Stack | Set 4 (Evaluation of Postfix Expression)
The Postfix notation is used to represent algebraic expressions. The expressions written in postfix form are evaluated faster compared to infix notation as parenthesis are not required in postfix. We have discussed infix to postfix conversion. In this post, evaluation of postfix expressions is discussed.
Following is an algorithm for evaluation postfix expressions.
1) Create a stack to store operands (or values).
2) Scan the given expression and do the following for every scanned element.
…..a) If the element is a number, push it into the stack
…..b) If the element is an operator, pop operands for the operator from the stack. Evaluate the operator and push the result back to the stack
3) When the expression is ended, the number in the stack is the final answer
Example:
Let the given expression be “2 3 1 * + 9 -“. We scan all elements one by one.
1) Scan ‘2’, it’s a number, so push it to stack. Stack contains ‘2’
2) Scan ‘3’, again a number, push it to stack, stack now contains ‘2 3’ (from bottom to top)
3) Scan ‘1’, again a number, push it to stack, stack now contains ‘2 3 1’
4) Scan ‘*’, it’s an operator, pop two operands from stack, apply the * operator on operands, we get 3*1 which results in 3. We push the result ‘3’ to stack. The stack now becomes ‘2 3’.
5) Scan ‘+’, it’s an operator, pop two operands from stack, apply the + operator on operands, we get 3 + 2 which results in 5. We push the result ‘5’ to stack. The stack now becomes ‘5’.
6) Scan ‘9’, it’s a number, we push it to the stack. The stack now becomes ‘5 9’.
7) Scan ‘-‘, it’s an operator, pop two operands from stack, apply the – operator on operands, we get 5 – 9 which results in -4. We push the result ‘-4’ to the stack. The stack now becomes ‘-4’.
8) There are no more elements to scan, we return the top element from the stack (which is the only element left in a stack).
Expression Evaluation
Evaluate an expression represented by a String. The expression can contain parentheses, you can assume parentheses are well-matched. For simplicity, you can assume only binary operations allowed are +, -, *, and /. Arithmetic Expressions can be written in one of three forms:
Infix Notation: Operators are written between the operands they operate on, e.g. 3 + 4.
Prefix Notation: Operators are written before the operands, e.g + 3 4
Postfix Notation: Operators are written after operands.
Infix Expressions are harder for Computers to evaluate because of the additional work needed to decide precedence. Infix notation is how expressions are written and recognized by humans and, generally, input to programs. Given that they are harder to evaluate, they are generally converted to one of the two remaining forms. A very well known algorithm for converting an infix notation to a postfix notation is Shunting Yard Algorithm by Edgar Dijkstra. This algorithm takes as input an Infix Expression and produces a queue that has this expression converted to postfix notation. The same algorithm can be modified so that it outputs the result of the evaluation of expression instead of a queue. The trick is using two stacks instead of one, one for operands, and one for operators. The algorithm was described succinctly on //www.cis.upenn.edu/ matuszek/cit594-2002/Assignments/5-expressions.htm, and is reproduced here. (Note that credit for succinctness goes to the author of said page)
It should be clear that this algorithm runs in linear time – each number or operator is pushed onto and popped from Stack only once. Also see //www2.lawrence.edu/fast/GREGGJ/CMSC270/Infix.html,
//faculty.cs.niu.edu/~hutchins/csci241/eval.htm.
