Minimum Swaps to Group All Ones Together: Part Two

Overview

The problem involves grouping all ones in an array (containing only zeros and ones) together using minimum swaps.
The challenge includes handling circular aspects of the array.

Given an array of zeros and ones, the goal is to group all ones together.
Example:
- For an array like [0, 1, 0, 1, 1], we need to swap elements to make ones contiguous.
- Initial swap example: swap [0, 1] to get [1, 1, 1, 0, 0] (1 swap).

Initial Thoughts:
- Consider finding the minimum length window that contains all ones and ignoring circular properties initially.
- Calculate: length of window - count of ones = number of swaps needed.
Counter Example:
- A longer string like [1, 0, 1, 0, 1, 1] with holes can lead to overestimation of swaps.
- Actual swaps needed could be less than calculated (2 instead of 4).
New Approach:
- Find a window of size equal to the number of ones (k) that contains the maximum number of ones.
- Example: In [1, 0, 1, 0, 1, 1] with 4 ones, a window of size 4 can help determine how many zeros need to be swapped.
- Resulting swaps = k - max number of ones in the window.

Create a new concatenated array (original + original) to handle circularity.
This way, we can treat the end and beginning of the array as one continuous segment.
Use sliding window technique to find the maximum number of ones in any valid subarray.

Instead of concatenating the array, use two pointers for a sliding window approach.
Right pointer goes until 2*n (where n is the original array length), using modulo to avoid out-of-bounds errors.
Count current window ones and keep track of the maximum found window.*

Initialize pointers and count total ones in the input.
Move the right pointer and increment window ones count if a one is found.
If the window size exceeds total ones, shift the left pointer and adjust the count accordingly.
- window size = right - left + 1
- Adjust the count of ones in the current window.