Evaluate the Value of an Arithmetic Expression in Reverse Polish Notation in Java
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
import java.io.*;
import java.util.*;
class solution {
public int stacky(String[] tokens)
{
Stack<String> stack = new Stack<String>();
int x, y;
String result = "" ;
int get = 0 ;
String choice;
int value = 0 ;
String p = "" ;
for ( int i = 0 ; i < tokens.length; i++) {
if (tokens[i] != "+" && tokens[i] != "-"
&& tokens[i] != "*" && tokens[i] != "/" ) {
stack.push(tokens[i]);
continue ;
}
else {
choice = tokens[i];
}
switch (choice) {
case "+" :
x = Integer.parseInt(stack.pop());
y = Integer.parseInt(stack.pop());
value = x + y;
result = p + value;
stack.push(result);
break ;
case "-" :
x = Integer.parseInt(stack.pop());
y = Integer.parseInt(stack.pop());
value = y - x;
result = p + value;
stack.push(result);
break ;
case "*" :
x = Integer.parseInt(stack.pop());
y = Integer.parseInt(stack.pop());
value = x * y;
result = p + value;
stack.push(result);
break ;
case "/" :
x = Integer.parseInt(stack.pop());
y = Integer.parseInt(stack.pop());
value = y / x;
result = p + value;
stack.push(result);
break ;
default :
continue ;
}
}
return Integer.parseInt(stack.pop());
}
}
class GFG {
public static void main(String[] args)
{
String[] s
= { "10" , "6" , "9" , "3" , "+" , "-11" , "*" ,
"/" , "*" , "17" , "+" , "5" , "+" };
solution str = new solution();
int result = str.stacky(s);
System.out.println(result);
}
}
|
Python3
def evaluate(expression):
expression = expression.split()
stack = []
for ele in expression:
if ele not in '/*+-' :
stack.append( int (ele))
else :
right = stack.pop()
left = stack.pop()
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 stack.pop()
arr = "10 6 9 3 + -11 * / * 17 + 5 +"
answer = evaluate(arr)
print (f "Value of given expression'{arr}' = {answer}" )
|
C#
using System;
using System.Collections.Generic;
public class GFG {
public class Solution {
public int stacky( string [] tokens)
{
Stack< string > stack = new Stack< string >();
int x, y;
string result = "" ;
string choice;
int value = 0;
string p = "" ;
for ( int i = 0; i < tokens.Length; i++) {
if (tokens[i] != "+" && tokens[i] != "-"
&& tokens[i] != "*"
&& tokens[i] != "/" ) {
stack.Push(tokens[i]);
continue ;
}
else {
choice = tokens[i];
}
switch (choice) {
case "+" :
x = int .Parse(stack.Pop());
y = int .Parse(stack.Pop());
value = x + y;
result = p + value;
stack.Push(result);
break ;
case "-" :
x = int .Parse(stack.Pop());
y = int .Parse(stack.Pop());
value = y - x;
result = p + value;
stack.Push(result);
break ;
case "*" :
x = int .Parse(stack.Pop());
y = int .Parse(stack.Pop());
value = x * y;
result = p + value;
stack.Push(result);
break ;
case "/" :
x = int .Parse(stack.Pop());
y = int .Parse(stack.Pop());
value = y / x;
result = p + value;
stack.Push(result);
break ;
default :
continue ;
}
}
return int .Parse(stack.Pop());
}
}
static public void Main()
{
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);
}
}
|
Javascript
function eval(A)
{
let st = [];
for (let i = 0; i < A.length; i++)
{
if (A[i] != "+" && A[i] != "-" && A[i] != "/"
&& A[i] != "*" ) {
st.push(parseInt(A[i]));
continue ;
}
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);
|
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)
{
int i = 0 ;
int lastNum = - 1 ;
while (i < tokens.length) {
if ( "/*+-" .contains(tokens[i])) {
int val1
= Integer.valueOf(tokens[lastNum - 1 ]);
int val2 = Integer.valueOf(tokens[lastNum]);
int ans = 0 ;
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;
tokens[lastNum - 1 ] = Integer.toString(ans);
lastNum--;
}
else {
lastNum++;
tokens[lastNum] = tokens[i];
}
i++;
}
return Integer.valueOf(
tokens[lastNum]);
}
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);
}
}
|
Output
Value of given expression '10 6 9 3 + -11 * / * 17 + 5 +' = 22
Time Complexity: O(N)
Space Complexity: O(1)
Last Updated :
23 Mar, 2023
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