Evaluate a space-separated postfix (Reverse Polish) expression with a stack in O(n), handling + - * / and operand order. Run tests in Python.
The problem
expression is a space-separated postfix expression of integers and the operators + - * /. Return its integer value.
- Operands come before their operator:
"2 3 + 4 *"means(2 + 3) * 4=20. /truncates toward zero (so-7 / 2is-3).
Examples
Example 1
Input
eval_postfix('2 3 + 4 *')Expected output
20
Example 2
Input
eval_postfix('5 1 2 + 4 * + 3 -')Expected output
14
+ 4 hidden tests on Submit — operand order, truncate toward zero.
Edge cases to ask about
- Single number
- Subtraction order
- Negative division
Hints
0/3How an interviewer scores this
0/9Your code runs in real CPython inside your browser — nothing is sent anywhere. The first run downloads the interpreter (about 6 MB, once). Your code is saved on this device as you type.
Complexity Lab
What does this cost as n grows?
Interviewers score the analysis as much as the code. Commit to an answer first — then check it, and measure your code against the optimal one at growing input sizes.
Pick both to reveal the answer.
Measure it
Runs the function on inputs of size 250 up to 16,000 and records the time and peak memory. Slow solutions stop early — a short curve is itself the answer.
From brute force to optimal
The progression an interviewer wants to hear, one step at a time.
| Approach | Time | Space | Idea |
|---|---|---|---|
| Stack of operands | O(n) | O(n) | Operators pop two operands — the SECOND pop is the left operand. |
Walkthrough of the optimal approach (try it yourself first)
Push numbers. On an operator, pop b first, then a, and push a op b — getting this order wrong breaks - and /. The final value is the only item left on the stack.
Python's // floors (-7 // 2 == -4), so use int(a / b) when truncation is required.
Complexity: O(n) time, O(n) space. Each token is processed once with O(1) stack work; the stack can hold up to n/2 operands.
Reveal the reference solution
def eval_postfix(expression): stack = [] for token in expression.split(): if token in ("+", "-", "*", "/"): b = stack.pop() a = stack.pop() if token == "+": stack.append(a + b) elif token == "-": stack.append(a - b) elif token == "*": stack.append(a * b) else: stack.append(int(a / b)) else: stack.append(int(token)) return stack[-1]
Follow-ups interviewers ask
- Convert infix to postfix (shunting-yard).
- Validate malformed input.
Frequently asked interview questions
Core interview concepts, complexities, and follow-ups scored by hiring teams.
What is the time complexity of Evaluate a Postfix (RPN) Expression in Python?
The optimal solution runs in O(n) time and O(n) auxiliary space. Each token is processed once with O(1) stack work; the stack can hold up to n/2 operands.
What follow-up questions do interviewers ask about Evaluate a Postfix (RPN) Expression?
Convert infix to postfix (shunting-yard). Validate malformed input.