Following is the implementation of above algorithm:
C++
|
// CPP program to evaluate a given // expression where tokens are // separated by space. #include <bits/stdc++.h> using namespace std; // Function to find precedence of // operators. int precedence(char op){ if(op == '+'||op == '-') return 1; if(op == '*'||op == '/') return 2; return 0; } // Function to perform arithmetic operations. int applyOp(int a, int b, char op){ switch(op){ case '+': return a + b; case '-': return a - b; case '*': return a * b; case '/': return a / b; } } // Function that returns value of // expression after evaluation. int evaluate(string tokens){ int i;
// stack to store integer values. stack <int> values;
// stack to store operators. stack <char> ops;
for(i = 0; i < tokens.length(); i++){
// Current token is a whitespace, // skip it. if(tokens[i] == ' ') continue;
// Current token is an opening // brace, push it to 'ops' else if(tokens[i] == '('){ ops.push(tokens[i]); }
// Current token is a number, push // it to stack for numbers. else if(isdigit(tokens[i])){ int val = 0;
// There may be more than one // digits in number. while(i < tokens.length() && isdigit(tokens[i])) { val = (val*10) + (tokens[i]-'0'); i++; }
values.push(val);
// right now the i points to // the character next to the digit, // since the for loop also increases // the i, we would skip one // token position; we need to // decrease the value of i by 1 to // correct the offset. i--; }
// Closing brace encountered, solve // entire brace. else if(tokens[i] == ')') { while(!ops.empty() && ops.top() != '(') { int val2 = values.top(); values.pop();
int val1 = values.top(); values.pop();
char op = ops.top(); ops.pop();
values.push(applyOp(val1, val2, op)); }
// pop opening brace. if(!ops.empty()) ops.pop(); }
// Current token is an operator. else { // While top of 'ops' has same or greater // precedence to current token, which // is an operator. Apply operator on top // of 'ops' to top two elements in values stack. while(!ops.empty() && precedence(ops.top()) >= precedence(tokens[i])){ int val2 = values.top(); values.pop();
int val1 = values.top(); values.pop();
char op = ops.top(); ops.pop();
values.push(applyOp(val1, val2, op)); }
// Push current token to 'ops'. ops.push(tokens[i]); } }
// Entire expression has been parsed at this // point, apply remaining ops to remaining // values. while(!ops.empty()){ int val2 = values.top(); values.pop();
int val1 = values.top(); values.pop();
char op = ops.top(); ops.pop();
values.push(applyOp(val1, val2, op)); }
// Top of 'values' contains result, return it. return values.top(); } int main() { cout << evaluate("10 + 2 * 6") << "\n"; cout << evaluate("100 * 2 + 12") << "\n"; cout << evaluate("100 * ( 2 + 12 )") << "\n"; cout << evaluate("100 * ( 2 + 12 ) / 14"); return 0; } // This code is contributed by Nikhil jindal. |
Java
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/* A Java program to evaluate a given expression where tokens are separated by space. */ import java.util.Stack; public class EvaluateString { public static int evaluate(String expression) { char[] tokens = expression.toCharArray(); // Stack for numbers: 'values' Stack<Integer> values = new Stack<Integer>(); // Stack for Operators: 'ops' Stack<Character> ops = new Stack<Character>(); for (int i = 0; i < tokens.length; i++) {
// Current token is a // whitespace, skip it if (tokens[i] == ' ') continue; // Current token is a number, // push it to stack for numbers if (tokens[i] >= '0' && tokens[i] <= '9') { StringBuffer sbuf = new StringBuffer();
// There may be more than one // digits in number while (i < tokens.length && tokens[i] >= '0' && tokens[i] <= '9') sbuf.append(tokens[i++]); values.push(Integer.parseInt(sbuf. toString()));
// right now the i points to // the character next to the digit, // since the for loop also increases // the i, we would skip one // token position; we need to // decrease the value of i by 1 to // correct the offset. i--; } // Current token is an opening brace, // push it to 'ops' else if (tokens[i] == '(') ops.push(tokens[i]); // Closing brace encountered, // solve entire brace else if (tokens[i] == ')') { while (ops.peek() != '(') values.push(applyOp(ops.pop(), values.pop(), values.pop())); ops.pop(); } // Current token is an operator. else if (tokens[i] == '+' || tokens[i] == '-' || tokens[i] == '*' || tokens[i] == '/') { // While top of 'ops' has same // or greater precedence to current // token, which is an operator. // Apply operator on top of 'ops' // to top two elements in values stack while (!ops.empty() && hasPrecedence(tokens[i], ops.peek())) values.push(applyOp(ops.pop(), values.pop(), values.pop())); // Push current token to 'ops'. ops.push(tokens[i]); } } // Entire expression has been // parsed at this point, apply remaining // ops to remaining values while (!ops.empty()) values.push(applyOp(ops.pop(), values.pop(), values.pop())); // Top of 'values' contains // result, return it return values.pop(); } // Returns true if 'op2' has higher // or same precedence as 'op1', // otherwise returns false. public static boolean hasPrecedence( char op1, char op2) { if (op2 == '(' || op2 == ')') return false; if ((op1 == '*' || op1 == '/') && (op2 == '+' || op2 == '-')) return false; else return true; } // A utility method to apply an // operator 'op' on operands 'a' // and 'b'. Return the result. public static int applyOp(char op, int b, int a) { switch (op) { case '+': return a + b; case '-': return a - b; case '*': return a * b; case '/': if (b == 0) throw new UnsupportedOperationException( "Cannot divide by zero"); return a / b; } return 0; } // Driver method to test above methods public static void main(String[] args) { System.out.println(EvaluateString. evaluate("10 + 2 * 6")); System.out.println(EvaluateString. evaluate("100 * 2 + 12")); System.out.println(EvaluateString. evaluate("100 * ( 2 + 12 )")); System.out.println(EvaluateString. evaluate("100 * ( 2 + 12 ) / 14")); } } |
Python3
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# Python3 program to evaluate a given # expression where tokens are # separated by space. # Function to find precedence # of operators. def precedence(op):
if op == '+' or op == '-': return 1 if op == '*' or op == '/': return 2 return 0 # Function to perform arithmetic # operations. def applyOp(a, b, op):
if op == '+': return a + b if op == '-': return a - b if op == '*': return a * b if op == '/': return a // b # Function that returns value of # expression after evaluation. def evaluate(tokens):
# stack to store integer values. values = []
# stack to store operators. ops = [] i = 0
while i < len(tokens):
# Current token is a whitespace, # skip it. if tokens[i] == ' ': i += 1 continue
# Current token is an opening # brace, push it to 'ops' elif tokens[i] == '(': ops.append(tokens[i])
# Current token is a number, push # it to stack for numbers. elif tokens[i].isdigit(): val = 0
# There may be more than one # digits in the number. while (i < len(tokens) and tokens[i].isdigit()):
val = (val * 10) + int(tokens[i]) i += 1
values.append(val)
# right now the i points to # the character next to the digit, # since the for loop also increases # the i, we would skip one # token position; we need to # decrease the value of i by 1 to # correct the offset. i-=1
# Closing brace encountered, # solve entire brace. elif tokens[i] == ')':
while len(ops) != 0 and ops[-1] != '(':
val2 = values.pop() val1 = values.pop() op = ops.pop()
values.append(applyOp(val1, val2, op))
# pop opening brace. ops.pop()
# Current token is an operator. else:
# While top of 'ops' has same or # greater precedence to current # token, which is an operator. # Apply operator on top of 'ops' # to top two elements in values stack. while (len(ops) != 0 and precedence(ops[-1]) >= precedence(tokens[i])):
val2 = values.pop() val1 = values.pop() op = ops.pop()
values.append(applyOp(val1, val2, op))
# Push current token to 'ops'. ops.append(tokens[i])
i += 1
# Entire expression has been parsed # at this point, apply remaining ops # to remaining values. while len(ops) != 0:
val2 = values.pop() val1 = values.pop() op = ops.pop()
values.append(applyOp(val1, val2, op))
# Top of 'values' contains result, # return it. return values[-1] # Driver Code if __name__ == "__main__":
print(evaluate("10 + 2 * 6")) print(evaluate("100 * 2 + 12")) print(evaluate("100 * ( 2 + 12 )")) print(evaluate("100 * ( 2 + 12 ) / 14")) # This code is contributed # by Rituraj Jain |
C#
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/* A C# program to evaluate a given expression where tokens are separated by space. */ using System; using System.Collections.Generic; using System.Text; public class EvaluateString { public static int evaluate(string expression) { char[] tokens = expression.ToCharArray(); // Stack for numbers: 'values' Stack<int> values = new Stack<int>(); // Stack for Operators: 'ops' Stack<char> ops = new Stack<char>(); for (int i = 0; i < tokens.Length; i++) { // Current token is a whitespace, skip it if (tokens[i] == ' ') { continue; } // Current token is a number, // push it to stack for numbers if (tokens[i] >= '0' && tokens[i] <= '9') { StringBuilder sbuf = new StringBuilder();
