Stack-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

***********************************Code************************************
import java.util.Stack;

	public static boolean checkRedundantBrackets(String expression) {
		
		Stack<Character> stack=new Stack<Character>();
        int count=0;
        for(int i=0;i<expression.length();i++)
        {
            char c=expression.charAt(i);
            if (c==')')
            {
                while(!stack.isEmpty() && stack.peek()!='(')
                {
                    count=count+1;
                    stack.pop();
                }
                if (count==0 || count==1)
                {
                    return true;
                }
                stack.pop();
                count=0;
            }
            else
            {
                stack.push(c);
            }
        }
        return false;
}