LeetCode #227 — MEDIUM

Basic Calculator II

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

The Problem

Problem Statement

Given a string s which represents an expression, evaluate this expression and return its value.

The integer division should truncate toward zero.

You may assume that the given expression is always valid. All intermediate results will be in the range of [-2³¹, 2³¹ - 1].

Note: You are not allowed to use any built-in function which evaluates strings as mathematical expressions, such as eval().

Example 1:

Input: s = "3+2*2"
Output: 7

Example 2:

Input: s = " 3/2 "
Output: 1

Example 3:

Input: s = " 3+5 / 2 "
Output: 5

Constraints:

1 <= s.length <= 3 * 10⁵
s consists of integers and operators ('+', '-', '*', '/') separated by some number of spaces.
s represents a valid expression.
All the integers in the expression are non-negative integers in the range [0, 2³¹ - 1].
The answer is guaranteed to fit in a 32-bit integer.

Step 01

Brute Force Baseline

Problem summary: Given a string s which represents an expression, evaluate this expression and return its value. The integer division should truncate toward zero. You may assume that the given expression is always valid. All intermediate results will be in the range of [-231, 231 - 1]. Note: You are not allowed to use any built-in function which evaluates strings as mathematical expressions, such as eval().

Baseline thinking

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

Pattern signal: Math · Stack

Example 1

"3+2*2"

Example 2

" 3/2 "

Example 3

" 3+5 / 2 "

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 #227: Basic Calculator II
class Solution {
    public int calculate(String s) {
        Deque<Integer> stk = new ArrayDeque<>();
        char sign = '+';
        int v = 0;
        for (int i = 0; i < s.length(); ++i) {
            char c = s.charAt(i);
            if (Character.isDigit(c)) {
                v = v * 10 + (c - '0');
            }
            if (i == s.length() - 1 || c == '+' || c == '-' || c == '*' || c == '/') {
                if (sign == '+') {
                    stk.push(v);
                } else if (sign == '-') {
                    stk.push(-v);
                } else if (sign == '*') {
                    stk.push(stk.pop() * v);
                } else {
                    stk.push(stk.pop() / v);
                }
                sign = c;
                v = 0;
            }
        }
        int ans = 0;
        while (!stk.isEmpty()) {
            ans += stk.pop();
        }
        return ans;
    }
}

// Accepted solution for LeetCode #227: Basic Calculator II
func calculate(s string) int {
	sign := '+'
	stk := []int{}
	v := 0
	for i, c := range s {
		digit := '0' <= c && c <= '9'
		if digit {
			v = v*10 + int(c-'0')
		}
		if i == len(s)-1 || !digit && c != ' ' {
			switch sign {
			case '+':
				stk = append(stk, v)
			case '-':
				stk = append(stk, -v)
			case '*':
				stk[len(stk)-1] *= v
			case '/':
				stk[len(stk)-1] /= v
			}
			sign = c
			v = 0
		}
	}
	ans := 0
	for _, v := range stk {
		ans += v
	}
	return ans
}

# Accepted solution for LeetCode #227: Basic Calculator II
class Solution:
    def calculate(self, s: str) -> int:
        v, n = 0, len(s)
        sign = '+'
        stk = []
        for i, c in enumerate(s):
            if c.isdigit():
                v = v * 10 + int(c)
            if i == n - 1 or c in '+-*/':
                match sign:
                    case '+':
                        stk.append(v)
                    case '-':
                        stk.append(-v)
                    case '*':
                        stk.append(stk.pop() * v)
                    case '/':
                        stk.append(int(stk.pop() / v))
                sign = c
                v = 0
        return sum(stk)

// Accepted solution for LeetCode #227: Basic Calculator II
struct Solution;

use std::iter::Peekable;
use std::slice::Iter;

