Minimum Cost to Merge Stones
<p>There are <code>n</code> piles of <code>stones</code> arranged in a row. The <code>i<sup>th</sup></code> pile has <code>stones[i]</code> stones.</p>

<p>A move consists of merging exactly <code>k</code> <strong>consecutive</strong> piles into one pile, and the cost of this move is equal to the total number of stones in these <code>k</code> piles.</p>

<p>Return <em>the minimum cost to merge all piles of stones into one pile</em>. If it is impossible, return <code>-1</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> stones = [3,2,4,1], k = 2
<strong>Output:</strong> 20
<strong>Explanation:</strong> We start with [3, 2, 4, 1].
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 [10].
The total cost was 20, and this is the minimum possible.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> stones = [3,2,4,1], k = 3
<strong>Output:</strong> -1
<strong>Explanation:</strong> After any merge operation, there are 2 piles left, and we can&#39;t merge anymore.  So the task is impossible.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> stones = [3,5,1,2,6], k = 3
<strong>Output:</strong> 25
<strong>Explanation:</strong> We start with [3, 5, 1, 2, 6].
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 [17].
The total cost was 25, and this is the minimum possible.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == stones.length</code></li>
	<li><code>1 &lt;= n &lt;= 30</code></li>
	<li><code>1 &lt;= stones[i] &lt;= 100</code></li>
	<li><code>2 &lt;= k &lt;= 30</code></li>
</ul>
Array; Dynamic Programming; Prefix Sum
func mergeStones(stones []int, K int) int {
	n := len(stones)
	if (n-1)%(K-1) != 0 {
		return -1
	}
	s := make([]int, n+1)
	for i, x := range stones {
		s[i+1] = s[i] + x
	}
	f := make([][][]int, n+1)
	for i := range f {
		f[i] = make([][]int, n+1)
		for j := range f[i] {
			f[i][j] = make([]int, K+1)
			for k := range f[i][j] {
				f[i][j][k] = 1 << 20
			}
		}
	}
	for i := 1; i <= n; i++ {
		f[i][i][1] = 0
	}
	for l := 2; l <= n; l++ {
		for i := 1; i <= n-l+1; i++ {
			j := i + l - 1
			for k := 2; k <= K; k++ {
				for h := i; h < j; h++ {
					f[i][j][k] = min(f[i][j][k], f[i][h][k-1]+f[h+1][j][1])
				}
			}
			f[i][j][1] = f[i][j][K] + s[j] - s[i-1]
		}
	}
	return f[1][n][1]
}