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

Last Updated : 23 Mar, 2023

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:

C++

#include <bits/stdc++.h>
using namespace std;
int eval(vector<string>& A)
{
    stack<int> st;
    for (int i = 0; i < A.size(); i++) {
        if (A[i] != "+" && A[i] != "-" && A[i] != "/"
            && A[i] != "*") {
            st.push(stoi(A[i]));
            continue;
        }
        else {
            int b = st.top();
            st.pop();
            int a = st.top();
            st.pop();
            if (A[i] == "+")
                st.push(a + b);
            else if (A[i] == "-")
                st.push(a - b);
            else if (A[i] == "*")
                st.push(a * b);
            else
                st.push(a / b);
        }
    }
    return st.top();
}
 
int main()
{
    vector<string> A
        = { "10", "6", "9",  "3", "+", "-11", "*",
            "/",  "*", "17", "+", "5", "+" };
 
    int res = eval(A);
    cout << res << endl;
}

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 popping
                // 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 popping
                // 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;
 
            case "*":
 
                // Performing the "*" operation
                // by popping 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 popping
                // 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);
    }
}

Python3

# Python 3 code to evaluate reverse polish notation
 
"""
This code is contributed by Harshal Gupta
"""
 
# function to evaluate reverse polish notation
def evaluate(expression):
  # splitting expression at whitespaces
  expression = expression.split()
   
  # stack
  stack = []
   
  # iterating expression
  for ele in expression:
     
    # ele is a number
    if ele not in '/*+-':
      stack.append(int(ele))
     
    # ele is an operator
    else:
      # getting operands
      right = stack.pop()
      left = stack.pop()
       
      # performing operation according to operator
      if ele == '+':
        stack.append(left + right)
         
      elif ele == '-':
        stack.append(left - right)
         
      elif ele == '*':
        stack.append(left * right)
         
      elif ele == '/':
        stack.append(int(left / right))
   
  # return final answer.
  return stack.pop()
 
# input expression
arr = "10 6 9 3 + -11 * / * 17 + 5 +"
 
# calling evaluate()
answer = evaluate(arr)
# printing final value of the expression
print(f"Value of given expression'{arr}' = {answer}")
         
         
         
         
         
        

C#

// C# Program to find the
// solution of the arithmetic
// using the stack
 
using System;
using System.Collections.Generic;
 
public class GFG {
 
  public class Solution {
    public int stacky(string[] tokens)
    {
      // Initialize the stack and the variable
      Stack<string> stack = new Stack<string>();
      int x, y;
      string result = "";
      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
            // popping put the first two character
            // and then again store back to the
            // stack
 
            x = int.Parse(stack.Pop());
            y = int.Parse(stack.Pop());
            value = x + y;
            result = p + value;
            stack.Push(result);
            break;
 
          case "-":
            // Performing the "-" operation by
            // popping put the first two character
            // and then again store back to the
            // stack
            x = int.Parse(stack.Pop());
            y = int.Parse(stack.Pop());
            value = y - x;
            result = p + value;
            stack.Push(result);
            break;
 
          case "*":
            // Performing the "*" operation
            // by popping put the first two
            // character and then again store back
            // to the stack
 
            x = int.Parse(stack.Pop());
            y = int.Parse(stack.Pop());
            value = x * y;
            result = p + value;
            stack.Push(result);
            break;
 
          case "/":
            // Performing the "/" operation by
            // popping put the first two character
            // and then again store back to the
            // stack
 
            x = int.Parse(stack.Pop());
            y = int.Parse(stack.Pop());
            value = y / x;
            result = p + value;
            stack.Push(result);
            break;
 
          default:
            continue;
        }
      }
 
      // Method to convert the String to integer
      return int.Parse(stack.Pop());
    }
  }
 
  static public void Main()
  {
 
    // String Input
    string[] s
      = { "10", "6", "9",  "3", "+", "-11", "*",
         "/",  "*", "17", "+", "5", "+" };
 
    Solution str = new Solution();
    int result = str.stacky(s);
    Console.WriteLine(
      "Value of given expression'10 6 9 3 + -11 * / * 17 + 5 +' = "
      + result);
  }
}
 
// This code is contributed by lokesh

Javascript

// Javascript code to evaluate reverse polish notation    
function eval(A)
{
 
    // Initialize the stack
    let st = [];
     
