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
64
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 the 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.
C++
// 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; else 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 "; i++; } else { cout << "end "; j--; } } 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; } |
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Java
// Program to calculate maximum price of wines import java.io.*; class GFG { static int N = 1000; static int [][]dp = new int[N][N]; // This array stores the "optimal action" // for each state i, j static int [][]sell = new int[N][N]; // Function to maximize profit static 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 = Math.max(x, y); dp[begin][end] = ans; if (x >= y) sell[begin][end] = 0; else sell[begin][end] = 1; return ans; } // Util Function to calculate maxProfit static 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){ System.out.print( "beg "); i++; } else { System.out.print( "end "); j--; } } System.out.println(); return ans; } // Driver code public static void main (String[] args) { // Price array int price[] = { 2, 4, 6, 2, 5 }; int n = price.length; int ans = maxProfit(price, n); System.out.println( ans ); } } // This code is contributed by anuj_67. |
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C#
// C# Program to calculate maximum // price of wines using System; class GFG { static int N = 1000; static int [,]dp = new int[N, N]; // This array stores the "optimal action" // for each state i, j static int [,]sell = new int[N,N]; // Function to maximize profit static 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 = Math.Max(x, y); dp[begin,end] = ans; if (x >= y) sell[begin,end] = 0; else sell[begin,end] = 1; return ans; } // Util Function to calculate maxProfit static int maxProfit(int []price, int n) { int i, j; // reseting the dp table for(i = 0; i < N; i++) for(j = 0; j < N; j++) dp[i, j] = -1; int ans = maxProfitUtil(price, 0, n - 1, n); 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){ Console.Write( "beg "); i++; } else { Console.Write( "end "); j--; } } Console.WriteLine(); return ans; } // Driver Code public static void Main () { // Price array int []price = {2, 4, 6, 2, 5}; int n = price.Length; int ans = maxProfit(price, n); Console.WriteLine( ans ); } } // This code is contributed by anuj_67. |
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Output:
beg end end beg beg 64
Time Complexity: O(n2)
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