The Minimum Swaps to Group All 1's Together II problem extends its predecessor to a circular binary array. The same goal applies — group all 1s into a contiguous segment — but now the array wraps around, so a valid segment can span the end and beginning of the array. This Minimum Swaps to Group All 1's Together II coding problem tests circular sliding window implementation.
IBM and Josh Technology use this to test whether candidates can adapt a known linear algorithm to a circular constraint. Handling circularity correctly — without just doubling the array, which wastes space — is the key skill. The array and sliding window interview pattern applies with the added circular twist.
Circular sliding window using modular indexing. Count total 1s (ones) to determine window size. Slide the window of size ones across the circular array using index i % n. Maintain a count of 1s (or 0s) inside the current window as it slides. Update by removing the outgoing element and adding the incoming element using modulo. Track the minimum zero-count across all window positions.
Array: [0, 1, 1, 1, 0, 0, 1, 1, 0]. Total 1s = 5. Window size = 5.
Circular array problems always bring the question: should I double the array or use modular arithmetic? Doubling is simpler to implement but less space-efficient; modulo is cleaner and more scalable. Practice both approaches, but prefer modulo in interviews to signal awareness of space complexity. After Minimum Swaps to Group All 1's Together II, practice other circular sliding window problems — they appear frequently in scheduling and buffer-management interview scenarios.
| Title | Difficulty | Topics | LeetCode |
|---|---|---|---|
| Minimum Swaps to Group All 1's Together II | Medium | Solve | |
| Grumpy Bookstore Owner | Medium | Solve | |
| Alternating Groups II | Medium | Solve | |
| Count Subarrays Where Max Element Appears at Least K Times | Medium | Solve | |
| Find the Power of K-Size Subarrays I | Medium | Solve |