Evaluate postfix expression

The given problem is to evaluate a postfix expression, which is a mathematical expression in which each operator appears after its operands. The postfix expression is also known as Reverse Polish Notation (RPN). The goal is to compute the final result of the expression.

Problem Statement

We are given a postfix expression as a string, and we need to evaluate it and print the final result.

Explanation with Suitable Example

Let's take a simple example to understand the process of evaluating a postfix expression: "34+2*".

1. Scan the expression from left to right:

   • Read '3': It is an operand, so push it onto the stack.
   • Read '4': It is another operand, so push it onto the stack.
   • Read '+': It is an operator. Pop the top two elements from the stack ('4' and '3'), perform the addition ('4+3'), and push the result ('7') back to the stack.

2. Read '2': It is an operand, so push it onto the stack.

3. Read '': It is an operator. Pop the top two elements from the stack ('2' and '7'), perform the multiplication ('27'), and push the result ('14') back to the stack.

4. The expression has been fully scanned. The final result is on the top of the stack, which is '14'.

Pseudocode

function evaluatePostfix(expression)
    create an empty stack

    for each character ch in expression
        if ch is an operand
            push ch to the stack
        else
            pop operand2 from the stack
            pop operand1 from the stack
            perform the operation (operand1  operand2)
            push the result back to the stack

    return the top element of the stack as the final result

Algorithm Explanation

1. Create an empty stack to store operands.
2. For each character in the expression, do the following:
   • If the character is an operand, push it onto the stack.
   • If the character is an operator, pop the top two elements from the stack (operand2 and operand1), perform the operation (operand1 <operator> operand2), and push the result back to the stack.
3. After processing the entire expression, the final result will be the top element of the stack.

Code Solution

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Time Complexity

The time complexity of the given algorithm is O(n), where n is the length of the input expression. This is because we scan the expression once, and for each character, the stack operations take constant time.

Resultant Output Explanation

For the given example expressions:

1. "12+35+*":

   • The postfix evaluation is: (1+2)*(3+5) = 3 * 8 = 24
   • The output will be: "12+35+* = 24"

2. "72*62/+4+-":

   • The postfix evaluation is: -((7*2) + (6/2) + 4) = -(14 + 3 + 4) = -21
   • The output will be: "72*62/+4+- = -21"