Problem Statement
Convert a simple infix expression (e.g., A+B) to postfix notation (e.g., AB+). In postfix notation, operators follow their operands, which allows for simpler evaluation without needing parentheses.
Example Usage
Additional Examples:
- Input:
"A+B*C"
- Input:
"A*B+C"
- Input:
"A*(B+C)"
JavaScript Stack - Using Custom Stack Class
We’ll use the following custom Stack class to handle operators during the conversion. This class includes methods like push, pop, peek, isEmpty, size, and clear.
Approach
- Initialize a Stack: We’ll use a stack to keep track of operators.
- Define Operator Precedence: Higher precedence operators (
*, /) should be processed before lower precedence ones (+, -).
- Traverse Each Character:
- If the character is an operand, add it directly to the output.
- If the character is an operator, pop operators from the stack with higher or equal precedence and add them to the output before pushing the current operator.
- Handle parentheses by pushing
( and popping until ).
- Pop Remaining Operators: Once all characters are processed, pop any operators left in the stack.
Steps
- Define an output string for the postfix expression.
- Loop through each character of the infix expression.
- Use
push to add operators to the stack and pop to manage precedence.
- Add remaining operators in the stack to the output.
Time & Space Complexity
- Time Complexity:
- Each character is processed once, so the complexity is O(n), where
n is the length of the expression.
- Space Complexity: O(n), as we may need to store up to
n operators in the stack.
Code Snippet
Dry Run
For the input "A+B*C":
- Process
A: Operand, add to postfix → postfix = "A"
- Process
+: Operator, push to stack → stack = ["+"]
- Process
B: Operand, add to postfix → postfix = "AB"
- Process
*: Higher precedence than +, push to stack → stack = ["+", "*"]
- Process
C: Operand, add to postfix → postfix = "ABC"
- Pop remaining operators: Pop
* and then + → postfix = "ABC*+"
Complexity Analysis
Custom Defined Stack Methods
- Time Complexity: Each character is processed once, so the solution is O(n) for
n characters in the expression.
- Space Complexity: O(n) for storing operators in the stack.
Conclusion
Using a stack to convert infix expressions to postfix notation simplifies the handling of operator precedence and parentheses. This method provides an efficient solution, especially for evaluating mathematical expressions.