<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 dp[50][50][50];
    
      int minCost(int i,int j,int piles,vector<int>&prefixsum,int &K){
            
          // Cost of converting segment [i..i] into 1 pile is zero
          if( i == j && piles == 1)
              return 0;
            
          // Cost of converting segment[i..i] into other than 1 pile is not possible , so placed MAX value
          if(i == j)
              return INT_MAX/4;
            
          // Check whether the subproblem has already been solved . 
          if(dp[i][j][piles]!=-1)
              return dp[i][j][piles];
            
          // If segment[i...j] is to converted into 1 pile 
          if(piles == 1){
              // Here dp(i,j,1) = dp(i,j,K) + sum[i...j]
                
              return dp[i][j][piles] = minCost(i,j,K,prefixsum,K) + (i==0 ? prefixsum[j] : prefixsum[j]-prefixsum[i-1]);
          }else {
                
              // dp(i,j,piles) = min( dp(i,j,piles), dp(i,t,1) + dp(t+1,j,piles-1)) for all t E i<=t<j
              int cost = INT_MAX/4;
              for(int t=i;t<j;t++){
                  cost = min(cost, minCost(i,t,1,prefixsum,K) + minCost(t+1,j,piles-1,prefixsum,K));                
              }
              return dp[i][j][piles] = cost;
          }
      }
    
      int mergeStones(vector<int>& stones, int k) {
            
            
          // the size of the stones 
          int n = stones.size();
            
          /* 
          Check whether we will be able to merge n stones into 1 stone . 
                
              In step-1 we merge k stones and then we are left with n-k+1 stones or n-(k-1);
              In Step-2 we again merge k stones and then we are left with ((n-k+1)-k)+1 or n-2*(k-1);
              In Step-3 we gain merge k stones and left with (((n-k+1)-k+1)-k)+1 or n-3*(k-1)
              .......
              .......
              .......
              After some m steps we should be left with 1 stones 
              Therefore , n-m*(k-1) == 1
                  (n-1)-m*(k-1)=0;
                  Since m needs to be an integer therefore , 
                  if (n-1)%(k-1) == 0 , 
                  then we can merge otherwise not possible.
          */
            
          if((n-1)%(k-1)!=0)
              return -1;
          int sum = 0;
          vector<int>prefixsum;
            
          // Calculating the prefix sum so that sum of any segment[i..j] can be calculated easily
          for(int i=0;i<stones.size();i++){
              sum+=stones[i];
              prefixsum.push_back(sum);
          }
            
          memset(dp,-1,sizeof(dp));
            
          // find the minimum cost to merge stones[0...n-1] into 1 piles .
          return minCost(0,n-1,1,prefixsum,k);
      }
  };

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