Check redundant brackets

 

For a given expression in the form of a string, find if there exist any redundant brackets or not. It is given that the expression contains only rounded brackets or parenthesis and the input expression will always be balanced.

A pair of the bracket is said to be redundant when a sub-expression is surrounded by unnecessary or needless brackets.

Example:

Expression: (a+b)+c
Since there are no needless brackets, hence, the output must be 'false'.

Expression: ((a+b))
The expression can be reduced to (a+b). Hence the expression has redundant brackets and the output will be 'true'.

Note:

You will not get a partial score for this problem. You will get marks only if all the test cases are passed.

Input Format :

The first and the only line of input contains a string expression, without any spaces in between.

Output Format :

The first and the only line of output will print either 'true' or 'false'(without the quotes) denoting whether the input expression contains redundant brackets or not.

Note:

You are not required to print the expected result. It has already been taken care of.

Constraints:

0 <= N <= 10^6
Where N is the length of the expression.

Time Limit: 1 second

Sample Input 1:

a+(b)+c 

Sample Output 1:

true

Explanation:

The expression can be reduced to a+b+c. Hence, the brackets are redundant.

Sample Input 2:

(a+b)

Sample Output 2:

false

#include<stack>
bool checkRedundantBrackets(string ex) {
    // Write your code here
    stack <char> s;
     int count;
    for(int i=0;ex[i]!='\0';i++){
        if(ex[i]==')')
        {
            count=0;
            while(s.top()!='(')
            {
                count++;
                s.pop();
            }
            s.pop();
            if(count<2)
                return true;
        }
        else
            s.push(ex[i]);
    }
    return false;
}