Create 0-1KnapsackProblem.java

Author: 17tanya

DynamicProgramming/0-1KnapsackProblem.java

@@ -0,0 +1,50 @@
+/*
+https://www.techiedelight.com/0-1-knapsack-problem/
+https://www.geeksforgeeks.org/0-1-knapsack-problem-dp-10/
+*/
+
+/*
+In the 0/1 Knapsack Problem we must include an item entirely or exclude the item with an objective of maximizing profit/value constrainted by the size of the knapsack.
+
+If the problem is not 0/1 knapsack and we are allowed to split items, we can take a greedy approach
+we can chose items by sorting items according to profit per unit weight.
+In doing so, only the last element or a single element will have to be split.
+*/
+
+/*RECURSION
+For each item we have 2 choices - 
+1. Include the item and recur for the remaining item. Here the capacity of the knapsack is reduced.
+2. Exclude the item and recur for the remaining item. Capacity of knapsack remains the same.
+
+*/
+
+public int knapsack(int W, int item, int wt[], int val[]){
+    //BASE CASE: we have exhausted all items or we have utilized the entire capacity of knapsack
+    if(n==0 || W==0) return 0;
+    
+    //If the weight of current item exceeds the capacity of knapsack it can not be included - EXCLUDE ITEM
+    else if(wt[n-1] > W) return knapsack(W, item-1, wt, val);
+    
+    //Try both - INCLUDE AND EXCLUDE - chose the best
+    else return Math.max(knapsack(W, item-1, wt, val), val[item-1] + knapsack(W-wt[item-1], item-1, wt, val))
+}
+
+
+/*
+Time Complexity - O(2^n)
+For each item we are exploring the two choices(We include and exclude each item) to find the best.
+*/
+
+/*
+DYNAMIC PROGRAMMING - BOTTOM UP APPROACH
+*/
+
+public int knapsack(int W, int item, int wt[], int val[]){
+    int k[][] = new int[item+1][W+1];
+    
+    for(int i=0;i<=item;i++){
+        for(int j=0)
+    }
+}
+
+