i], str_list[non_space_index] = str_list[non_space_index], str_list[i]
non_space_index -= 1
return ''.join(str_list)
print(move_spaces_to_front("move spaces to front"))Output:
movespacestofront
By modifying the list in-place, we can save on space complexity. The non-space characters “bubble up” to the correct position while spaces get pushed to the beginning.
Method 4: Using Regular Expressions
Regular expression can be leveraged to identify all spaces in the string and then reconstruct the string with spaces placed at the beginning followed by the other characters.
Here’s an example:
import re
def move_spaces_to_front(s):
space_count = len(re.findall(r' ', s))
non_space_chars = re.sub(r' ', '', s)
return ' ' * space_count + non_space_chars
print(move_spaces_to_front("move spaces to front"))Output:
movespacestofront
By using regular expressions, this method performs a concise search-and-replace to separate spaces from non-space characters, followed by string reconstruction.
Bonus One-Liner Method 5: List Comprehension and Join
A Python one-liner utilizing list comprehension and join can compactly perform the operation by separating spaces and characters and then concatenating the results.
Here’s an example:
move_spaces_to_front = lambda s: ' ' * s.count(' ') + ''.join(c for c in s if c != ' ')
print(move_spaces_to_front("move spaces to front"))Output:
movespacestofront
This one-liner makes effective use of list comprehension and lambda to count spaces and filter out the non-space characters, and then joins them in the desired order.
Summary/Discussion
- Method 1: Count and Rebuild Approach. Simple and easy to understand. Utilizes efficient string operations but involves the creation of a new string instead of modifying in place.
- Method 2: Two-Pointer Technique. More sophisticated and can be harder to grasp. Does not involve creating a new data structure and is very efficient in terms of both time and space complexity.
- Method 3: In-place Swap with Python List. Efficient with space as it modifies in place. Can be slightly slower than methods that do not involve swapping elements.
- Method 4: Using Regular Expressions. Quick to write and may be faster for short to moderately long strings. However, regex can be less efficient for very long strings and is relatively harder to read and maintain.
- Method 5: List Comprehension and Join. Very concise and Pythonic, but as a one-liner, it may be harder to read and debug for some developers.