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Today in this article I will show you how to solve 0-1 Knapsack problem using the concept of dynamic programming in Java. This problem can also be solved by recursion but the problem is that it will result in solving of already solved sub-problem. So we will be using dynamic programming to solve by storing the results of solved sub-problems. Also this technique can be used to solve the problem in polynomial time. The complexity of solving it using this algorithm is O(n*W).
What is 0/1 Knapsack problem ?
The knapsack or rucksack problem is a problem in combinatorial optimization. Here there are a set of items(n) which has a profit value(v) and weight(w). There is a bag which has a maximum capacity(W). Now the problem is to fill the bag with the following restrictions :
   1. You can either choose an item completely or reject it. (0 or 1)
   2. The total weight of bag must not exceed its maximum capacity W. i.e. (Sum of w[i]) <= W
   3.  The total value or profit of items chosen must be maximum. i.e. Sum of p[i] is maximum
where 0 <  i <=n.
How to solve 0/1 Knapsack problem in Java ?
Step1: Structure: Characterize the structure of an optimal solution.
    – Decompose the problem into smaller problems, and find a relation between the structure of the optimal solution of the original problem and the solutions of the smaller problems.

Today I am going to show how to convert a postfix expression to an infix expression using stack in Java. In an earlier post here we ahve shown how to convert an infix expression to postfix using stack. Here we will just try to do the reverse. Here in our example we will be able to convert any postfix expression to infix irrespective of the operators. A number of parenthesis may be generated extra which were not there in original infix expression. The parenthesis that may be generated extra will have no impact on actual expression, they are just for better understanding. The process of postfix to infix conversion is summarised below -