# Solving Leetcode Interviews in Seconds with AI: Minimum Swaps to Group All 1's Together II


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2134" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> A swap is defined as taking two distinct positions in an array and swapping the values in them. A circular array is defined as an array where we consider the first element and the last element to be adjacent. Given a binary circular array nums, return the minimum number of swaps required to group all 1's present in the array together at any location.   Example 1:  Input: nums = [0,1,0,1,1,0,0] Output: 1 Explanation: Here are a few of the ways to group all the 1's together: [0,0,1,1,1,0,0] using 1 swap. [0,1,1,1,0,0,0] using 1 swap. [1,1,0,0,0,0,1] using 2 swaps (using the circular property of the array). There is no way to group all 1's together with 0 swaps. Thus, the minimum number of swaps required is 1.  Example 2:  Input: nums = [0,1,1,1,0,0,1,1,0] Output: 2 Explanation: Here are a few of the ways to group all the 1's together: [1,1,1,0,0,0,0,1,1] using 2 swaps (using the circular property of the array). [1,1,1,1,1,0,0,0,0] using 2 swaps. There is no way to group all 1's together with 0 or 1 swaps. Thus, the minimum number of swaps required is 2.  Example 3:  Input: nums = [1,1,0,0,1] Output: 0 Explanation: All the 1's are already grouped together due to the circular property of the array. Thus, the minimum number of swaps required is 0.    Constraints:  1 <= nums.length <= 105 nums[i] is either 0 or 1.  

	# Explanation
	Here's the breakdown of the problem and the solution:

*   **High-Level Approach:**
    *   Count the total number of 1s in the array. This determines the window size we need to consider.
    *   Iterate through all possible subarrays (windows) of that size, leveraging the circular property by treating the array as contiguous.
    *   For each window, count the number of 0s. The minimum number of 0s in any window represents the minimum number of swaps needed to group all 1s together.

*   **Complexity:**
    *   Runtime Complexity: O(n), where n is the length of the input array.
    *   Storage Complexity: O(1)

	
	# Code
	```python
	def minSwaps(nums):
    """
    Calculates the minimum number of swaps to group all 1s in a circular binary array.

    Args:
        nums: A list of integers representing the binary circular array.

    Returns:
        The minimum number of swaps required to group all 1s together.
    """

    n = len(nums)
    ones = sum(nums)  # Count the total number of 1s

    if ones == 0 or ones == n:
        return 0  # No swaps needed if all 0s or all 1s

    window_size = ones
    min_zeros = float('inf')

    for i in range(n):
        # Calculate the number of zeros in the current window
        zeros = 0
        for j in range(window_size):
            index = (i + j) % n  # Handle circular array indexing
            if nums[index] == 0:
                zeros += 1

        min_zeros = min(min_zeros, zeros)

    return min_zeros
	```