Prefix and Postfix expressions can be evaluated faster than an infix expression. This is because we don’t need to process any brackets or follow operator precedence rule. In postfix and prefix expressions which ever operator comes before will be evaluated first, irrespective of its priority. Also, there are no brackets in these expressions. As long as we can guarantee that a valid prefix or postfix expression is used, it can be evaluated with correctness.
We can convert infix to postfix and can convert infix to prefix.
In this article, we will discuss how to evaluate an expression written in prefix notation. The method is similar to evaluating a postfix expression. Please read Evaluation of Postfix Expression to know how to evaluate postfix expressions
Algorithm:
EVALUATE_PREFIX(STRING) Step 1: Put a pointer P at the end of the end Step 2: If character at P is an operand push it to Stack Step 3: If the character at P is an operator pop two elements from the Stack. Operate on these elements according to the operator, and push the result back to the Stack Step 4: Decrement P by 1 and go to Step 2 as long as there are characters left to be scanned in the expression. Step 5: The Result is stored at the top of the Stack, return it Step 6: End
Example to demonstrate working of the algorithm
Expression: +9*26 Character | Stack | Explanation Scanned | (Front to | | Back) | ------------------------------------------- 6 6 6 is an operand, push to Stack 2 6 2 2 is an operand, push to Stack * 12 (6*2) * is an operator, pop 6 and 2, multiply them and push result to Stack 9 12 9 9 is an operand, push to Stack + 21 (12+9) + is an operator, pop 12 and 9 add them and push result to Stack Result: 21
Examples:
Input : -+8/632 Output : 8 Input : -+7*45+20 Output : 25
Complexity The algorithm has linear complexity since we scan the expression once and perform at most O(N) push and pop operations which take constant time.
Implementation of the algorithm is given below.
Implementation:
C++
// C++ program to evaluate a prefix expression. #include <bits/stdc++.h> using namespace std; bool isOperand( char c) { // If the character is a digit then it must // be an operand return isdigit (c); } double evaluatePrefix(string exprsn) { stack< double > Stack; for ( int j = exprsn.size() - 1; j >= 0; j--) { // Push operand to Stack // To convert exprsn[j] to digit subtract // '0' from exprsn[j]. if (isOperand(exprsn[j])) Stack.push(exprsn[j] - '0' ); else { // Operator encountered // Pop two elements from Stack double o1 = Stack.top(); Stack.pop(); double o2 = Stack.top(); Stack.pop(); // Use switch case to operate on o1 // and o2 and perform o1 Or o2. switch (exprsn[j]) { case '+' : Stack.push(o1 + o2); break ; case '-' : Stack.push(o1 - o2); break ; case '*' : Stack.push(o1 * o2); break ; case '/' : Stack.push(o1 / o2); break ; } } } return Stack.top(); } // Driver code int main() { string exprsn = "+9*26" ; cout << evaluatePrefix(exprsn) << endl; return 0; } |
Java
// Java program to evaluate // a prefix expression. import java.io.*; import java.util.*; class GFG { static Boolean isOperand( char c) { // If the character is a digit // then it must be an operand if (c >= 48 && c <= 57 ) return true ; else return false ; } static double evaluatePrefix(String exprsn) { Stack<Double> Stack = new Stack<Double>(); for ( int j = exprsn.length() - 1 ; j >= 0 ; j--) { // Push operand to Stack // To convert exprsn[j] to digit subtract // '0' from exprsn[j]. if (isOperand(exprsn.charAt(j))) Stack.push(( double )(exprsn.charAt(j) - 48 )); else { // Operator encountered // Pop two elements from Stack double o1 = Stack.peek(); Stack.pop(); double o2 = Stack.peek(); Stack.pop(); // Use switch case to operate on o1 // and o2 and perform o1 Or o2. switch (exprsn.charAt(j)) { case '+' : Stack.push(o1 + o2); break ; case '-' : Stack.push(o1 - o2); break ; case '*' : Stack.push(o1 * o2); break ; case '/' : Stack.push(o1 / o2); break ; } } } return Stack.peek(); } /* Driver program to test above function */ public static void main(String[] args) { String exprsn = "+9*26" ; System.out.println(evaluatePrefix(exprsn)); } } // This code is contributed by Gitanjali |
