Evaluate the Value of an Arithmetic Expression in Reverse Polish Notation in Java

Difficulty Level : Easy
Last Updated : 12 Jul, 2021

Reverse Polish ‘Notation is postfix notation which in terms of mathematical notion signifies operators following operands. Let’s take a problem statement to implement RPN

Problem Statement: The task is to find the value of the arithmetic expression present in the array using valid operators like +, -, *, /. Each operand may be an integer or another expression.

Note:

The division between two integers should truncate toward zero.
The given RPN expression is always valid. That means the expression would always evaluate to a result and there won’t be any divide by zero operation.

Layman Working of RPN as shown

Input: ["2", "1", "+", "3", "*"]
Output: 9
Explanation: ((2 + 1) * 3) = 9

Input: ["4", "13", "5", "/", "+"]
Output: 6
Explanation: (4 + (13 / 5)) = 6

Input: ["10", "6", "9", "3", "+", "-11", "*", "/", "*", "17", "+", "5", "+"]
Output: 22
Explanation: 
  ((10 * (6 / ((9 + 3) * -11))) + 17) + 5
= ((10 * (6 / (12 * -11))) + 17) + 5
= ((10 * (6 / -132)) + 17) + 5
= ((10 * 0) + 17) + 5
= (0 + 17) + 5
= 17 + 5
= 22

Approach:

The basic approach for the problem is using the stack.

Accessing all elements in the array, if the element is not matching with the special character (‘+’, ‘-‘,’*’, ‘/’) then push the element to the stack.
Then whenever the special character is found then pop the first two-element from the stack and perform the action and then push the element to stack again.
Repeat the above two process to all elements in the array
At last pop the element from the stack and print the Result

Implementation:

Java

// Java Program to find the
// solution of the arithmetic
// using the stack
import java.io.*;
import java.util.*;
 
class solution {
    public int stacky(String[] tokens)
    {
 
        // Initialize the stack and the variable
        Stack<String> stack = new Stack<String>();
        int x, y;
        String result = "";
        int get = 0;
        String choice;
        int value = 0;
        String p = "";
 
        // Iterating to the each character
        // in the array of the string
        for (int i = 0; i < tokens.length; i++) {
 
            // If the character is not the special character
            // ('+', '-' ,'*' , '/')
            // then push the character to the stack
            if (tokens[i] != "+" && tokens[i] != "-"
                && tokens[i] != "*" && tokens[i] != "/") {
                stack.push(tokens[i]);
                continue;
            }
            else {
                // else if the character is the special
                // character then use the switch method to
                // perform the action
                choice = tokens[i];
            }
 
            // Switch-Case
            switch (choice) {
            case "+":
 
                // Performing the "+" operation by poping
                // put the first two character
                // and then again store back to the stack
 
                x = Integer.parseInt(stack.pop());
                y = Integer.parseInt(stack.pop());
                value = x + y;
                result = p + value;
                stack.push(result);
                break;
 
            case "-":
 
                // Performing the "-" operation by poping
                // put the first two character
                // and then again store back to the stack
                x = Integer.parseInt(stack.pop());
                y = Integer.parseInt(stack.pop());
                value = x - y;
                result = p + value;
                stack.push(result);
                break;
 
            case "*":
 
                // Performing the "*" operation
                // by poping put the first two character
                // and then again store back to the stack
 
                x = Integer.parseInt(stack.pop());
                y = Integer.parseInt(stack.pop());
                value = x * y;
                result = p + value;
                stack.push(result);
                break;
 
            case "/":
 
                // Performing the "/" operation by poping
                // put the first two character
                // and then again store back to the stack
 
                x = Integer.parseInt(stack.pop());
                y = Integer.parseInt(stack.pop());
                value = y / x;
                result = p + value;
                stack.push(result);
                break;
 
            default:
                continue;
            }
        }
 
        // Method to convert the String to integer
        return Integer.parseInt(stack.pop());
    }
}
 
class GFG {
   
    public static void main(String[] args)
    {
        // String Input
        String[] s
            = { "10", "6", "9",  "3", "+", "-11", "*",
                "/",  "*", "17", "+", "5", "+" };
       
        solution str = new solution();
        int result = str.stacky(s);
        System.out.println(result);
    }
}

Output :

Time complexity: O(n)

Attention reader! Don’t stop learning now. Get hold of all the important Java Foundation and Collections concepts with the Fundamentals of Java and Java Collections Course at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.

My Personal Notes

Related Articles

Evaluate the Value of an Arithmetic Expression in Reverse Polish Notation in Java

Java