Maximum profit from sale of wines

Given n wines in a row, with integers denoting the cost of each wine respectively. Each year you can sale the first or the last wine in the row. However, the price of wines increases over time. Let the initial profits from the wines be P1, P2, P3…Pn. On the Yth year the profit from the ith wine will be Y*Pi. For each year, your task is to print “beg” or “end” denoting whether first or last wine should be sold. Also, calculate the maximum profit from all the wines.

Examples :

Input : Price of wines : 2 4 6 2 5
Output : beg end end beg beg 
Explanation :

Approach : It is a standard Dynamic Programming problem. It initially looks like a greedy problem in which we should sell the cheaper of the wines each year but the example case (year 2) clearly proves the approach is wrong. Sometimes we need to sell an expensive wine earlier to save relatively costly wines for later years (Here, if 4 was sold in 2nd year in the 4th year we had to sell 2 which would be waste of a heavy coefficient).

The second problem is to “store the strategy” to obtain the calculated price which has a fairly standard method that can be used in other problems as well. The idea is to store the optimal action for each state and use that to navigate through the optimal states starting from the initial state.

[sourcecode language=”CPP”]
// Program to calculate maximum price of wines
#include <bits/stdc++.h>
using namespace std;

#define N 1000

int dp[N][N];

// This array stores the "optimal action"
// for each state i, j
int sell[N][N];

// Function to maximize profit
int maxProfitUtil(int price[], int begin,
int end, int n)
if (dp[begin][end] != -1)
return dp[begin][end];

int year = n – (end – begin);

if (begin == end)
return year * price[begin];

// x = maximum profit on selling the
// wine from the front this year
int x = price[begin] * year +
maxProfitUtil(price, begin + 1, end, n);

// y = maximum profit on selling the
// wine from the end this year
int y = price[end] * year +
maxProfitUtil(price, begin, end – 1, n);

int ans = max(x, y);
dp[begin][end] = ans;

if (x >= y)
sell[begin][end] = 0;
sell[begin][end] = 1;

return ans;

// Util Function to calculate maxProfit
int maxProfit(int price[], int n)
// reseting the dp table
for (int i = 0; i < N; i++)
for (int j = 0; j < N; j++)
dp[i][j] = -1;

int ans = maxProfitUtil(price, 0, n – 1, n);

int i = 0, j = n – 1;

while (i <= j) {

// sell[i][j]=0 implies selling the
// wine from beginning will be more
// profitable in the long run
if (sell[i][j] == 0)
cout << "beg ";
cout << "end ";

cout << endl;

return ans;

// Driver code
int main()
// Price array
int price[] = { 2, 4, 6, 2, 5 };

int n = sizeof(price) / sizeof(price[0]);

int ans = maxProfit(price, n);

cout << ans << endl;

return 0;


beg end end beg beg 

Time Complexity: O(n2)

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