<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:
    vector<int> prefixSum;

int f(int i, int j, vector<vector<int>>& dp, vector<int>& stones, int k) {
    if (j - i + 1 < k)
        return 0;
    if (dp[i][j] != -1)
        return dp[i][j];

    int minCost = INT_MAX;

    for (int mid = i; mid < j; mid += (k - 1)) {
        int leftCost = f(i, mid, dp, stones, k);
        int rightCost = f(mid + 1, j, dp, stones, k);

        if (leftCost != INT_MAX && rightCost != INT_MAX) {
            minCost = min(minCost, leftCost + rightCost);
        }
    }

    if ((j - i) % (k - 1) == 0) {
        minCost += (prefixSum[j + 1] - prefixSum[i]);
    }

    return dp[i][j] = minCost;
}

int mergeStones(vector<int>& stones, int k) {
    int n = stones.size();
    prefixSum = vector<int>(1, 0);
    for (auto& x : stones)
        prefixSum.push_back(x + prefixSum.back());

    vector<vector<int>> dp(n + 2, vector<int>(n + 2, -1));

    if ((n - 1) % (k - 1) != 0) return -1;

    return f(0, n - 1, dp, stones, k);
}
};
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