Optimizing Python to Find the Longest Substring of 1s in a Binary String With a Single 0 Flip

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πŸ’‘ Problem Formulation: We aim to solve the challenge of identifying the longest continuous substring composed of 1s within a binary string after flipping exactly one 0 to 1. For instance, if our input string is “11001101111”, flipping the second 0 would result in “11011101111”. The length of the longest substring of 1s in this modified string is 5.

Method 1: Brute Force Approach

The brute force approach scans through each 0 in a binary string, flips it, and calculates the length of the resulting longest 1s substring. It is straightforward but potentially inefficient for large strings due to its O(n^2) time complexity. This method exhaustively searches all possibilities to find the optimal flip location.

Here’s an example:

def longest_substring_after_flip(binary_str):
        max_length = 0
        for i in range(len(binary_str)):
            if binary_str[i] == "0":
                flipped_str = binary_str[:i] + "1" + binary_str[i+1:]
                max_length = max(max_length, len(max(flipped_str.split("0"), key=len)))
        return max_length
    print(longest_substring_after_flip("11001101111"))

The output of the code snippet is: 5

This code iterates over each character in the binary string, flips each ‘0’ to ‘1’, and then splits the string by ‘0’ to find the longest continuous ‘1’s. It returns the length of the longest sequence after each flip, keeping track of the maximum length found.

Method 2: Sliding Window Technique

The sliding window technique involves moving a window across the binary string to identify potential substrings while keeping a count of zeroes. Upon encountering a zero, you can flip it and continue extending the window until you exceed the permitted number of flips. It’s more efficient than brute force, with a time complexity of O(n).

Here’s an example:

def longest_substring_sliding_window(binary_str):
        left = count = max_length = 0
        for right in range(len(binary_str)):
            if binary_str[right] == '0':
                count += 1
            while count > 1:
                if binary_str[left] == '0':
                    count -= 1
                left += 1
            max_length = max(max_length, right - left + 1)
        return max_length
    print(longest_substring_sliding_window("11001101111"))

The output of the code snippet is: 5

This snippet uses two pointers to create a window and slides it across the binary string, tracking the longest substring of 1s that can be obtained by flipping a single 0 to 1. It is more efficient than the brute force method and provides an optimized solution for larger inputs.

Method 3: Using Dynamic Programming

Dynamic programming can optimize the process by storing intermediate results and using them to find the longest substring after a 0 flip. It provides a compromise between the brute force and sliding window methods, with better performance in certain cases than brute force.

Here’s an example:

def longest_substring_dynamic(binary_str):
        prev_count = curr_count = max_length = 0
        for char in binary_str:
            if char == '1':
                curr_count += 1
            else:
                prev_count, curr_count = (curr_count + 1 if curr_count + 1 > prev_count else prev_count), 0
            max_length = max(max_length, curr_count + prev_count)
        return max_length
    print(longest_substring_dynamic("11001101111"))

The output of the code snippet is: 5

This code snippet uses dynamic programming to avoid recalculating the length of substrings for each position in the string. By keeping a tally of contiguous 1s before and after each zero, the function can quickly determine the maximum possible substring length when flipping any zero to a one.

Method 4: Using Regular Expressions

Regular expressions can be used to pattern-match substrings of 1s separated by a single 0, then identify and calculate the length of these substrings. This method allows for terse and expressive code but may be less intuitive for those not familiar with regex syntax.

Here’s an example:

import re
    def longest_substring_regex(binary_str):
        return max(len(x) + len(y) for x, y in re.findall(r'(1*)(0)(1*)', binary_str))
    print(longest_substring_regex("11001101111"))

The output of the code snippet is: 5

In this snippet, the regex pattern ‘(1*)(0)(1*)’ identifies all sequences of 1s that are separated by a single 0. The ‘findall’ function lists all such occurrences, and the longest combined length of pairs of 1 sequences around a zero is determined.

Bonus One-Liner Method 5: Using List Comprehension and Built-in Functions

A one-liner can also achieve the same result by exploiting Python’s built-in functions and list comprehension. This method trades readability for brevity and is best reserved for those who prefer concise code.

Here’s an example:

print(max(map(sum, zip([0]+list(map(len, '11001101111'.split('1'))), list(map(len, '11001101111'.split('0')))+[0]))))

The output of the code snippet is: 5

This one-liner splits the binary string by ‘1’ and ‘0’, zips the resulting lengths with an offset, takes the sum of pairs – effectively computing the length of substrings if a 0 is flipped – and returns the maximum sum through the use of map and max functions.

Summary/Discussion

  • Method 1: Brute Force. Straightforward to implement but has poor performance with large strings.
  • Method 2: Sliding Window Technique. Efficient time complexity and suitable for large datasets.
  • Method 3: Dynamic Programming. Better performance than brute force in certain cases by avoiding redundant calculations.
  • Method 4: Regular Expressions. Concise and clever, but not as performant or readable as Method 2 and 3 for those not versed in regex.
  • Method 5: One-Liner. Extremely concise but at the cost of readability and potentially harder to debug or maintain.