Minimize the Cost of Merging Exactly K Adjacent Piles at a Time
Company: Amazon
Role: Software Engineer
Category: Coding & Algorithms
Difficulty: hard
Interview Round: Other
### Problem
Piles in `stones` remain in row order. Each move merges exactly k consecutive current piles into one and costs their combined stone count. Return the minimum cost of obtaining one pile, or-1 if that is impossible.
```cpp
class Solution {
public:
int mergeStones(vector<int>& stones, int k);
};
```
### Examples
**Example 1:**
```text
Input: stones = [1,2,3], k = 2
Output: 9
Explanation: Merge1 and2 for3, then merge the resulting3 with the last3 for6. The other first merge costs5+6=11.
```
**Example 2:**
```text
Input: stones = [2,2,2,2,2], k = 3
Output: 16
Explanation: Any first triple costs6 and leaves three piles totaling10; their final merge costs10, for16.
```
### Constraints
- `n == stones.length`; `1 <= n <= 30`.
- `1 <= stones[i] <= 100`.
- `2 <= k <= 30`.
### Contract considerations
Only consecutive current piles may merge. The required output is the minimum total cost, not a merge sequence.
```hint Read the boundary condition
An interval reduced to p piles is a useful state because the final merge always requires exactly k piles.
```
### Boundary reminders
Only consecutive current piles may merge. The required output is the minimum total cost, not a merge sequence.
Overview: Minimize the Cost of Merging Exactly K Adjacent Piles at a Time. Piles in stones remain in row order.
