Last updated on January 20th, 2025 at 11:11 pm
Here, we see a Basic Calculator II LeetCode Solution. This Leetcode problem is solved using different approaches in many programming languages, such as C++, Java, JavaScript, Python, etc.
List of all LeetCode Solution
Topics
String
Companies
Airbnb
Level of Question
Medium

Basic Calculator II LeetCode Solution
Table of Contents
1. 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].
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
2. Coding Pattern Used in Solution
The coding pattern used in all the provided implementations is “Expression Evaluation”. This pattern involves parsing a mathematical expression (given as a string) and evaluating it based on operator precedence and associativity.
3. Code Implementation in Different Languages
3.1 Basic Calculator II C++
class Solution { public: int calculate(string s) { istringstream in('+' + s + '+'); long long total = 0, term = 0, n; char op; while (in >> op) { if (op == '+' or op == '-') { total += term; in >> term; term *= 44 - op; } else { in >> n; if (op == '*') term *= n; else term /= n; } } return total; } };
3.2 Basic Calculator II Java
class Solution { public int calculate(String s) { if (s == null || s.length() == 0) return 0; int num = 0, tmp = 0, res = 0; char op = '+'; for (char c : s.toCharArray()) { if (Character.isDigit(c)) { tmp = tmp*10 + c - '0'; } else if (c != ' ') { //process the numerical value of string so far; based on what 'op' we have before it num = cal(num, tmp, op); if (c == '+' || c == '-') { res += num; num = 0; } // reset tmp = 0; op = c; } } return res + cal(num, tmp, op); } private int cal(int num, int tmp, char op) { if (op == '+') return num + tmp; else if (op == '-') return num - tmp; else if (op == '*') return num * tmp; else return num / tmp; } }
3.3 Basic Calculator II JavaScript
var calculate = function(s) { let stack = []; let num = ''; let sign = null for(let i = 0; i <= s.length; i++){ const curr = s[i]; if(curr === ' ') continue; if(!isNaN(curr)) num+=curr; if(isNaN(curr)){ num = Number(num) switch(sign){ case '+': case null: stack.push(num) break; case '-': stack.push(-num) break; case '*': stack.push(stack.pop()*num) break; case '/': stack.push(parseInt(stack.pop()/num, 10)) break; } sign = curr; num = ''; } } return stack.reduce((a,b)=>{ return a+b },0) };
3.4 Basic Calculator II Python
class Solution: def calculate(self, s: str) -> int: num = 0 res = 0 pre_op = '+' s+='+' stack = [] for c in s: if c.isdigit(): num = num*10 + int(c) elif c == ' ': pass else: if pre_op == '+': stack.append(num) elif pre_op == '-': stack.append(-num) elif pre_op == '*': operant = stack.pop() stack.append((operant*num)) elif pre_op == '/': operant = stack.pop() stack.append(math.trunc(operant/num)) num = 0 pre_op = c return sum(stack)
4. Time and Space Complexity
Time Complexity | Space Complexity | |
C++ | O(n) | O(1) |
Java | O(n) | O(1) |
JavaScript | O(n) | O(n) |
Python | O(n) | O(n) |
- It efficiently evaluates a mathematical expression string using a single pass through the string.
- The time complexity is O(n) for all implementations, while the space complexity varies between O(1) (C++/Java) and O(n) (JavaScript/Python) depending on the use of a stack.