Python3
""" Python3 program to evaluate a prefix expression. """ def is_operand(c): """ Return True if the given char c is an operand, e.g. it is a number """ return c.isdigit() def evaluate(expression): """ Evaluate a given expression in prefix notation. Asserts that the given expression is valid. """ stack = [] # iterate over the string in reverse order for c in expression[:: - 1 ]: # push operand to stack if is_operand(c): stack.append( int (c)) else : # pop values from stack can calculate the result # push the result onto the stack again o1 = stack.pop() o2 = stack.pop() if c = = '+' : stack.append(o1 + o2) elif c = = '-' : stack.append(o1 - o2) elif c = = '*' : stack.append(o1 * o2) elif c = = '/' : stack.append(o1 / o2) return stack.pop() # Driver code if __name__ = = "__main__" : test_expression = "+9*26" print (evaluate(test_expression)) # This code is contributed by Leon Morten Richter (GitHub: M0r13n) |
C#
// C# program to evaluate // a prefix expression. using System; using System.Collections.Generic; class GFG { static Boolean isOperand( char c) { // If the character is a digit // then it must be an operand if (c >= 48 && c <= 57) return true ; else return false ; } static double evaluatePrefix(String exprsn) { Stack<Double> Stack = new Stack<Double>(); for ( int j = exprsn.Length - 1; j >= 0; j--) { // Push operand to Stack // To convert exprsn[j] to digit subtract // '0' from exprsn[j]. if (isOperand(exprsn[j])) Stack.Push(( double )(exprsn[j] - 48)); else { // Operator encountered // Pop two elements from Stack double o1 = Stack.Peek(); Stack.Pop(); double o2 = Stack.Peek(); Stack.Pop(); // Use switch case to operate on o1 // and o2 and perform o1 Or o2. switch (exprsn[j]) { case '+' : Stack.Push(o1 + o2); break ; case '-' : Stack.Push(o1 - o2); break ; case '*' : Stack.Push(o1 * o2); break ; case '/' : Stack.Push(o1 / o2); break ; } } } return Stack.Peek(); } /* Driver code */ public static void Main(String[] args) { String exprsn = "+9*26" ; Console.WriteLine(evaluatePrefix(exprsn)); } } /* This code contributed by PrinciRaj1992 */ |
Javascript
<script> // Javascript program to evaluate a prefix expression. function isOperand(c) { // If the character is a digit // then it must be an operand if (c.charCodeAt() >= 48 && c.charCodeAt() <= 57) return true ; else return false ; } function evaluatePrefix(exprsn) { let Stack = []; for (let j = exprsn.length - 1; j >= 0; j--) { // Push operand to Stack // To convert exprsn[j] to digit subtract // '0' from exprsn[j]. if (isOperand(exprsn[j])) Stack.push((exprsn[j].charCodeAt() - 48)); else { // Operator encountered // Pop two elements from Stack let o1 = Stack[Stack.length - 1]; Stack.pop(); let o2 = Stack[Stack.length - 1]; Stack.pop(); // Use switch case to operate on o1 // and o2 and perform o1 Or o2. switch (exprsn[j]) { case '+' : Stack.push(o1 + o2); break ; case '-' : Stack.push(o1 - o2); break ; case '*' : Stack.push(o1 * o2); break ; case '/' : Stack.push(o1 / o2); break ; } } } return Stack[Stack.length - 1]; } let exprsn = "+9*26" ; document.write(evaluatePrefix(exprsn)); // This code is contributed by suresh07. </script> |
21
Time Complexity: O(N)
Auxiliary Space: O(N)
Note:
To perform more types of operations only the switch case table needs to be modified. This implementation works only for single digit operands. Multi-digit operands can be implemented if some character-like space is used to separate the operands and operators.
Below given is the extended program which allows operands to have multiple digits.
C++
// C++ program to evaluate a prefix expression. #include <bits/stdc++.h> using namespace std; double evaluatePrefix(string exprsn) { stack< double > Stack; for ( int j = exprsn.size() - 1; j >= 0; j--) { // if jth character is the delimiter ( which is // space in this case) then skip it if (exprsn[j] == ' ' ) continue ; // Push operand to Stack // To convert exprsn[j] to digit subtract // '0' from exprsn[j]. if ( isdigit (exprsn[j])) { // there may be more than // one digits in a number double num = 0, i = j; while (j < exprsn.size() && isdigit (exprsn[j])) j--; j++; // from [j, i] exprsn contains a number for ( int k = j; k <= i; k++) num = num * 10 + double (exprsn[k] - '0' ); Stack.push(num); } else { // Operator encountered // Pop two elements from Stack double o1 = Stack.top(); Stack.pop(); double o2 = Stack.top(); Stack.pop(); // Use switch case to operate on o1 // and o2 and perform o1 O o2. switch (exprsn[j]) { case '+' : Stack.push(o1 + o2); break ; case '-' : Stack.push(o1 - o2); break ; case '*' : Stack.push(o1 * o2); break ; case '/' : Stack.push(o1 / o2); break ; } } } return Stack.top(); } // Driver code int main() { string exprsn = "+ 9 * 12 6" ; cout << evaluatePrefix(exprsn) << endl; return 0; // this code is contributed by Mohd Shaad Khan } |
Java
// Java program to evaluate a prefix expression. import java.util.*; public class Main { static boolean isdigit( char ch) { if (ch >= 48 && ch <= 57 ) { return true ; } return false ; } static double evaluatePrefix(String exprsn) { Stack<Double> stack = new Stack<Double>(); for ( int j = exprsn.length() - 1 ; j >= 0 ; j--) { // if jth character is the delimiter ( which is // space in this case) then skip it if (exprsn.charAt(j) == ' ' ) continue ; // Push operand to Stack // To convert exprsn[j] to digit subtract // '0' from exprsn[j]. if (isdigit(exprsn.charAt(j))) { // there may be more than // one digits in a number double num = 0 , i = j; while (j < exprsn.length() && isdigit(exprsn.charAt(j))) j--; j++; // from [j, i] exprsn contains a number for ( int k = j; k <= i; k++) { num = num * 10 + ( double )(exprsn.charAt(k) - '0' ); } stack.push(num); } else { // Operator encountered // Pop two elements from Stack double o1 = ( double )stack.peek(); stack.pop(); double o2 = ( double )stack.peek(); stack.pop(); // Use switch case to operate on o1 // and o2 and perform o1 O o2. switch (exprsn.charAt(j)) { case '+' : stack.push(o1 + o2); break ; case '-' : stack.push(o1 - o2); break ; case '*' : stack.push(o1 * o2); break ; case '/' : stack.push(o1 / o2); break ; } } } return stack.peek(); } // Driver code public static void main(String[] args) { String exprsn = "+ 9 * 12 6" ; System.out.print(( int )evaluatePrefix(exprsn)); } } // This code is contributed by mukesh07. |
Python3
# Python3 program to evaluate a prefix expression. def isdigit(ch): if ( ord (ch) > = 48 and ord (ch) < = 57 ): return True return False def evaluatePrefix(exprsn): Stack = [] for j in range ( len (exprsn) - 1 , - 1 , - 1 ): # if jth character is the delimiter ( which is # space in this case) then skip it if (exprsn[j] = = ' ' ): continue # Push operand to Stack # To convert exprsn[j] to digit subtract # '0' from exprsn[j]. if (isdigit(exprsn[j])): # there may be more than # one digits in a number num, i = 0 , j while (j < len (exprsn) and isdigit(exprsn[j])): j - = 1 j + = 1 # from [j, i] exprsn contains a