Basic Calculator II

Published at23 Jul 2026

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You are given a string containing a valid arithmetic expression with integers and the operators +, -, *, /. There are no parentheses. Evaluate the expression and return the result.

Operator precedence matters: multiplication and division must be evaluated before addition and subtraction. A stack makes this easy. Iterate through the string, parse numbers, and when you see a + or -, push the number (positive or negative). When you see * or /, pop the last number, compute the result, and push it back. At the end, sum all values on the stack.

Integer division should truncate toward zero. The expression is always valid.Operator precedence makes this problem more complex than a simple left-to-right evaluation. Multiplication and division must be computed before addition and subtraction. The stack handles this by deferring addition and subtraction until all higher-precedence operations are resolved.The sign variable tracks the operator immediately before the current number. When a + or - is encountered, the current number (with its sign) is pushed onto the stack. When * or / is encountered, the last number on the stack is popped, the operation is performed immediately, and the result is pushed back.At the end, the stack contains only positive and negative numbers that can be summed directly. This approach works because multiplication and division are binary operations that produce a single result, while addition and subtraction simply accumulate signed values.Integer division must truncate toward zero. The expression is guaranteed to be valid and will not contain division by zero. Spaces in the input should be ignored.

Example Input & Output

Example 1

Input

3+5/2

Output

Explanation

5/2=2, then 3+2=5

Example 2

Input

3+2*2

Output

Explanation

2*2=4, then 3+4=7

Example 3

Input

3-5*2+8/4

Output

-5

Explanation

-5*2=-10, then 3-10=-7, then -7+2=-5

Example 4

Input

3/2

Output

Explanation

Integer division truncates toward zero: 3/2=1

Algorithm Flow

Recommendation Algorithm Flow for Basic Calculator II

Solution Approach

Iterate through the string character by character, building the current number. When you encounter an operator or reach the end, check the previous operator. If it was + or -, push the current number (with sign) onto the stack. If it was * or /, pop the last number, compute the result, and push it back. At the end, sum all values on the stack.

Time complexity is O(n). Space complexity is O(n).

Best Answers
These are the verified best solutions for this Challenge. You can use them as a reference to improve your solution.

java

import java.util.*;
class Solution {
    public int solution(String s) {
        Stack<Integer> stack = new Stack<>();
        int num = 0; char sign = '+';
        for (int i = 0; i < s.length(); i++) {
            char ch = s.charAt(i);
            if (ch >= '0' && ch <= '9') num = num * 10 + (ch - '0');
            if (ch == '+' || ch == '-' || ch == '*' || ch == '/' || i == s.length() - 1) {
                if (sign == '+') stack.push(num); else if (sign == '-') stack.push(-num);
                else if (sign == '*') stack.push(stack.pop() * num);
                else if (sign == '/') stack.push(stack.pop() / num);
                num = 0; sign = ch;
            }
        }
        int sum = 0; while (!stack.isEmpty()) sum += stack.pop();
        return sum;
    }
}

javascript

function solution(s) {
  var stack = [];
  var num = 0;
  var sign = '+';
  for (var i = 0; i < s.length; i++) {
    var ch = s[i];
    if (ch >= '0' && ch <= '9') num = num * 10 + Number(ch);
    if (ch === '+' || ch === '-' || ch === '*' || ch === '/' || i === s.length - 1) {
      if (sign === '+') stack.push(num); else if (sign === '-') stack.push(-num);
      else if (sign === '*') stack.push(stack.pop() * num);
      else if (sign === '/') stack.push(Math.trunc(stack.pop() / num));
      num = 0; sign = ch;
    }
  }
  var sum = 0; while (stack.length) sum += stack.pop();
  return sum;
}

php

<?php
function solution($s) {
    $stack = []; $num = 0; $sign = '+';
    for ($i = 0; $i < strlen($s); $i++) {
        $ch = $s[$i];
        if ($ch >= '0' && $ch <= '9') $num = $num * 10 + intval($ch);
        if (in_array($ch, ['+','-','*','/']) || $i == strlen($s) - 1) {
            if ($sign == '+') array_push($stack, $num); elseif ($sign == '-') array_push($stack, -$num);
            elseif ($sign == '*') array_push($stack, array_pop($stack) * $num);
            elseif ($sign == '/') array_push($stack, intdiv(array_pop($stack), $num));
            $num = 0; $sign = $ch;
        }
    }
    return array_sum($stack);
}

python

def solution(s):
    stack = []
    num = 0
    sign = '+'
    for i, ch in enumerate(s):
        if ch.isdigit():
            num = num * 10 + int(ch)
        if ch in '+-*/' or i == len(s) - 1:
            if sign == '+': stack.append(num)
            elif sign == '-': stack.append(-num)
            elif sign == '*': stack.append(stack.pop() * num)
            elif sign == '/': stack.append(int(stack.pop() / num))
            num = 0
            sign = ch
    return sum(stack)

rust

fn solution(s: &str) -> i32 {
    let mut stack: Vec<i32> = Vec::new();
    let mut num = 0; let mut sign = '+';
    let chars: Vec<char> = s.chars().collect();
    for i in 0..chars.len() {
        let ch = chars[i];
        if ch.is_digit(10) { num = num * 10 + (ch as i32 - 48); }
        if ch == '+' || ch == '-' || ch == '*' || ch == '/' || i == chars.len() - 1 {
            match sign { '+' => stack.push(num), '-' => stack.push(-num),
                '*' => { let a = stack.pop().unwrap(); stack.push(a * num); },
                '/' => { let a = stack.pop().unwrap(); stack.push(a / num); }, _ => {} }
            num = 0; sign = ch;
        }
    }
    stack.iter().sum()
}

typescript

function solution(s: string): number {
    var stack: number[] = [];
    var num = 0; var sign = '+';
    for (var i = 0; i < s.length; i++) {
        var ch = s[i];
        if (ch >= '0' && ch <= '9') num = num * 10 + Number(ch);
        if (ch === '+' || ch === '-' || ch === '*' || ch === '/' || i === s.length - 1) {
            if (sign === '+') stack.push(num); else if (sign === '-') stack.push(-num);
            else if (sign === '*') stack.push(stack.pop() as number * num);
            else if (sign === '/') stack.push(~~((stack.pop() as number) / num));
            num = 0; sign = ch;
        }
    }
    var sum = 0; while (stack.length) sum += stack.pop() as number;
    return sum;
}

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Basic Calculator II

Example Input & Output

Algorithm Flow

Solution Approach

Best AnswersThese are the verified best solutions for this Challenge. You can use them as a reference to improve your solution.

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Best Answers
These are the verified best solutions for this Challenge. You can use them as a reference to improve your solution.