<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:
    
    // declare a dp
    
    int dp[35][35];
    
    int helper(vector<int>& arr, int start, int end, vector<int>& prefix, int k)
    {
        // if no. of elements is less than k
        
        if(end - start <= k - 2)
        {
            return 0;
        }
        
        // if already calculated
        
        if(dp[start][end] != -1)
        {
            return dp[start][end];
        }
        
        int mini = INT_MAX;
        
        // partition the array at j
        
        for(int j = start; j < end; j += k - 1)
        {
            int left = helper(arr, start, j, prefix, k);
            
            int right = helper(arr, j + 1, end, prefix, k);
            
            mini = min({mini, left + right});
        }
        
        // add the cost of merging two part
        
        if((end - start) % (k - 1) == 0)
        {
            int sum = 0;
            
            sum += prefix[end];
            
            if(start > 0)
            {
                sum -= prefix[start - 1];
            }
                
            mini += sum;
        }
        
        // store the res and return
     
        return dp[start][end] = mini;
    }
    
    int mergeStones(vector<int>& arr, int k) {
        
        int n = arr.size();
        
        // if it is not possible
        
        if((n - 1) % (k - 1))
        {
            return -1;
        }
        
        // find the prefix sum
        
        vector<int> prefix(n, 0);
        
        prefix[0] = arr[0];
        
        for(int i = 1; i < n; i++)
        {
            prefix[i] = prefix[i - 1] + arr[i];
        }
        
        // initialize dp with -1
        
        memset(dp, -1, sizeof(dp));
        
        return helper(arr, 0, n - 1, prefix, k);
    }
};
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