<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) {
        int n = stones.size();
        
        // If it's impossible to end with exactly 1 pile, return -1
        if ((n - 1) % (k - 1) != 0) return -1;
        
        // 3D DP table to store minimum costs
        vector<vector<vector<int>>> dp(n, vector<vector<int>>(n, vector<int>(k + 1, 1e9)));
        
        // Prefix sum array to quickly calculate the sum of any subarray
        vector<int> prefixSum(n + 1, 0);
        for (int i = 0; i < n; ++i) {
            dp[i][i][1] = 0;  // Cost to merge one pile is 0
            prefixSum[i + 1] = prefixSum[i] + stones[i];
        }
        
        // Fill the DP table
        for (int length = 2; length <= n; ++length) {
            for (int i = 0; i + length <= n; ++i) {
                int j = i + length - 1;
                for (int m = 2; m <= k; ++m) {
                    for (int mid = i; mid < j; mid += k - 1) {
                        dp[i][j][m] = min(dp[i][j][m], dp[i][mid][1] + dp[mid + 1][j][m - 1]);
                    }
                }
                // To merge subarray stones[i...j] into 1 pile
                dp[i][j][1] = dp[i][j][k] + prefixSum[j + 1] - prefixSum[i];
            }
        }
        
        return dp[0][n-1][1];
    }
};

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