# Redundant Braces

Given a string A denoting an expression. It contains the following operators ’+’, ‘-‘, ‘*’, ‘/’.

Check whether A has redundant braces or not.

Return 1 if A has redundant braces, else return 0.

Note: A will be always a valid expression.

Input Format

```The only argument given is string A.
```
Output Format

```Return 1 if string has redundant braces, else return 0.
```
For Example

```Input 1:
A = "((a + b))"
Output 1:
1
Explanation 1:
((a + b)) has redundant braces so answer will be 1.

Input 2:
A = "(a + (a + b))"
Output 2:
0
Explanation 2:
(a + (a + b)) doesn't have have any redundant braces so answer will be 0.```
Method:

A brace is redundant if it does not contain an operator.

Solution:

Time: O(n)
Space: O(n)

public class Solution {
public int braces(String A) {
Deque<Character> stack = new ArrayDeque<>();
char[] arr = A.toCharArray();
for (int i = 0; i < arr.length;) {
char c = arr[i];
if (c == '(') {
stack.push(c);
}
if (c == '+' || c == '-' || c == '*' || c == '/') {
if (!stack.isEmpty()) {
stack.pop();
while (c != ')' && c != '(') {
i ++;
c = arr[i];
}
continue;
}
}
i ++;
}
if (stack.isEmpty()) {
return 0;
} else {
return 1;
}
}
}