<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>

class Solution {
public:
    int mergeStones(vector<int>& stones, int k) {
        const int kInf = 1 << 30;
        if (stones.size() == 1) {
            return 0;
        }

        if ((stones.size() - 1) % (k - 1)) {
            return -1;
        }

        vector<vector<int>> dp(stones.size(), vector<int>(stones.size(), kInf));
        vector<int> prefix_sum(stones.size() + 1);

        for (int i = 0; i < stones.size(); i++) {
            prefix_sum[i + 1] = prefix_sum[i] + stones[i];
        }

        for (int i = 0; i < stones.size() - k + 1; i++) {
            dp[i][i + k - 1] = prefix_sum[i + k] - prefix_sum[i];
        }

        auto dfs = [&](int start, int end, int cut, auto Dfs) -> int {
            if (start > end) {
                return kInf;
            }

            if (cut == 0) {
                return start == end ? 0 : dp[start][end];
            }

            int ans = kInf;
            for (int i = start; i < end; i += k - 1) {
                int cost = (i == start ? 0 : dp[start][i]);
                ans = min(ans, cost + Dfs(i + 1, end, cut - 1, Dfs));
            }
            return ans;
        };

        for (int len = 2 * k - 1; len <= stones.size(); len += k - 1) {
            for (int i = 0; i < stones.size() - len + 1; i++) {
                int seg_sum = prefix_sum[i + len] - prefix_sum[i];
                dp[i][i + len - 1] = dfs(i, i + len - 1, k - 1, dfs) + seg_sum;
            }
        }
        return dp[0][stones.size() - 1];
    }
};
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