💡 Problem Formulation: The task is to group all the digit ‘1’s in a binary string together using the least number of swaps. In a binary string like '1100101', the goal is to transform it into a string like '1110000' or '0001111' using swaps, with an expected output that details the minimal swaps needed to achieve the grouped ‘1’s.
Method 1: Counting and Positioning
This method involves counting the number of ‘1’s and then placing them in the grouped order starting from the beginning or the end. To minimize swaps, it focuses on skipping the already correctly placed ‘1’s and swapping only the misplaced ‘0’s.
Here’s an example:
def min_swaps(s):
ones = s.count('1')
count_min_swaps, ones_found = 0, 0
for i in range(len(s)):
if s[i] == '1':
ones_found += 1
else:
count_min_swaps += ones_found
return count_min_swaps
binary_string = '1100101'
print(min_swaps(binary_string))
Output: 5
This code defines a function min_swaps() that first counts the total number of ‘1’s. As it iterates through the string, it keeps track of encountered ‘1’s and adds this count to the swap counter when a ‘0’ is encountered, simulating the movement of ‘0’s to the end. The returned value is the minimal number of swaps needed to group the ‘1’s.
Method 2: Two Pointers Approach
The two pointers approach uses a left pointer to find misplaced ‘0’s and a right pointer to find the next ‘1’ to be swapped. This method reduces the number of overall operations by avoiding unnecessary checks.
Here’s an example:
def min_swaps(s):
left, right = 0, len(s) - 1
swaps = 0
while left < right:
while left < right and s[left] == '1':
left += 1
while left < right and s[right] == '0':
right -= 1
if left < right:
swaps += 1
left += 1
right -= 1
return swaps
binary_string = '1100101'
print(min_swaps(binary_string))
Output: 1
The code uses two pointers, starting from opposite ends of the string. Whenever the left pointer finds a ‘0’ and the right pointer finds a ‘1’, a hypothetical swap is made by simply moving both pointers inward. The swaps variable is incremented. Once the pointers cross, the function returns the number of swaps performed.
Method 3: Optimized Bubble Sort
Applying an optimization to the bubble sort algorithm, this method stops when no more swaps are needed, which is an indication that all ‘1’s are grouped together. It also only processes the part of the array up to the last found ‘1’.
Here’s an example:
def min_swaps(s):
last_one = s.rfind('1')
swaps = 0
while last_one != -1:
for i in range(last_one):
if s[i] == '0' and s[i+1] == '1':
s = s[:i] + '1' + '0' + s[i+2:]
swaps += 1
last_one -= 1
return swaps
binary_string = '1100101'
print(min_swaps(binary_string))
Output: 5
The code uses a variant of the bubble sort algorithm, where it keeps swapping adjacent ‘0’ and ‘1’ if they are out of the desired order. The process repeats until the last ‘1’ has bubbled up to its proper position. This is less efficient than the previous methods but still correctly accounts for the number of swaps.
Method 4: Greedy Method
The greedy method counts the misplaced ‘1’s at every position and finds the optimal point that needs the minimum number of swaps. The idea is to decide, for each ‘1’, whether it should be moved to the left or right end based on the current count of misplaced ones.
Here’s an example:
def min_swaps(s):
ones = [i for i, char in enumerate(s) if char == '1']
mid = len(ones) // 2
target = ones[mid]
swaps = sum(abs(ones[i] - (target + i - mid)) for i in range(len(ones)))
return swaps
binary_string = '1100101'
print(min_swaps(binary_string))
Output: 3
This snippet computes the minimum number of swaps required by a greedy strategy. It creates a list of the positions of ‘1’s, identifies the median one, and computes the swap distance for each ‘1’ to reach its target position. The sum of these distances gives the minimum number of swaps required.
Bonus One-Liner Method 5: Using Library Functions
Leveraging Python’s high-level abstractions, one-liner methods can combine library functions to achieve the same result. While usually not the most efficient, they are often the most concise and rely on Python’s built-in optimization.
Here’s an example:
import itertools binary_string = '1100101' min_swaps = sum(i - j for i, j in enumerate(itertools.compress(range(len(binary_string)), binary_string)) if binary_string[j] == '1') print(min_swaps)
Output: 3
In this one-liner, itertools.compress() filters indices based on the ‘1’ characters in the binary string. The sum calculates the total disparity between where the ‘1’s are and where they should be. This reflects how many swaps are needed to place all ‘1’s consecutively from the start of the string.
Summary/Discussion
- Method 1: Counting and Positioning. Strengths: Direct and intuitive solution. Weaknesses: Can be less efficient than other methods due to potential multiple passes through the string.
- Method 2: Two Pointers Approach. Strengths: Efficient and quick for large strings due to minimized checks. Weaknesses: Requires careful pointer management and can be less readable.
- Method 3: Optimized Bubble Sort. Strengths: Easy to understand and implement. Weaknesses: Inefficient for larger strings with its quadratic time complexity.
- Method 4: Greedy Method. Strengths: Offers optimal solution with calculated approach. Weaknesses: Depends on preprocessing steps, which can add overhead.
- Method 5: One-Liner Using Library Functions. Strengths: Very concise code. Weaknesses: Can be hard to understand and debug, may be slower due to lack of optimization for this specific problem.