<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>

// TC: O((N^3)/K) here as we increment by k piles in every step
// SC: O(N) for prefix sum + O(N^2) for memo + O(N) auxilary stack space
class Solution {
public:
    vector<int> prefixSum = {0};
    int memo[31][31];
    int rec(vector<int> &stones, int i, int j, int piles){
        // Base Case: Invalid Index or For single element answer is always 0
        if(i >= j) return 0;
        
        if(memo[i][j] != -1) return memo[i][j];
        
        int mini = INT_MAX;
        for(int k = i; k < j; k = k+piles-1){
            // Sub Problems
            int tempAns = rec(stones, i, k, piles) + rec(stones, k+1, j, piles);
            mini = min(tempAns, mini);
        }
        
        // If we can take current segment into our answer: 
        if((j-i)%(piles-1) == 0){
            mini += prefixSum[j+1] - prefixSum[i];
        }
        
        return memo[i][j] = mini;
    }
    int mergeStones(vector<int>& stones, int k){
        int n = stones.size();
        
        if((n-1)%(k-1) != 0) return -1;
        
        for(int i : stones) prefixSum.emplace_back(prefixSum.back() + i);
        
        memset(memo, -1, sizeof(memo));
        return rec(stones, 0, n-1, k);
    }
};  
10000