LeetCode #224 — HARD

Basic Calculator

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

Given a string s representing a valid expression, implement a basic calculator to evaluate it, and return the result of the evaluation.

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

Example 1:

Input: s = "1 + 1"
Output: 2

Example 2:

Input: s = " 2-1 + 2 "
Output: 3

Example 3:

Input: s = "(1+(4+5+2)-3)+(6+8)"
Output: 23

Constraints:

1 <= s.length <= 3 * 10⁵
s consists of digits, '+', '-', '(', ')', and ' '.
s represents a valid expression.
'+' is not used as a unary operation (i.e., "+1" and "+(2 + 3)" is invalid).
'-' could be used as a unary operation (i.e., "-1" and "-(2 + 3)" is valid).
There will be no two consecutive operators in the input.
Every number and running calculation will fit in a signed 32-bit integer.

Step 01

Brute Force Baseline

Problem summary: Given a string s representing a valid expression, implement a basic calculator to evaluate it, and return the result of the evaluation. 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

"1 + 1"

Example 2

" 2-1 + 2 "

Example 3

"(1+(4+5+2)-3)+(6+8)"

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

Largest constraint values

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

// Accepted solution for LeetCode #224: Basic Calculator
func calculate(s string) (ans int) {
	stk := []int{}
	sign := 1
	n := len(s)
	for i := 0; i < n; i++ {
		switch s[i] {
		case ' ':
		case '+':
			sign = 1
		case '-':
			sign = -1
		case '(':
			stk = append(stk, ans)
			stk = append(stk, sign)
			ans, sign = 0, 1
		case ')':
			ans *= stk[len(stk)-1]
			stk = stk[:len(stk)-1]
			ans += stk[len(stk)-1]
			stk = stk[:len(stk)-1]
		default:
			x := 0
			j := i
			for ; j < n && '0' <= s[j] && s[j] <= '9'; j++ {
				x = x*10 + int(s[j]-'0')
			}
			ans += sign * x
			i = j - 1
		}
	}
	return
}

# Accepted solution for LeetCode #224: Basic Calculator
class Solution:
    def calculate(self, s: str) -> int:
        stk = []
        ans, sign = 0, 1
        i, n = 0, len(s)
        while i < n:
            if s[i].isdigit():
                x = 0
                j = i
                while j < n and s[j].isdigit():
                    x = x * 10 + int(s[j])
                    j += 1
                ans += sign * x
                i = j - 1
            elif s[i] == "+":
                sign = 1
            elif s[i] == "-":
                sign = -1
            elif s[i] == "(":
                stk.append(ans)
                stk.append(sign)
                ans, sign = 0, 1
            elif s[i] == ")":
                ans = stk.pop() * ans + stk.pop()
            i += 1
        return ans

// Accepted solution for LeetCode #224: Basic Calculator
struct Solution;

use std::iter::Peekable;
use std::str::Chars;
use std::vec::IntoIter;

#[derive(Debug)]
enum Tok {
    Num(i32),
    Op(char),
}

impl Solution {
    fn calculate(s: String) -> i32 {
        let mut it = s.chars().peekable();
        let tokens = Self::parse_tokens(&mut it);
        let mut it = tokens.into_iter().peekable();
        Self::parse(&mut it)
    }

    fn parse_term(it: &mut Peekable<IntoIter<Tok>>) -> i32 {
        if Self::eat(it, '(') {
            let res = Self::parse(it);
            Self::eat(it, ')');
            res
        } else {
            if let Some(Tok::Num(x)) = it.next() {
                x
            } else {
                panic!()
            }
        }
    }

    fn parse(it: &mut Peekable<IntoIter<Tok>>) -> i32 {
        let mut left = Self::parse_term(it);
        loop {
            if Self::eat(it, '+') {
                left += Self::parse_term(it);
                continue;
            }
            if Self::eat(it, '-') {
                left -= Self::parse_term(it);
                continue;
            }
            break;
        }
        left
    }

    fn eat(it: &mut Peekable<IntoIter<Tok>>, c: char) -> bool {
        if let Some(&Tok::Op(first)) = it.peek() {
            if first == c {
                it.next();
                return true;
            }
        }
        false
    }

    fn parse_tokens(it: &mut Peekable<Chars>) -> Vec<Tok> {
        let mut res = vec![];
        while let Some(c) = it.next() {
            match c {
                '0'..='9' => {
                    let mut x = (c as u8 - b'0') as i32;
                    while let Some('0'..='9') = it.peek() {
                        x *= 10;
                        x += (it.next().unwrap() as u8 - b'0') as i32;
                    }
                    res.push(Tok::Num(x));
                }
                '(' | ')' | '+' | '-' => {
                    res.push(Tok::Op(c));
                }
                _ => {}
            }
        }
        res
    }
}

#[test]
fn test() {
    let s = "1 + 1".to_string();
    let res = 2;
    assert_eq!(Solution::calculate(s), res);
    let s = " 2-1 + 2 ".to_string();
    let res = 3;
    assert_eq!(Solution::calculate(s), res);
    let s = "(1+(4+5+2)-3)+(6+8)".to_string();
    let res = 23;
    assert_eq!(Solution::calculate(s), res);
}

// Accepted solution for LeetCode #224: Basic Calculator
function calculate(s: string): number {
    const stk: number[] = [];
    let sign = 1;
    let ans = 0;
    const n = s.length;
    for (let i = 0; i < n; ++i) {
        if (s[i] === ' ') {
            continue;
        }
        if (s[i] === '+') {
            sign = 1;
        } else if (s[i] === '-') {
            sign = -1;
        } else if (s[i] === '(') {
            stk.push(ans);
            stk.push(sign);
            ans = 0;
            sign = 1;
        } else if (s[i] === ')') {
            ans *= stk.pop() as number;
            ans += stk.pop() as number;
        } else {
            let x = 0;
            let j = i;
            for (; j < n && !isNaN(Number(s[j])) && s[j] !== ' '; ++j) {
                x = x * 10 + (s[j].charCodeAt(0) - '0'.charCodeAt(0));
            }
            ans += sign * x;
            i = j - 1;
        }
    }
    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.