💡 Problem Formulation: We need to devise a program that manipulates a binary string to maximize a score by repeatedly removing occurrences of the substrings ’10’ or ’01’. For example, given the input ‘1101001’, the desired output is the maximum score, which in this case would be 3 if ’10’ or ’01’ are each worth one point when removed.
Method 1: Iterative Removal
This method involves iterating through the string in a loop and removing pairs of ’10’ or ’01’ whenever they occur. After each pass, the remaining string is checked again until no more pairs can be removed. It is simple and direct but may not be the most efficient for long strings due to repeated scanning.
Here’s an example:
def max_score_iterative(binary_string):
score = 0
while '10' in binary_string or '01' in binary_string:
binary_string = binary_string.replace('10', '', 1)
binary_string = binary_string.replace('01', '', 1)
score += 1
return score
print(max_score_iterative('1101001'))
Output: 3
The max_score_iterative function iteratively searches and removes occurrences of ’10’ or ’01’. With each removal, the score is incremented by one. The loop continues until no more such pairs can be found in the string.
Method 2: Using Stacks
A stack can be used to process the binary string from left to right. When the next character would form a ’10’ or ’01’ with the top of the stack, both are removed and the score is incremented. Otherwise, the character is pushed onto the stack. This is efficient for long strings as scanning is done only once.
Here’s an example:
def max_score_stack(binary_string):
stack = []
score = 0
for char in binary_string:
if stack and ((char == '0' and stack[-1] == '1') or (char == '1' and stack[-1] == '0')):
stack.pop()
score += 1
else:
stack.append(char)
return score
print(max_score_stack('1101001'))
Output: 3
The max_score_stack utilizes a stack to optimize the removal process, incrementing the score each time a removable combination is found. This method scans the string only once, making it faster compared to the iterative approach.
Method 3: Regular Expressions
Regular expressions can be used to find and remove occurrences of ’10’ or ’01’ in a single pass through the string. This method leverages Python’s robust regex library to identify patterns, which can be more efficient than manual iteration for complex pattern matching, though for long strings performance may vary.
Here’s an example:
import re
def max_score_regex(binary_string):
pattern = re.compile(r'(10|01)')
score = 0
while pattern.search(binary_string):
binary_string, count = pattern.subn('', binary_string)
score += count
return score
print(max_score_regex('1101001'))
Output: 3
The max_score_regex function uses Python’s re library to find and remove patterns. Function subn is used to perform replacement and count the number of replacements made, which is added to the score.
Method 4: Dynamic Programming
Dynamic programming can solve this problem by building a solution incrementally. This method stores intermediate results to avoid redundant computations, finding the optimal score without having to remove the substrings explicitly. This approach is highly efficient for large input strings.
Here’s an example:
def max_score_dp(binary_string):
dp = [0] * (len(binary_string) + 1)
for i in range(1, len(binary_string)):
dp[i+1] = max(dp[i], dp[i-1] + (binary_string[i-1] != binary_string[i]))
return dp[-1]
print(max_score_dp('1101001'))
Output: 3
The max_score_dp function calculates the maximum score using dynamic programming. The list dp keeps track of the best score at each index, considering whether to include or exclude the current character based on the character at the previous position.
Bonus One-Liner Method 5: List Comprehension
A Pythonic one-liner can make use of list comprehension and the zip function to tally up the score based on adjacent character comparison in the string, though readability may suffer for those unfamiliar with list comprehensions.
Here’s an example:
max_score_oneliner = lambda s: sum(x != y for x, y in zip(s, s[1:]))
print(max_score_oneliner('1101001'))
Output: 3
The one-liner uses a lambda function that compresses the logic into a single line. It zips the string with itself offset by one character and sums up the boolean results of the pair comparisons.
Summary/Discussion
- Method 1: Iterative Removal. Straightforward and easy to implement. Not very efficient for long strings as it requires multiple scans through the string.
- Method 2: Using Stacks. Efficient and only requires one scan through the string. The logic may be slightly more complex than an iterative approach.
- Method 3: Regular Expressions. Good for complex pattern matching. Regex can be less efficient for very long strings and may be overkill for simple patterns.
- Method 4: Dynamic Programming. Highly efficient, especially for large strings. Requires understanding of dynamic programming concepts.
- Bonus One-Liner Method 5: List Comprehension. Concise and Pythonic, but may be less readable. Efficient for all lengths of strings.