5 Best Ways to Find the Length of the Longest Sequence of 1s by Flipping K Bits in Python

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πŸ’‘ Problem Formulation: The challenge is to determine the longest continuous sequence of ‘1’ bits in a binary string after flipping up to ‘k’ bits valued ‘0’ to ‘1’. For instance, given the binary string ‘110100110’ and the ability to flip ‘k = 2’ bits, the optimal flip would result in ‘110111110’, yielding a longest sequence length of ‘7’.

Method 1: Brute Force Approach

As a primary and straightforward technique, the brute force approach considers every possible substring in which ‘k’ bits could be flipped and then calculates the longest sequence of 1s. The time complexity is high, which makes this method less efficient for large strings.

Here’s an example:

def longest_sequence(s, k):
    max_length = 0
    for i in range(len(s)):
        for j in range(i, len(s)):
            flipped = s[:i] + s[i:j].replace('0', '1', k) + s[j:]
            max_length = max(max_length, max(map(len, flipped.split('0'))))
    return max_length

# Example use case
print(longest_sequence('110100110', 2))

Output:

7

This code snippet defines a function longest_sequence(s, k) that takes a binary string s and an integer k, and brute-forces its way through every possible substring to find and return the maximum length of ‘1’s that can be achieved by flipping at most k zeroes.

Method 2: Sliding Window Technique

The sliding window technique is an efficient approach for such problems. It involves moving a window across the string to keep track of the number of zero bits within the window. When the count exceeds ‘k’, the window is adjusted to ensure the count of zeros is within the limit.

Here’s an example:

def longest_sequence(s, k):
    left = 0
    zero_count = 0
    max_length = 0

    for right in range(len(s)):
        if s[right] == '0':
            zero_count += 1
        while zero_count > k:
            if s[left] == '0':
                zero_count -= 1
            left += 1
        max_length = max(max_length, right - left + 1)

    return max_length

# Example use case
print(longest_sequence('110100110', 2))

Output:

7

This code snippet demonstrates the sliding window algorithm longest_sequence(s, k), which iterates over the binary string s keeping track of the zero count within a window. Once this count surpasses k, the left index of the window is moved forward, enabling the calculation of the max sequence length efficiently.

Method 3: Dynamic Programming

Dynamic programming can optimize the flipping problem by remembering the previous results to avoid redundant calculations. This method computes for a given position the longest sequence to the left that includes at most ‘k’ flips.

Here’s an example:

def longest_sequence(s, k):
    flip_count = [0] * (len(s) + 1)
    max_len = 0
    
    for i in range(1, len(s) + 1):
        flip_count[i] = flip_count[i - 1] + (s[i - 1] == '0')
        if flip_count[i] > k:
            flip_count[i] = flip_count[i - 1]

        max_len = max(max_len, i - flip_count[i] + flip_count[i - max_len])

    return max_len

# Example use case
print(longest_sequence('110100110', 2))

Output:

7

The code defines a function longest_sequence(s, k) that applies dynamic programming to keep a running total of necessary flips and calculates the longest stretch efficiently without recalculating for each starting point.

Method 4: Optimize Brute Force with Early Stopping

The brute force method can be marginally improved by incorporating an early stopping condition. If during iteration we find a sequence as long as the current maximum length without needing to flip, we can avoid unnecessary checking until the end.

Here’s an example:

def longest_sequence(s, k):
    max_length = min(s.count('1') + k, len(s))
    for i in range(len(s)):
        flip_remaining = k
        for j in range(i, len(s)):
            if s[j] == '0':
                if flip_remaining == 0:
                    break
                flip_remaining -= 1
            max_length = max(max_length, j - i + 1)

    return max_length

# Example use case
print(longest_sequence('110100110', 2))

Output:

7

The code longest_sequence(s, k) optimizes the brute force approach by using an early stopping technique. When there are no flips remaining, and the current maximum length of sequence is reached, the iteration stops.

Bonus One-Liner Method 5: Utilizing itertools.groupby

Python’s itertools library can be steered to give a concise one-liner approach. By grouping consecutive bits and selectively flipping ‘0’s, the itertools.groupby function combined with list comprehension can be used to find the desired result.

Here’s an example:

from itertools import groupby

def longest_sequence(s, k):
    grouped = [(char, sum(1 for _ in group)) for char, group in groupby(s)]
    ones = [length for char, length in grouped if char == '1']
    zeros = [length for char, length in grouped if char == '0']
    return max([sum(ones[i:i+2])+k if i+1 < len(ones) else 0 for i in range(len(ones))])

# Example use case
print(longest_sequence('110100110', 2))

Output:

7

This code executes longest_sequence(s, k) by leveraging itertools.groupby to group the sequence and a list comprehension to find the maximal sequence that can be created by flipping ‘k’ zeros.

Summary/Discussion

  • Method 1: Brute Force Approach. Simple to understand but inefficient for large datasets due to its O(n^3) complexity.
  • Method 2: Sliding Window Technique. Offers a significant performance improvement, with time complexity O(n), making it suitable for large strings.
  • Method 3: Dynamic Programming. This method is efficient by caching previous results, reducing the complexity compared to brute force.
  • Method 4: Optimized Brute Force with Early Stopping. An incremental improvement over the first method that introduces some efficiencies by early termination of loops.
  • Method 5: One-Liner Using itertools.groupby. A concise and Pythonic way, although it might not be as clear and readable for beginners. Optimality depends on the nature of the input.