Add solution for Wine problem (TheAlgorithms#2443)

nikitakapoor1919 · web-flow · commit f1ec159d85be · 2021-10-09T19:22:17.000+03:00
diff --git a/DynamicProgramming/WineProblem.java b/DynamicProgramming/WineProblem.java
@@ -0,0 +1,88 @@
+package DynamicProgramming;
+
+/**
+ * Imagine you have a collection of N wines placed next to each other on the
+ * shelf. The price of ith wine is pi(Prices of different wines are different).
+ * Because wine gets better every year supposing today is year 1, on year y the
+ * price would be y*pi i.e y times the value of the initial year. You want to
+ * sell all wines but you have to sell one wine per year. One more constraint on
+ * each year you are allowed to sell either leftmost or rightmost wine on the
+ * shelf. You are not allowed to reorder. You have to find the maximum profit
+ **/
+ 
+public class WineProblem {
+
+	// Method 1: Using Recursion
+	// Time Complexity=0(2^N) Space Complexity=Recursion extra space
+	public static int WPRecursion(int[] arr, int si, int ei) {
+		int n = arr.length;
+		int year = (n - (ei - si + 1)) + 1;
+		if (si == ei) {
+			return arr[si] * year;
+		}
+
+		int start = WPRecursion(arr, si + 1, ei) + arr[si] * year;
+		int end = WPRecursion(arr, si, ei - 1) + arr[ei] * year;
+
+		int ans = Math.max(start, end);
+
+		return ans;
+	}
+
+	// Method 2: Top-Down DP(Memoization)
+	// Time Complexity=0(N*N) Space Complexity=0(N*N)+Recursion extra space
+	public static int WPTD(int[] arr, int si, int ei, int[][] strg) {
+		int n = arr.length;
+		int year = (n - (ei - si + 1)) + 1;
+		if (si == ei) {
+			return arr[si] * year;
+		}
+
+		if (strg[si][ei] != 0) {
+			return strg[si][ei];
+		}
+		int start = WPTD(arr, si + 1, ei, strg) + arr[si] * year;
+		int end = WPTD(arr, si, ei - 1, strg) + arr[ei] * year;
+
+		int ans = Math.max(start, end);
+
+		strg[si][ei] = ans;
+
+		return ans;
+	}
+
+	// Method 3: Bottom-Up DP(Tabulation)
+	// Time Complexity=0(N*N/2)->0(N*N) Space Complexity=0(N*N)
+	public static int WPBU(int[] arr) {
+		int n = arr.length;
+		int[][] strg = new int[n][n];
+
+		for (int slide = 0; slide <= n - 1; slide++) {
+			for (int si = 0; si <= n - slide - 1; si++) {
+				int ei = si + slide;
+				int year = (n - (ei - si + 1)) + 1;
+				if (si == ei) {
+					strg[si][ei] = arr[si] * year;
+				} else {
+					int start = strg[si + 1][ei] + arr[si] * year;
+					int end = strg[si][ei - 1] + arr[ei] * year;
+
+					strg[si][ei] = Math.max(start, end);
+
+				}
+			}
+		}
+		return strg[0][n - 1];
+	}
+
+	public static void main(String[] args) {
+		int[] arr = { 2, 3, 5, 1, 4 };
+		System.out.println("Method 1: " + WPRecursion(arr, 0, arr.length - 1));
+		System.out.println("Method 2: " + WPTD(arr, 0, arr.length - 1, new int[arr.length][arr.length]));
+		System.out.println("Method 3: " + WPBU(arr));
+
+	}
+
+}
+// Memoization vs Tabulation : https://www.geeksforgeeks.org/tabulation-vs-memoization/
+// Question Link : https://www.geeksforgeeks.org/maximum-profit-sale-wines/
