MinimumCostToMerge.cpp

#include<bits/stdc++.h>
using namespace std;

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);
    }
};