// There may be more than // one digits in number while (i < tokens.Length && tokens[i] >= '0' && tokens[i] <= '9') { sbuf.Append(tokens[i++]); } values.Push(int.Parse(sbuf.ToString()));
// Right now the i points to // the character next to the digit, // since the for loop also increases // the i, we would skip one // token position; we need to // decrease the value of i by 1 to // correct the offset. i--; } // Current token is an opening // brace, push it to 'ops' else if (tokens[i] == '(') { ops.Push(tokens[i]); } // Closing brace encountered, // solve entire brace else if (tokens[i] == ')') { while (ops.Peek() != '(') { values.Push(applyOp(ops.Pop(), values.Pop(), values.Pop())); } ops.Pop(); } // Current token is an operator. else if (tokens[i] == '+' || tokens[i] == '-' || tokens[i] == '*' || tokens[i] == '/') {
// While top of 'ops' has same // or greater precedence to current // token, which is an operator. // Apply operator on top of 'ops' // to top two elements in values stack while (ops.Count > 0 && hasPrecedence(tokens[i], ops.Peek())) { values.Push(applyOp(ops.Pop(), values.Pop(), values.Pop())); } // Push current token to 'ops'. ops.Push(tokens[i]); } } // Entire expression has been // parsed at this point, apply remaining // ops to remaining values while (ops.Count > 0) { values.Push(applyOp(ops.Pop(), values.Pop(), values.Pop())); } // Top of 'values' contains // result, return it return values.Pop(); } // Returns true if 'op2' has // higher or same precedence as 'op1', // otherwise returns false. public static bool hasPrecedence(char op1, char op2) { if (op2 == '(' || op2 == ')') { return false; } if ((op1 == '*' || op1 == '/') && (op2 == '+' || op2 == '-')) { return false; } else { return true; } } // A utility method to apply an // operator 'op' on operands 'a' // and 'b'. Return the result. public static int applyOp(char op, int b, int a) { switch (op) { case '+': return a + b; case '-': return a - b; case '*': return a * b; case '/': if (b == 0) { throw new System.NotSupportedException( "Cannot divide by zero"); } return a / b; } return 0; } // Driver method to test above methods public static void Main(string[] args) { Console.WriteLine(EvaluateString. evaluate("10 + 2 * 6")); Console.WriteLine(EvaluateString. evaluate("100 * 2 + 12")); Console.WriteLine(EvaluateString. evaluate("100 * ( 2 + 12 )")); Console.WriteLine(EvaluateString. evaluate("100 * ( 2 + 12 ) / 14")); } } // This code is contributed by Shrikant13 |
Javascript
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<script> /* A Javascript program to evaluate a given expression where tokens are separated by space. */
function evaluate(expression) { let tokens = expression.split('');
// Stack for numbers: 'values' let values = [];
// Stack for Operators: 'ops' let ops = [];
for (let i = 0; i < tokens.length; i++) { // Current token is a whitespace, skip it if (tokens[i] == ' ') { continue; }
// Current token is a number, // push it to stack for numbers if (tokens[i] >= '0' && tokens[i] <= '9') { let sbuf = "";
// There may be more than // one digits in number while (i < tokens.length && tokens[i] >= '0' && tokens[i] <= '9') { sbuf = sbuf + tokens[i++]; } values.push(parseInt(sbuf, 10));
// Right now the i points to // the character next to the digit, // since the for loop also increases // the i, we would skip one // token position; we need to // decrease the value of i by 1 to // correct the offset. i--; }
// Current token is an opening // brace, push it to 'ops' else if (tokens[i] == '(') { ops.push(tokens[i]); }
// Closing brace encountered, // solve entire brace else if (tokens[i] == ')') { while (ops[ops.length - 1] != '(') { values.push(applyOp(ops.pop(), values.pop(), values.pop())); } ops.pop(); }
// Current token is an operator. else if (tokens[i] == '+' || tokens[i] == '-' || tokens[i] == '*' || tokens[i] == '/') {
// While top of 'ops' has same // or greater precedence to current // token, which is an operator. // Apply operator on top of 'ops' // to top two elements in values stack while (ops.length > 0 && hasPrecedence(tokens[i], ops[ops.length - 1])) { values.push(applyOp(ops.pop(), values.pop(), values.pop())); }
