Check if expression contains redundant bracket or not | Set 2

Given a string of balanced expressions, find if it contains a redundant parenthesis or not. A set of parenthesis is redundant if the same sub-expression is surrounded by unnecessary or multiple brackets. Print ‘Yes’ if redundant else ‘No’.
Note: Expression may contain ‘+’, ‘*‘, ‘–‘ and ‘/‘ operators. Given expression is valid and there are no white spaces present.
Note: The problem is intended to solve in O(1) extra space.

Examples:

Input: ((a+b)) 
Output: YES
((a+b)) can reduced to (a+b)
Input: (a+(b)/c) 
Output: YES
(a+(b)/c) can reduced to (a+b/c) because b is surrounded by () which is redundant
Input: (a+b*(c-d)) 
Output: NO
(a+b*(c-d)) doesn’t have any redundant or multiple brackets

Approach: 
The idea is very similar to the idea discussed in the previous article but here in place of stack we are counting the symbol ( ‘+’, ‘*‘, ‘–‘ and ‘/‘ ) and the total number of brackets used in the expression.
If the count of brackets is not equal to the count of the symbols then the function will return false. 

C++

// C++ program to check for/
// redundant braces in the string
#include <iostream>
using namespace std;

// Function to check for
// redundant braces
bool IsRedundantBraces(string A)
{
    // count of no of signs
    int a = 0, b = 0;
    for (int i = 0; i < A.size(); i++) {
        if (i + 2 < A.size() && A[i] == '('
            && A[i + 2] == ')')
            return 1;
        if (A[i] == '*'
            || A[i] == '+'
            || A[i] == '-'
            || A[i] == '/')
            a++;
        if (A[i] == '(')
            b++;
    }
    if (b > a)
        return 1;
    return 0;
}

// Driver function
int main()
{
    string A = "(((a+b) + c) + d)";
    if (IsRedundantBraces(A)) {
        cout << "YES\n";
    }
    else {
        cout << "NO";
    }
}

Java

// Java program to check for
// redundant braces in the string
class GFG
{

    // Function to check for
    // redundant braces
    static boolean IsRedundantBraces(String A)
    {
        // count of no of signs
        int a = 0, b = 0;
        for (int i = 0; i < A.length(); i++)
        {
            if (A.charAt(i) == '(' &&
                A.charAt(i + 2) == ')')
                return true;

            if (A.charAt(i) == '*' ||
                A.charAt(i) == '+' ||
                A.charAt(i) == '-' ||
                A.charAt(i) == '/')
                a++;
            if (A.charAt(i) == '(')
                b++;
        }
        if (b > a)
            return true;
        return false;
    }

    // Driver Code
    public static void main (String[] args)
    {
        String A = "(((a+b) + c) + d)";
        if (IsRedundantBraces(A))
        {
            System.out.println("YES");
        }
        else
        {
            System.out.println("NO");
        }
    }
}

// This code is contributed by AnkitRai01

Python3

# Python3 program to check for/
# redundant braces in the string

# Function to check for
# redundant braces
def IsRedundantBraces(A):

    # count of no of signs
    a, b = 0, 0;
    for i in range(len(A)):
        if (A[i] == '(' and A[i + 2] == ')'):
            return True;
        if (A[i] == '*' or A[i] == '+' or
            A[i] == '-' or A[i] == '/'):
            a += 1;
        if (A[i] == '('):
            b += 1;

    if (b > a):
        return True;
    return False;

# Driver Code
if __name__ == '__main__':
    A = "(((a+b) + c) + d)";
    if (IsRedundantBraces(A)):
        print("YES");

    else:
        print("NO");

# This code is contributed by PrinciRaj1992

C#

// C# program to check for
// redundant braces in the string
using System;

class GFG
{

// Function to check for
// redundant braces
static bool IsRedundantBraces(string A)
{
    // count of no of signs
    int a = 0, b = 0;
    for (int i = 0; i < A.Length; i++)
    {
        if (A[i] == '(' && A[i + 2] == ')')
            return true;
        if (A[i] == '*' || A[i] == '+' ||
            A[i] == '-' || A[i] == '/')
            a++;
        if (A[i] == '(')
            b++;
    }
    if (b > a)
        return true;
    return false;
}

// Driver Code
public static void Main (String[] args)
{
    String A = "(((a+b) + c) + d)";
    if (IsRedundantBraces(A))
    {
        Console.WriteLine("YES");
    }
    else
    {
        Console.WriteLine("NO");
    }
}
}

// This code is contributed by PrinciRaj1992

Javascript

<script>
    // Javascript program to check for
    // redundant braces in the string

    // Function to check for
    // redundant braces
    function IsRedundantBraces(A)
    {
        // count of no of signs
        let a = 0, b = 0;
        for (let i = 0; i < A.length; i++)
        {
            if (A[i] == '(' && A[i + 2] == ')')
                return true;
            if (A[i] == '*' || A[i] == '+' ||
                A[i] == '-' || A[i] == '/')
                a++;
            if (A[i] == '(')
                b++;
        }
        if (b > a)
            return true;
        return false;
    }

    let A = "(((a+b) + c) + d)";
    if (IsRedundantBraces(A))
    {
        document.write("YES");
    }
    else
    {
        document.write("NO");
    }

// This code is contributed by divyeshrabadiya07.
</script>

Output:

NO

Time Complexity: O(n), where n is the length of the given string.
Auxiliary Space: O(1), no extra space is required, so it is a constant.