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.
C++
// CPP 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; } |
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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 |
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Output:
e/d-c*b+a
This article is contributed by Raghav Sharma. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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