// Push current token to 'ops'. ops.push(tokens[i]); } }
// Entire expression has been // parsed at this point, apply remaining // ops to remaining values while (ops.length > 0) { values.push(applyOp(ops.pop(), values.pop(), values.pop())); }
// Top of 'values' contains // result, return it return values.pop(); }
// Returns true if 'op2' has // higher or same precedence as 'op1', // otherwise returns false. function hasPrecedence(op1, op2) { if (op2 == '(' || op2 == ')') { return false; } if ((op1 == '*' || op1 == '/') && (op2 == '+' || op2 == '-')) { return false; } else { return true; } }
// A utility method to apply an // operator 'op' on operands 'a' // and 'b'. Return the result. function applyOp(op, b, a) { switch (op) { case '+': return a + b; case '-': return a - b; case '*': return a * b; case '/': if (b == 0) { document.write("Cannot divide by zero"); } return parseInt(a / b, 10); } return 0; }
document.write(evaluate("10 + 2 * 6") + "</br>"); document.write(evaluate("100 * 2 + 12") + "</br>"); document.write(evaluate("100 * ( 2 + 12 )") + "</br>"); document.write(evaluate("100 * ( 2 + 12 ) / 14") + "</br>"); // This code is contributed by decode2207. </script> |
Output:
22 212 1400 100Time Complexity: O(n)
Space Complexity: O(n)
See this for a sample run with more test cases.
This article is compiled by Ciphe. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
Article Tags :
Mathematical
Stack
Oracle
Practice Tags :
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Mathematical
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Evaluation of Postfix Expressions Using Stack [with C program]
Learn: How to evaluate postfix expression using stack in C language program? This article explains the basic idea, algorithm (with systematic diagram and table) and program to evaluate postfix expression using stack.
Submitted by Abhishek Jain, on June 19, 2017
As discussed in Infix To Postfix Conversion Using Stack, the compiler finds it convenient to evaluate an expression in its postfix form. The virtues of postfix form include elimination of parentheses which signify priority of evaluation and the elimination of the need to observe rules of hierarchy, precedence and associativity during evaluation of the expression.
As Postfix expression is without parenthesis and can be evaluated as two operands and an operator at a time, this becomes easier for the compiler and the computer to handle.
Evaluation rule of a Postfix Expression states:
- While reading the expression from left to right, push the element in the stack if it is an operand.
- Pop the two operands from the stack, if the element is an operator and then evaluate it.
- Push back the result of the evaluation. Repeat it till the end of the expression.
Algorithm
1) Add ) to postfix expression.
2) Read postfix expression Left to Right until ) encountered
3) If operand is encountered, push it onto Stack
[End If]
4) If operator is encountered, Pop two elements
i) A -> Top element
ii) B-> Next to Top element
iii) Evaluate B operator A
push B operator A onto Stack
5) Set result = pop
6) END
Let's see an example to better understand the algorithm:
Expression: 456*+
Result: 34
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Introduction to Arithmetic expressions
Arithmetic expressions can be represented in 3 forms:
- Infix notation
- Postfix notation (Reverse Polish Notation)
- Prefix notation (Polish Notation)
Infix Notation is of the form operand1 operator operator2.
Eg: 5 + 3
Postfix Notation can be represented as operand1 operand2 operator.
Eg: 5 3 +
Prefix notation can be represented as operator operand1 operand2.
Eg: + 5 3
We use the infix notation most frequently in our day to day tasks. However, machines find infix notations tougher to process than prefic/postfix notations. Hence, compilers convert infix notations to prefix/postfix before the expression is evaluated.
The precedence of operators needs to be taken case of:
Exponentiation(^) > Multiplication( * ) or Division(/) > Addition(+) or Subtraction(-)Brackets have the highest priority and their presence can override the precedence order.
To evaluate an infix expression, We need to perform 2 main tasks:
- Convert infix to postfix
- Evaluate postfix
Let's discuss both the steps one by one.
For step 1, Refer this article on converting infix to postfix expression using Stack.
Once the expression is converted to postfix notation, step 2 can be performed: