## Algorithm

Problem Name: 1000. Minimum Cost to Merge Stones

There are `n` piles of `stones` arranged in a row. The `ith` pile has `stones[i]` stones.

A move consists of merging exactly `k` consecutive piles into one pile, and the cost of this move is equal to the total number of stones in these `k` piles.

Return the minimum cost to merge all piles of stones into one pile. If it is impossible, return `-1`.

Example 1:

```Input: stones = [3,2,4,1], k = 2
Output: 20
We merge [3, 2] for a cost of 5, and we are left with [5, 4, 1].
We merge [4, 1] for a cost of 5, and we are left with [5, 5].
We merge [5, 5] for a cost of 10, and we are left with .
The total cost was 20, and this is the minimum possible.
```

Example 2:

```Input: stones = [3,2,4,1], k = 3
Output: -1
Explanation: After any merge operation, there are 2 piles left, and we can't merge anymore.  So the task is impossible.
```

Example 3:

```Input: stones = [3,5,1,2,6], k = 3
Output: 25
We merge [5, 1, 2] for a cost of 8, and we are left with [3, 8, 6].
We merge [3, 8, 6] for a cost of 17, and we are left with .
The total cost was 25, and this is the minimum possible.
```

Constraints:

• `n == stones.length`
• `1 <= n <= 30`
• `1 <= stones[i] <= 100`
• `2 <= k <= 30`

## Code Examples

### #1 Code Example with Javascript Programming

```Code - Javascript Programming```

``````
const mergeStones = function(stones, K) {
const KMO = K - 1
const N = stones.length
if ((N - 1) % KMO !== 0) return -1
const sum = 
const dp = stones.map(s => stones.map(s1 => 0))
stones.forEach(s => {
sum.push(sum[sum.length - 1] + s)
})
for (let e = KMO; e < N; e++) {
for (let b = e - KMO; b >= 0; b--) {
for (let split = e - 1; split >= b; split -= KMO) {
let cost = dp[b][split] + dp[split + 1][e]
dp[b][e] = dp[b][e] === 0 ? cost : Math.min(dp[b][e], cost)
}
if ((e - b) % KMO === 0) {
dp[b][e] += sum[e + 1] - sum[b]
}
}
}
return dp[N - 1]
}
``````
Copy The Code &

Input

cmd
stones = [3,2,4,1], k = 2

Output

cmd
20

### #2 Code Example with Python Programming

```Code - Python Programming```

``````
class Solution:
def mergeStones(self, stones, K):
n = len(stones)
if (n - 1) % (K - 1): return -1
prefix =  * (n + 1)
for i in range(n):
prefix[i + 1] = prefix[i] + stones[i]

import functools
@functools.lru_cache(None)
def dp(i, j):
if j - i + 1 < K: return 0
res = min(dp(i, mid) + dp(mid + 1, j) for mid in range(i, j, K - 1))
if (j - i) % (K - 1) == 0:
res += prefix[j + 1] - prefix[i]
return res
return dp(0, n - 1)
``````
Copy The Code &

Input

cmd
stones = [3,2,4,1], k = 2

Output

cmd
20