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***********************Explanation***********************
- nikitakumari01021998 July 21, 2020We have two options from given stack -> i) Do not take any element from ith stack
ii) Take elements from ith stack.
We will use dp here.
Definition of dp -> choose j elements from ith stack.
State of dp -> dp[i][j] = max(dp[i][j], dp[i-1][0,1,2...k]+(sum of top j elements from stack i)
**********************Code*************************
int sum(int stackI[], int p, int n)
{
int j=n-1;
int sumOfTopElement=0;
while(p)
{
sumOfTopElement += stackI[j--];
p--;
}
}
void solution(int stack[][],int K)
{
int dp[stack.size()][K+1];
memset(dp, 0, sizeof dp);
for(int k=1;k<=K;k++)
{
if(stack[0].size()>=k)
dp[i][k]=sum(stack[0],k,stack[0].size());
}
for(int i=1;i<n;i++)
{
for(int k=1;k<=K;k++)
{
dp[i][k]=dp[i-1][k];
for(int j=1;j<=k;j++)
{
if(stack[i].size()>=j)
dp[i][k]=max(dp[i][k],dp[i-1][k-j]+sum(stack[i],j,stack[i].size()));
}
}
}
}