number for k in range (j, i + 1 , 1 ): num = num * 10 + ( ord (exprsn[k]) - ord ( '0' )) Stack.append(num) else : # Operator encountered # Pop two elements from Stack o1 = Stack[ - 1 ] Stack.pop() o2 = Stack[ - 1 ] Stack.pop() # Use switch case to operate on o1 # and o2 and perform o1 O o2. if exprsn[j] = = '+' : Stack.append(o1 + o2 + 12 ) elif exprsn[j] = = '-' : Stack.append(o1 - o2) elif exprsn[j] = = '*' : Stack.append(o1 * o2 * 5 ) elif exprsn[j] = = '/' : Stack.append(o1 / o2) return Stack[ - 1 ] exprsn = "+ 9 * 12 6" print (evaluatePrefix(exprsn)) # This code is contributed by divyesh072019. |
C#
// C# program to evaluate a prefix expression. using System; using System.Collections; class GFG { static bool isdigit( char ch) { if (ch >= 48 && ch <= 57) { return true ; } return false ; } static double evaluatePrefix( string exprsn) { Stack stack = new Stack(); for ( int j = exprsn.Length - 1; j >= 0; j--) { // if jth character is the delimiter ( which is // space in this case) then skip it if (exprsn[j] == ' ' ) continue ; // Push operand to Stack // To convert exprsn[j] to digit subtract // '0' from exprsn[j]. if (isdigit(exprsn[j])) { // there may be more than // one digits in a number double num = 0, i = j; while (j < exprsn.Length && isdigit(exprsn[j])) j--; j++; // from [j, i] exprsn contains a number for ( int k = j; k <= i; k++) { num = num * 10 + ( double )(exprsn[k] - '0' ); } stack.Push(num); } else { // Operator encountered // Pop two elements from Stack double o1 = ( double )stack.Peek(); stack.Pop(); double o2 = ( double )stack.Peek(); stack.Pop(); // Use switch case to operate on o1 // and o2 and perform o1 O o2. switch (exprsn[j]) { case '+' : stack.Push(o1 + o2); break ; case '-' : stack.Push(o1 - o2); break ; case '*' : stack.Push(o1 * o2); break ; case '/' : stack.Push(o1 / o2); break ; } } } return ( double )stack.Peek(); } static void Main() { string exprsn = "+ 9 * 12 6" ; Console.Write(evaluatePrefix(exprsn)); } } // This code is contributed by divyeshrabadiya07. |
Javascript
<script> // Javascript program to evaluate a prefix expression. function isdigit(ch) { if (ch.charCodeAt() >= 48 && ch.charCodeAt() <= 57) { return true ; } return false ; } function evaluatePrefix(exprsn) { let Stack = []; for (let j = exprsn.length - 1; j >= 0; j--) { // if jth character is the delimiter ( which is // space in this case) then skip it if (exprsn[j] == ' ' ) continue ; // Push operand to Stack // To convert exprsn[j] to digit subtract // '0' from exprsn[j]. if (isdigit(exprsn[j])) { // there may be more than // one digits in a number let num = 0, i = j; while (j < exprsn.length && isdigit(exprsn[j])) j--; j++; // from [j, i] exprsn contains a number for (let k = j; k <= i; k++) num = num * 10 + (exprsn[k].charCodeAt() - '0' .charCodeAt()); Stack.push(num); } else { // Operator encountered // Pop two elements from Stack let o1 = Stack[Stack.length - 1]; Stack.pop(); let o2 = Stack[Stack.length - 1]; Stack.pop(); // Use switch case to operate on o1 // and o2 and perform o1 O o2. switch (exprsn[j]) { case '+' : Stack.push(o1 + o2); break ; case '-' : Stack.push(o1 - o2); break ; case '*' : Stack.push(o1 * o2); break ; case '/' : Stack.push(o1 / o2); break ; } } } return Stack[Stack.length - 1]; } let exprsn = "+ 9 * 12 6" ; document.write(evaluatePrefix(exprsn)); // This code is contributed by rameshtravel07. </script> |
81
Time Complexity: O(n)
Auxiliary Space: O(n)
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