Divide an Array Into Subarrays With Minimum Cost II

Divide an Array Into Subarrays With Minimum Cost II

What is this problem about?

This is the "Hard" version of the minimum cost subarray problem. You need to divide an array into $k$ contiguous subarrays such that the sum of the first elements of these subarrays is minimized. The first subarray must start at index 0. Additionally, there is a constraint on the "distance" between the first and last subarray heads: the index of the $k$-th subarray head must be within a certain distance dist from the first subarray head (index 0).

Why is this asked in interviews?

Microsoft and American Express use this to test your mastery of Sliding Window and Priority Queues. It evaluation whether you can maintain a set of "top $K$" elements within a moving window. This is a significant step up in complexity, requiring you to efficiently add and remove elements from a "lowest $k-1$ values" collection as the window slides across the array.

Algorithmic pattern used

This problem uses a Sliding Window combined with a Multi-Set or two Heaps.

1. The first cost is nums[0]. You need to pick $k-1$ more elements from a window of size dist + 1.
2. As you slide the window from left to right, you maintain two heaps: one for the $k-1$ smallest elements currently in the window, and another for the rest.
3. When a new element enters the window, add it to the "small" heap and balance.
4. When an element leaves the window, remove it from whichever heap it was in and balance.
5. The sum of the "small" heap plus nums[0] is a candidate for the minimum cost.

Example explanation

nums = [1, 3, 2, 6, 4, 2], $k=3$, dist=3.

1. Fixed cost = 1. We need 2 more elements.
2. Window of indices [1, 4] (since dist=3, index must be $\le 3$? actually the rule is $idx_{last} - idx_{first} \le dist$).
3. Window [3, 2, 6, 4]: Smallest two are 2 and 3. Sum = $1 + 2 + 3 = 6$.
4. Slide window...

Common mistakes candidates make

• Inefficient Updates: Sorting the window every time ($O(N \times dist \log dist)$), which is too slow.
• Removal from Heap: Forgetting that standard heaps don't support $O(1)$ or $O(\log N)$ removal of an arbitrary element. You must use a TreeMap or a lazy-removal heap.
• Window Size: Miscalculating the bounds of the window relative to dist.

Interview preparation tip

Practice the Sliding Window with Top-K pattern. This requires two balanced structures (like two TreeSets in Java or two heaps with counts) to keep track of the sum of the smallest $K$ elements in $O(\log N)$ time per update.