#[derive(Debug, PartialEq, Eq, Clone, Copy)]
enum Tok {
    Num(i32),
    Op(char),
}
use Tok::*;

impl Tok {
    fn is_expr_op(self) -> bool {
        matches!(self, Op('+') | Op('-'))
    }

    fn is_factor_op(self) -> bool {
        matches!(self, Op('*') | Op('/'))
    }

    fn val(self) -> Option<i32> {
        match self {
            Num(x) => Some(x),
            _ => None,
        }
    }
    fn eval(self, lhs: i32, rhs: i32) -> Option<i32> {
        match self {
            Op('+') => Some(lhs + rhs),
            Op('-') => Some(lhs - rhs),
            Op('*') => Some(lhs * rhs),
            Op('/') => {
                if rhs != 0 {
                    Some(lhs / rhs)
                } else {
                    None
                }
            }
            _ => None,
        }
    }
}

impl Solution {
    fn calculate(s: String) -> i32 {
        let tokens = Self::tokens(&s);
        let mut it = tokens.iter().peekable();
        if let Some(val) = Self::parse_expr(&mut it) {
            val
        } else {
            0
        }
    }

    fn parse_expr(it: &mut Peekable<Iter<Tok>>) -> Option<i32> {
        let mut lhs = Self::parse_factor(it)?;
        while let Some(&tok) = it.peek() {
            if tok.is_expr_op() {
                let op = *it.next().unwrap();
                let rhs = Self::parse_factor(it)?;
                lhs = op.eval(lhs, rhs)?;
            } else {
                break;
            }
        }
        Some(lhs)
    }

    fn parse_factor(it: &mut Peekable<Iter<Tok>>) -> Option<i32> {
        let mut lhs = Self::parse_num(it)?;
        while let Some(&tok) = it.peek() {
            if tok.is_factor_op() {
                let op = *it.next().unwrap();
                let rhs = Self::parse_num(it)?;
                lhs = op.eval(lhs, rhs)?;
            } else {
                break;
            }
        }
        Some(lhs)
    }

    fn parse_num(it: &mut Peekable<Iter<Tok>>) -> Option<i32> {
        match it.next() {
            Some(Tok::Num(x)) => Some(*x),
            _ => None,
        }
    }

    fn tokens(s: &str) -> Vec<Tok> {
        let mut v: Vec<Tok> = vec![];
        let mut it = s.chars().peekable();
        while let Some(c) = it.next() {
            match c {
                '0'..='9' => {
                    let mut x: i32 = (c as u8 - b'0') as i32;
                    while let Some(c) = it.peek() {
                        if c.is_numeric() {
                            x *= 10;
                            x += (it.next().unwrap() as u8 - b'0') as i32;
                        } else {
                            break;
                        }
                    }
                    v.push(Tok::Num(x));
                }
                '+' | '-' | '*' | '/' => {
                    v.push(Tok::Op(c));
                }
                _ => {}
            }
        }
        v
    }
}

#[test]
fn test() {
    let s = "3+2*2".to_string();
    assert_eq!(Solution::calculate(s), 7);
    let s = "0-0".to_string();
    assert_eq!(Solution::calculate(s), 0);
    let s = "0-2147483647".to_string();
    assert_eq!(Solution::calculate(s), -2_147_483_647);
}

// Accepted solution for LeetCode #227: Basic Calculator II
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #227: Basic Calculator II
// class Solution {
//     public int calculate(String s) {
//         Deque<Integer> stk = new ArrayDeque<>();
//         char sign = '+';
//         int v = 0;
//         for (int i = 0; i < s.length(); ++i) {
//             char c = s.charAt(i);
//             if (Character.isDigit(c)) {
//                 v = v * 10 + (c - '0');
//             }
//             if (i == s.length() - 1 || c == '+' || c == '-' || c == '*' || c == '/') {
//                 if (sign == '+') {
//                     stk.push(v);
//                 } else if (sign == '-') {
//                     stk.push(-v);
//                 } else if (sign == '*') {
//                     stk.push(stk.pop() * v);
//                 } else {
//                     stk.push(stk.pop() / v);
//                 }
//                 sign = c;
//                 v = 0;
//             }
//         }
//         int ans = 0;
//         while (!stk.isEmpty()) {
//             ans += stk.pop();
//         }
//         return ans;
//     }
// }

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.

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.