# 224. Basic Calculator

### Description

Implement a basic calculator to evaluate a simple expression string.

The expression string may contain open `(` and closing parentheses `)`, the plus `+` or minus sign `-`, **non-negative** integers and empty spaces .

 **Note:**

* You may assume that the given expression is always valid.
* **Do not** use the `eval` built-in library function.

### Constraints

### Approach

### Links

* GeeksforGeeks
* [Leetcode](https://leetcode.com/problems/basic-calculator/)
* ProgramCreek
* YouTube

### **Examples**

{% tabs %}
{% tab title="Example 1" %}
**Input:** "1 + 1"

**Output:** 2
{% endtab %}

{% tab title="Example 2" %}
**Input:** " 2-1 + 2 "

**Output:** 3
{% endtab %}

{% tab title="Example 3" %}
**Input:** "(1+(4+5+2)-3)+(6+8)"

**Output:** 23
{% endtab %}
{% endtabs %}

### **Solutions**

{% tabs %}
{% tab title="Solution 1" %}

```java
/**
 * Time complexity : 
 * Space complexity : 
 */

class Solution {
    public int calculate(String s) {
        Stack<Integer> stack = new Stack<>();
        int n = s.length(), sign = 1, result = 0;
        
        for(int i = 0; i < n; i++) {
            if(Character.isDigit(s.charAt(i))) {
                int sum = s.charAt(i) - '0';
                while(i+1 < n && Character.isDigit(s.charAt(i+1))) {
                    sum = sum * 10 + (s.charAt(i+1) - '0');
                    i++;
                }
                result += sum * sign;
            } else if(s.charAt(i) == '+') {
                sign = 1;
            } else if(s.charAt(i) == '-') {
                sign = -1;
            } else if(s.charAt(i) == '(') {
                stack.push(result);
                stack.push(sign);
                result = 0;
                sign = 1;
            } else if(s.charAt(i) == ')') {
                result = result * stack.pop() + stack.pop();
            }
        }
        return result;
    }
}
```

{% endtab %}
{% endtabs %}

### **Follow up**

*


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