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Create Minimum Cost to Merge Stones.cpp
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anotherwebguy authored Nov 4, 2022
1 parent 620c0f1 commit f371431
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69 changes: 69 additions & 0 deletions Dynamic Programming/Minimum Cost to Merge Stones.cpp
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//recursion
class Solution {
public:
int recurr(int i,int j,int k,vector<int> &prefix,vector<int> &stones){
if(j-i<=k-2) return 0;
int mini = INT_MAX;
for(int t=i;t<j;t+=(k-1)){
int left = recurr(i,t,k,prefix,stones);
int right = recurr(t+1,j,k,prefix,stones);
mini = min(mini,(left+right));
}
if((j-i)%(k-1)==0){
int sum=0;
sum+=prefix[j];
if(i>0){
sum-=prefix[i-1];
}
mini+=sum;
}
return mini;
}

int mergeStones(vector<int>& stones, int k) {
int n = stones.size();
if((n-1)%(k-1)) return -1;
vector<int> prefix(n,0);
prefix[0] = stones[0];
for(int i=1;i<n;i++){
prefix[i] = prefix[i-1]+stones[i];
}
return recurr(0,n-1,k,prefix,stones);
}
};

//memoization
class Solution {
public:
int recurr(int i,int j,int k,vector<int> &prefix,vector<int> &stones,vector<vector<int>> &dp){
if(j-i<=k-2) return 0;
int mini = INT_MAX;
if(dp[i][j]!=-1) return dp[i][j];
for(int t=i;t<j;t+=(k-1)){
int left = recurr(i,t,k,prefix,stones,dp);
int right = recurr(t+1,j,k,prefix,stones,dp);
mini = min(mini,(left+right));
}
if((j-i)%(k-1)==0){
int sum=0;
sum+=prefix[j];
if(i>0){
sum-=prefix[i-1];
}
mini+=sum;
}
return dp[i][j]=mini;
}

int mergeStones(vector<int>& stones, int k) {
int n = stones.size();
if((n-1)%(k-1)) return -1;
vector<int> prefix(n,0);
prefix[0] = stones[0];
for(int i=1;i<n;i++){
prefix[i] = prefix[i-1]+stones[i];
}
vector<vector<int>> dp(n,vector<int>(n,-1));
return recurr(0,n-1,k,prefix,stones,dp);
}
};

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