Minimum Swaps to Group Circular Ones

Given a circular binary array nums, return the minimum number of swaps between any two positions needed to group all 1s into one contiguous circular block.

Function
minSwapsCircularOnes(nums: int[]) → int

Examples
Example 1
nums = [0,1,0,1,1,0,0]
return = 1
Example 2
nums = [1,1,0,0,1]
return = 0

Constraints
1 <= nums.length <= 100000.
Every value is 0 or 1.

Count the total ones, call it k. The final block is a window of length k, so the answer is the minimum number of zeros inside any circular window of size k. Each zero in the window needs exactly one swap with a one outside it. Slide the window across the array and track zeros as you add and drop elements. For the circular part, use index i mod n, or concatenate the array with itself and scan windows starting at 0 to n-1. Common pitfalls: forgetting the wraparound, so example 1 gives the wrong answer on edge windows, and not handling k = 0 or k = n, where the answer is 0. Don't reach for a hash table here, it adds nothing. Runtime is O(n) with O(1) extra space, which matters at 100000 elements. If the window logic slips under pressure, StealthCoder can hand you the working solution live.

Drill it cold or hedge it with StealthCoder. Either way, don't walk into the OA hoping you remember the trick.