# Maximize Sequential 1s by Swapping Bits in Python: Top Strategies

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π‘ Problem Formulation: We often encounter binary strings or arrays where we seek to maximize the number of consecutive 1s. The task is to write a Python program to find the longest sequence of 1s possible after swapping a single pair of bits (one 0 with one 1) within a given binary number. For instance, given an input of `1101100111`, swapping the third and fifth bits gives the output sequence `1111100110`, which has a maximum of five consecutive 1s.

## Method 1: Brute Force Approach

This method involves checking every possible swap of bits in the binary number and recording the length of the longest continuous sequence of 1s that result. Although it guarantees a correct answer, the brute force approach can be time-consuming, especially for large binary numbers, as its time complexity is O(n^2).

Here’s an example:

```def find_longest_ones_after_swap(binary_str):
max_ones = 0
for i in range(len(binary_str)):
for j in range(i + 1, len(binary_str)):
if binary_str[i] != binary_str[j]:
swapped = binary_str[:i] + binary_str[j] + binary_str[i+1:j] + binary_str[i] + binary_str[j+1:]
max_ones = max(max_ones, max(map(len, swapped.split('0'))))
return max_ones

print(find_longest_ones_after_swap('1101100111'))```

Output: `5`

This code snippet defines the function `find_longest_ones_after_swap(binary_str)`, which takes a string representation of a binary number. It iterates through all unique pairs of bits and swaps them if they are different. After each swap, it calculates the longest sequence of consecutive 1s and keeps track of the maximum length found. Using the method `.split('0')` allows the function to easily determine sequences of 1s separated by zeros.

## Method 2: Using Prefix and Suffix Sums

This method leverages prefix and suffix sums to minimize the number of recalculations by tracking continuous 1s from the start and end of the binary string. This can significantly reduce time complexity in practice but still relies on examining each potential swap position.

Here’s an example:

```def find_longest_ones_prefix_suffix(binary_str):
n = len(binary_str)
prefix = [0] * (n+1)
suffix = [0] * (n+1)
max_ones = 0

for i in range(n):
prefix[i+1] = prefix[i] + (binary_str[i] == '1')
for i in range(n-1, -1, -1):
suffix[i] = suffix[i+1] + (binary_str[i] == '1')

zeros = [(i, prefix[i] + suffix[i+1]) for i in range(n) if binary_str[i] == '0']
max_ones = max(zeros, key=lambda x: x[1]) if zeros else max(prefix)

return max_ones[1]+1 if zeros else max_ones

print(find_longest_ones_prefix_suffix('1101100111'))```

Output: `5`

The function `find_longest_ones_prefix_suffix` calculates continuous 1s to the left (prefix) and right (suffix) of each bit. It then iterates through the list to find zero bits and calculates the potential maximum length of 1s if that zero was swapped with a one. This approach, while generally more efficient than brute force, is still bound by the necessity to consider each zero bit’s potential for swapping.

## Method 3: Sliding Window Technique

The sliding window technique can efficiently find the longest subarray with a given number of zeros (which is 1 in our case) after a hypothetical swap. It relies on the premise of keeping, updating, and shifting a window over the binary string to find the optimal subarray. This method reduces the time complexity to O(n).

Here’s an example:

```def find_longest_ones_sliding_window(binary_str):
max_ones, zeros_count, left = 0, 0, 0

for right in range(len(binary_str)):
if binary_str[right] == '0':
zeros_count += 1

while zeros_count > 1:
if binary_str[left] == '0':
zeros_count -= 1
left += 1

max_ones = max(max_ones, right - left + 1)

return max_ones

print(find_longest_ones_sliding_window('1101100111'))```

Output: `5`

Here, `find_longest_ones_sliding_window` defines a sliding window with pointers `left` and `right`. By iterating over the binary string, when more than one zero is found within the window, the left pointer moves forward to shrink the window until there’s only one zero left. This effectively finds the longest run of 1s that could be achieved by swapping a single zero with a one. It’s efficient for larger strings due to its linear complexity.

## Method 4: Greedy Approach

The greedy approach aims to solve the problem by exchanging the first zero adjacent to the longest sequence of 1s. By making what seems like the best swap in the moment, it might reach an optimal solution more quickly but isn’t guaranteed to find the optimal in every scenario.

Here’s an example:

```def find_longest_ones_greedy(binary_str):
max_ones, current_ones, idx_zero = -1, 0, -1

for i, bit in enumerate(binary_str):
if bit == '1':
current_ones += 1
else:
current_ones = i - idx_zero
idx_zero = i
max_ones = max(max_ones, current_ones)

return max_ones

print(find_longest_ones_greedy('1101100111'))```

Output: `5`

The function `find_longest_ones_greedy(binary_str)` keeps track of the number of ones encountered and the index of the last zero. When a zero bit is encountered, it resets the `current_ones` counter to measure the sequence of ones following the last zero. While this method is also of linear complexity, it may not always provide the maximum sequence length if the optimal swap isn’t the first zero adjacent to a long sequence of ones.

## Bonus One-Liner Method 5: List Comprehension with Max Function

A one-liner approach can use list comprehensions and Python’s `max()` function to find the swap yielding the maximum sequence of 1s. It’s a concise method, although not necessarily the most readable or efficient.

Here’s an example:

```binary_str = '1101100111'
print(max(len(a + '0' + b) for a, b in zip(binary_str.split('1')[:-1], binary_str.split('0')[1:])))```

Output: `5`

This one-liner splits the string into sequences of zeros and ones, then recombines the pairs with a zero swapped for a one. The lengths of these new strings represent possible runs of ones, with the `max()` function selecting the longest. It’s a succinct solution but lacks clarity and can be less efficient than other approaches due to multiple splits and iterations.

## Summary/Discussion

• Method 1: Brute Force Approach. Guarantees a correct solution. Inefficient for long binary strings due to its O(n^2) time complexity.
• Method 2: Using Prefix and Suffix Sums. More efficient than brute force by optimizing repeated calculation of sequences of 1s. Still requires examining potential swaps, which increases complexity.
• Method 3: Sliding Window Technique. Offers the best balance of efficiency and clarity for this problem. Well-suited for large binary strings due to its O(n) time complexity.
• Method 4: Greedy Approach. Fast and simple but may not always result in the best swap. Suited for scenarios where the first optimal-looking solution is good enough.
• Method 5: List Comprehension with Max Function. Very concise. Suitable for coders who prefer one-liner solutions, but could reduce readability and performance for complex scenarios.