Given a mathematical equation using numbers/variables and +, -, *, /. Print the equation in reverse.
Examples:
Input : 20 - 3 + 5 * 2
Output : 2 * 5 + 3 - 20
Input : 25 + 3 - 2 * 11
Output : 11 * 2 - 3 + 25
Input : a + b * c - d / e
Output : e / d - c * b + a
Approach : The approach to this problem is simple. We iterate the string from left to right and as soon we strike a symbol we insert the number and the symbol in the beginning of the resultant string.
Implementation:
// C++ program to reverse an equation #include <bits/stdc++.h> using namespace std;
// Function to reverse order of words string reverseEquation(string s) { // Resultant string
string result;
int j = 0;
for ( int i = 0; i < s.length(); i++) {
// A space marks the end of the word
if (s[i] == '+' || s[i] == '-' ||
s[i] == '/' || s[i] == '*' ) {
// insert the word at the beginning
// of the result string
result.insert(result.begin(),
s.begin() + j, s.begin() + i);
j = i + 1;
// insert the symbol
result.insert(result.begin(), s[i]);
}
}
// insert the last word in the string
// to the result string
result.insert(result.begin(), s.begin() + j,
s.end());
return result;
} // driver code int main()
{ string s = "a+b*c-d/e" ;
cout << reverseEquation(s) << endl;
return 0;
} |
// Java program to reverse an equation import java.util.*;
class GFG{
// Function to reverse order of words public static String reverseEquation(String s)
{ // Resultant string
String result = "" , str = "" ;
int j = 0 ;
for ( int i = 0 ; i < s.length(); i++)
{
// A space marks the end of the word
if (s.charAt(i) == '+' ||
s.charAt(i) == '-' ||
s.charAt(i) == '/' ||
s.charAt(i) == '*' )
{
// Insert the word at the beginning
// of the result string
result = s.charAt(i) + str + result;
str = "" ;
}
else
{
str += s.charAt(i);
}
}
result = str + result;
return result;
} // Driver code public static void main(String args[])
{ String s = "a+b*c-d/e" ;
System.out.println(reverseEquation(s));
} } // This code is contributed by bolliranadheer |
# Python3 Program to reverse an equation # Function to reverse order of words def reverseEquation(s):
# Reverse String
result = ""
for i in range ( len (s)):
# A space marks the end of the word
if (s[i] = = '+' or s[i] = = '-' or s[i] = = '/' or s[i] = = '*' ):
# insert the word at the beginning
# of the result String
result = s[i] + result
# insert the symbol
else :
result = s[i] + result
return result
# Driver Code s = "a+b*c-d/e"
print (reverseEquation(s))
# This code is contributed by simranjenny84 |
// C# program to reverse an equation using System;
using System.Text;
public class GFG{
// Function to reverse order of words
public static string reverseEquation( string s)
{
// Resultant string
string result = "" , str = "" ;
for ( int i = 0; i < s.Length; i++)
{
// A space marks the end of the word
if (s[i] == '+' || s[i] == '-' || s[i] == '/' || s[i] == '*' )
{
// Insert the word at the beginning
// of the result string
result = s[i] + str + result;
str = "" ;
}
else
{
str += s[i];
}
}
result = str + result;
return result;
}
// Driver Code
static public void Main (){
string s = "a+b*c-d/e" ;
Console.Write(reverseEquation(s));
}
} // This code is contributed by shruti456rawal |
// JavaScript program to reverse an equation // Function to reverse order of words function reverseEquation(s) {
// Resultant string
let result = "" , str = "" ;
for (let i = 0; i < s.length; i++) {
// A space marks the end of the word
if (s[i] === "+" || s[i] === "-" || s[i] === "/" || s[i] === "*" ) {
// Insert the word at the beginning
// of the result string
result = s[i] + str + result;
str = "" ;
}
else {
str += s[i];
}
}
result = str + result;
return result;
} let s = "a+b*c-d/e" ;
console.log(reverseEquation(s)); // This code is contributed by lokeshmvs21. |
e/d-c*b+a
Time Complexity: O(N)
Auxiliary Space: O(N) because it is using extra space for string result
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