What is the optimal time complexity of solving Infix Prefix Postfix?

The optimized solution achieves a time complexity of O(n) by using sorting, mathematical shortcuts, or hash table lookups.

What is the auxiliary space complexity for Infix Prefix Postfix?

The space complexity is O(1) since we only use a minimal/constant amount of extra memory for tracking variables (or auxiliary space if dynamic structures are allocated).

What are the typical edge cases we must handle in Infix Prefix Postfix?

Key edge cases include empty collections, negative or large bounds causing integer overflow, arrays with only one element, or inputs where target criteria are not met.

How can we optimize the brute force approach for Infix Prefix Postfix?

Brute force methods often inspect all possible combinations. We optimize this by sorting the input list to enable binary search, or tracking processed items in a set to avoid redundant nested loop lookups.

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DSA SectionEasy

Infix Prefix Postfix

Learn how to solve the 'Infix Prefix Postfix' problem. This detailed resource details brute force and optimized approaches.

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Problem Statement

Easy

Write a function infix_to_postfix(expression) that takes a string expression representing an infix mathematical expression and returns the equivalent postfix expression. You may assume operators are +, -, *, /, ^ and parentheses are (, ).

Constraints

•1 <= len(expression) <= 100
•Expression contains only valid variables, operators, and parentheses.

Examples

Example 1

Input

expression = "A*(B+C)/D"

Output

"ABC+*D/"

Explanation

Infix 'A*(B+C)/D' converted to postfix is 'ABC+*D/'.

Need a Hint?

Analyze the input constraints. Try sorting first (O(n log n)) or using a hash map/set to track seen elements in O(n) time.

Edge Cases to Watch

Empty list or null input variables
Single item lists/arrays
Extremely large input bounds causing integer or stack overflow

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