Reversing an Equation
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++
// 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
// 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 |
C#
// 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 |
Python3
# 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 |
Javascript
// 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. |
Output
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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