    // Iterating to the each character
    // in the array of the string
    for (let i = 0; i < A.length; i++)
    {
     
        // If the character is not the special character
        // ('+', '-' ,'*' , '/')
        // then push the character to the stack
        if (A[i] != "+" && A[i] != "-" && A[i] != "/"
            && A[i] != "*") {
            st.push(parseInt(A[i]));
            continue;
        }
         
         // else if the character is the special
         // character then use them to
         // perform the action
        else {
            let b = parseInt(st.pop());
            let a = parseInt(st.pop());
            if (A[i] == "+")
                st.push(a + b);
            else if (A[i] == "-")
                st.push(a - b);
            else if (A[i] == "*")
                st.push(a * b);
            else
                st.push(parseInt(a / b));
        }
    }
    return parseInt(st[st.length-1]);
}
 
    let A = [ "10", "6", "9", "3", "+", "-11", "*",
            "/", "*", "17", "+", "5", "+" ];
 
    let res = eval(A);
    console.log(res);
     
    // This code is contributed by Pushpesh Raj.

Output

Value of given expression'10 6 9 3 + -11 * / * 17 + 5 +' = 22

Time complexity: O(n)

Auxiliary Space: O(n)

Approach – Space Optimized

Follow the below steps to Implement the idea:

Create two integer variables i = 0 and lastNum (variable for keeping track of the index of the last number seen) = -1.
Check if the current element is an arithmetic operator (+, -, *, /), If the current element is an arithmetic operator, then do the following:
a. Convert the previous two elements (the operands) to integers and store them in val1(get the second-to-last number seen) and val2(get the last number seen).
b. Perform the operation based on the operator, update the second-to-last number with the result, update the index of the last number seen.
if the current element is a number dp the following:
a.Increment the index of the last number seen.
b.Add the number to the next available index in the array.

Java

import java.util.*;
 
public class Main {
    public static int eval(String[] tokens)
    {
        // initialize a counter for iterating
        // over the tokens
        int i = 0;
        // initialize a variable for keeping track
        // of the index of the last number seen
        int lastNum = -1;
        while (i < tokens.length) {
            // if the current token is
            // an operator
            if ("/*+-".contains(tokens[i])) {
                // get the second-to-last
                // number seen
                int val1
                    = Integer.valueOf(tokens[lastNum - 1]);
                // get the last number
                // seen
                int val2 = Integer.valueOf(tokens[lastNum]);
                // initialize a variable for
                // storing the result of the
                // operation
                int ans = 0;
                // perform the operation based
                // on the operator
                if (tokens[i].equals("*"))
                    ans = val1 * val2;
                else if (tokens[i].equals("/"))
                    ans = val1 / val2;
                else if (tokens[i].equals("+"))
                    ans = val1 + val2;
                else if (tokens[i].equals("-"))
                    ans = val1 - val2;
                // update the second-to-last
                // number with the result
                tokens[lastNum - 1] = Integer.toString(ans);
                // update the index of the last
                // number seen
                lastNum--;
            }
            // if the current token is a number
            else {
                // increment the index of the
                // last number seen
                lastNum++;
                // add the number to the
                // next available index in
                // the array
                tokens[lastNum] = tokens[i];
            }
            i++; // increment the counter for iterating over
                 // the tokens
        }
        return Integer.valueOf(
            tokens[lastNum]); // return the final result
    }
 
    public static void main(String[] args)
    {
 
        String[] tokens
            = { "10", "6", "9",  "3", "+", "-11", "*",
                "/",  "*", "17", "+", "5", "+" };
 
        int res = eval(tokens);
        System.out.println(
            "Value of given expression '10 6 9 3 + -11 * / * 17 + 5 +' = "
            + res);
    }
}
// this code is contributed by Ravi Singh

Output

Value of given expression '10 6 9 3 + -11 * / * 17 + 5 +' = 22

Time Complexity: O(N)
Space Complexity: O(1)

Arithmetic Expression Having Only + and * Operators in Java

rohit2sahu

Improve

Article Tags :

Practice Tags :

Java

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

C++

Java

Python3

C#

Javascript

Approach – Space Optimized

Java

Similar Reads

Thank You!

What kind of Experience do you want to share?