LeetCode #150 — MEDIUM

Evaluate Reverse Polish Notation

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode

The Problem

Problem Statement

You are given an array of strings tokens that represents an arithmetic expression in a Reverse Polish Notation.

Evaluate the expression. Return an integer that represents the value of the expression.

Note that:

The valid operators are '+', '-', '*', and '/'.
Each operand may be an integer or another expression.
The division between two integers always truncates toward zero.
There will not be any division by zero.
The input represents a valid arithmetic expression in a reverse polish notation.
The answer and all the intermediate calculations can be represented in a 32-bit integer.

Example 1:

Input: tokens = ["2","1","+","3","*"]
Output: 9
Explanation: ((2 + 1) * 3) = 9

Example 2:

Input: tokens = ["4","13","5","/","+"]
Output: 6
Explanation: (4 + (13 / 5)) = 6

Example 3:

Input: tokens = ["10","6","9","3","+","-11","*","/","*","17","+","5","+"]
Output: 22
Explanation: ((10 * (6 / ((9 + 3) * -11))) + 17) + 5
= ((10 * (6 / (12 * -11))) + 17) + 5
= ((10 * (6 / -132)) + 17) + 5
= ((10 * 0) + 17) + 5
= (0 + 17) + 5
= 17 + 5
= 22

Constraints:

1 <= tokens.length <= 10⁴
tokens[i] is either an operator: "+", "-", "*", or "/", or an integer in the range [-200, 200].

Step 01

Brute Force Baseline

Problem summary: You are given an array of strings tokens that represents an arithmetic expression in a Reverse Polish Notation. Evaluate the expression. Return an integer that represents the value of the expression. Note that: The valid operators are '+', '-', '*', and '/'. Each operand may be an integer or another expression. The division between two integers always truncates toward zero. There will not be any division by zero. The input represents a valid arithmetic expression in a reverse polish notation. The answer and all the intermediate calculations can be represented in a 32-bit integer.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Math · Stack

Example 1

["2","1","+","3","*"]

Example 2

["4","13","5","/","+"]

Example 3

["10","6","9","3","+","-11","*","/","*","17","+","5","+"]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #150: Evaluate Reverse Polish Notation
class Solution {
    public int evalRPN(String[] tokens) {
        Deque<Integer> stk = new ArrayDeque<>();
        for (String t : tokens) {
            if (t.length() > 1 || Character.isDigit(t.charAt(0))) {
                stk.push(Integer.parseInt(t));
            } else {
                int y = stk.pop();
                int x = stk.pop();
                switch (t) {
                case "+":
                    stk.push(x + y);
                    break;
                case "-":
                    stk.push(x - y);
                    break;
                case "*":
                    stk.push(x * y);
                    break;
                default:
                    stk.push(x / y);
                    break;
                }
            }
        }
        return stk.pop();
    }
}

// Accepted solution for LeetCode #150: Evaluate Reverse Polish Notation
func evalRPN(tokens []string) int {
	// https://github.com/emirpasic/gods#arraystack
	stk := arraystack.New()
	for _, token := range tokens {
		if len(token) > 1 || token[0] >= '0' && token[0] <= '9' {
			num, _ := strconv.Atoi(token)
			stk.Push(num)
		} else {
			y := popInt(stk)
			x := popInt(stk)
			switch token {
			case "+":
				stk.Push(x + y)
			case "-":
				stk.Push(x - y)
			case "*":
				stk.Push(x * y)
			default:
				stk.Push(x / y)
			}
		}
	}
	return popInt(stk)
}

func popInt(stack *arraystack.Stack) int {
	v, _ := stack.Pop()
	return v.(int)
}

# Accepted solution for LeetCode #150: Evaluate Reverse Polish Notation
import operator


class Solution:
    def evalRPN(self, tokens: List[str]) -> int:
        opt = {
            "+": operator.add,
            "-": operator.sub,
            "*": operator.mul,
            "/": operator.truediv,
        }
        s = []
        for token in tokens:
            if token in opt:
                s.append(int(opt[token](s.pop(-2), s.pop(-1))))
            else:
                s.append(int(token))
        return s[0]

// Accepted solution for LeetCode #150: Evaluate Reverse Polish Notation
impl Solution {
    pub fn eval_rpn(tokens: Vec<String>) -> i32 {
        let mut stack = vec![];
        for token in tokens {
            match token.parse() {
                Ok(num) => stack.push(num),
                Err(_) => {
                    let a = stack.pop().unwrap();
                    let b = stack.pop().unwrap();
                    stack.push(match token.as_str() {
                        "+" => b + a,
                        "-" => b - a,
                        "*" => b * a,
                        "/" => b / a,
                        _ => 0,
                    });
                }
            }
        }
        stack[0]
    }
}

// Accepted solution for LeetCode #150: Evaluate Reverse Polish Notation
function evalRPN(tokens: string[]): number {
    const stack = [];
    for (const token of tokens) {
        if (/\d/.test(token)) {
            stack.push(Number(token));
        } else {
            const a = stack.pop();
            const b = stack.pop();
            switch (token) {
                case '+':
                    stack.push(b + a);
                    break;
                case '-':
                    stack.push(b - a);
                    break;
                case '*':
                    stack.push(b * a);
                    break;
                case '/':
                    stack.push(~~(b / a));
                    break;
            }
        }
    }
    return stack[0];
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

For each element, scan left (or right) to find the next greater/smaller element. The inner scan can visit up to n elements per outer iteration, giving O(n²) total comparisons. No extra space needed beyond loop variables.

MONOTONIC STACK

O(n) time

O(n) space

Each element is pushed onto the stack at most once and popped at most once, giving 2n total operations = O(n). The stack itself holds at most n elements in the worst case. The key insight: amortized O(1) per element despite the inner while-loop.

Shortcut: Each element pushed once + popped once → O(n) amortized. The inner while-loop does not make it O(n²).

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

Breaking monotonic invariant

Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.

Usually fails on: Indices point to blocked elements and outputs shift.

Fix: Pop while invariant is violated before pushing current element.