Check if expression contains redundant bracket or not | Set 2

Given a string of balanced expression, find if it contains a redundant parenthesis or not. A set of parenthesis are 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++

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// 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 (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";
    }
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:

NO


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Improved By : AnkitRai01, princiraj1992



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