Create wine_problem.cpp

Advanced/CPP/wine_problem.cpp

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+/* 
+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.
+*/
+
+#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;
+}
+
+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;
+}