Expression contains redundant bracket or not

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

Example:

Input: 
((a+b))
(a+(b)/c)
(a+b*(c-d))
Output: 
Yes
Yes
No

Explanation:
1. ((a+b)) can reduced to (a+b), this Redundant
2. (a+(b)/c) can reduced to (a+b/c) because b is
surrounded by () which is redundant.
3. (a+b*(c-d)) doesn't have any redundant or multiple
brackets.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

The idea is to use stack which is discussed in this article. For any sub-expression of expression, if we able to pick any sub-expression of expression surrounded by (), then we again left with () as part of string, we have redundant braces.

We iterate through the given expression and for each character in the expression, if the character is a open parenthesis ‘(‘ or any of the operators or operands, we push it to the stack. If the character is close parenthesis ‘)’, then pop characters from the stack till matching open parenthesis ‘(‘ is found.
Now for redundancy two condition will arise while popping-

If immediate pop hits a open parenthesis ‘(‘, then we have found a duplicate parenthesis. For example, (((a+b))+c) has duplicate brackets around a+b. When we reach second “)” after a+b, we have “((” in the stack. Since the top of stack is a opening bracket, we conclude that there are duplicate brackets.
If immediate pop doesn’t hit any operand(‘*’, ‘+’, ‘/’, ‘-‘) then it indicates the presence of unwanted brackets surrounded by expression. For instance, (a)+b contain unwanted () around a thus it is redundant.

C++

/* C++ Program to check whether valid  
 expression is redundant or not*/
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to check redundant brackets in a 
// balanced expression 
bool checkRedundancy(string& str) 
{ 
    // create a stack of characters 
    stack<char> st; 
  
    // Iterate through the given expression 
    for (auto& ch : str) { 
  
        // if current character is close parenthesis ')' 
        if (ch == ')') { 
            char top = st.top(); 
            st.pop(); 
  
            // If immediate pop have open parenthesis '(' 
            // duplicate brackets found 
            bool flag = true; 
  
            while (top != '(') { 
  
                // Check for operators in expression 
                if (top == '+' || top == '-' ||  
                    top == '*' || top == '/') 
                    flag = false; 
  
                // Fetch top element of stack 
                top = st.top(); 
                st.pop(); 
            } 
  
            // If operators not found 
            if (flag == true) 
                return true; 
        } 
  
        else
            st.push(ch); // push open parenthesis '(', 
                  // operators and operands to stack 
    } 
    return false; 
} 
  
// Function to check redundant brackets 
void findRedundant(string& str) 
{ 
    bool ans = checkRedundancy(str); 
    if (ans == true) 
        cout << "Yes\n"; 
    else
        cout << "No\n"; 
} 
  
// Driver code 
int main() 
{ 
    string str = "((a+b))"; 
    findRedundant(str); 
  
    str = "(a+(b)/c)"; 
    findRedundant(str); 
  
    str = "(a+b*(c-d))"; 
    findRedundant(str); 
  
    return 0; 
} 

Java

/* Java Program to check whether valid  
expression is redundant or not*/
import java.util.Stack; 
public class GFG { 
// Function to check redundant brackets in a  
// balanced expression  
  
    static boolean checkRedundancy(String s) { 
        // create a stack of characters  
        Stack<Character> st = new Stack<>(); 
        char[] str = s.toCharArray(); 
        // Iterate through the given expression  
        for (char ch : str) { 
  
            // if current character is close parenthesis ')'  
            if (ch == ')') { 
                char top = st.peek(); 
                st.pop(); 
  
                // If immediate pop have open parenthesis '('  
                // duplicate brackets found  
                boolean flag = true; 
  
                while (top != '(') { 
  
                    // Check for operators in expression  
                    if (top == '+' || top == '-'
                            || top == '*' || top == '/') { 
                        flag = false; 
                    } 
  
                    // Fetch top element of stack  
                    top = st.peek(); 
                    st.pop(); 
                } 
  
                // If operators not found  
                if (flag == true) { 
                    return true; 
                } 
            } else { 
                st.push(ch); // push open parenthesis '(',  
            }                // operators and operands to stack  
        } 
        return false; 
    } 
  
// Function to check redundant brackets  
    static void findRedundant(String str) { 
        boolean ans = checkRedundancy(str); 
        if (ans == true) { 
            System.out.println("Yes"); 
        } else { 
            System.out.println("No"); 
        } 
    } 
  
// Driver code  
    public static void main(String[] args) { 
        String str = "((a+b))"; 
        findRedundant(str); 
  
        str = "(a+(b)/c)"; 
        findRedundant(str); 
  
        str = "(a+b*(c-d))"; 
        findRedundant(str); 
    } 
} 

Python3

# Python3 Program to check whether valid  
# expression is redundant or not 
  
# Function to check redundant brackets  
# in a balanced expression  
def checkRedundancy(Str): 
      
    # create a stack of characters  
    st = []  
  
    # Iterate through the given expression  
    for ch in Str:  
  
        # if current character is close  
        # parenthesis ')'  
        if (ch == ')'):  
            top = st[-1]  
            st.pop()  
  
            # If immediate pop have open parenthesis  
            # '(' duplicate brackets found  
            flag = True
  
            while (top != '('):  
  
                # Check for operators in expression  
                if (top == '+' or top == '-' or
                    top == '*' or top == '/'):  
                    flag = False
  
                # Fetch top element of stack  
                top = st[-1]  
                st.pop() 
  
            # If operators not found  
            if (flag == True): 
                return True
  
        else: 
            st.append(ch) # append open parenthesis '(',  
                          # operators and operands to stack 
    return False
  
# Function to check redundant brackets  
def findRedundant(Str): 
    ans = checkRedundancy(Str)  
    if (ans == True):  
        print("Yes")  
    else: 
        print("No") 
  
# Driver code  
if __name__ == '__main__': 
    Str = "((a+b))"
    findRedundant(Str)  
  
    Str = "(a+(b)/c)"
    findRedundant(Str)  
  
    Str = "(a+b*(c-d))"
    findRedundant(Str) 
  
# This code is contributed by PranchalK 

C#

/* C# Program to check whether valid  
expression is redundant or not*/
using System;  
using System.Collections.Generic; 
  
class GFG 
{ 
    // Function to check redundant brackets in a  
    // balanced expression  
    static bool checkRedundancy(String s) 
    { 
        // create a stack of characters  
        Stack<char> st = new Stack<char>(); 
        char[] str = s.ToCharArray(); 
          
        // Iterate through the given expression  
        foreach (char ch in str) 
        { 
  
            // if current character is close parenthesis ')'  
            if (ch == ')')  
            { 
                char top = st.Peek(); 
                st.Pop(); 
  
                // If immediate pop have open parenthesis '('  
                // duplicate brackets found  
                bool flag = true; 
  
                while (top != '(')  
                { 
  
                    // Check for operators in expression  
                    if (top == '+' || top == '-'
                            || top == '*' || top == '/')  
                    { 
                        flag = false; 
                    } 
  
                    // Fetch top element of stack  
                    top = st.Peek(); 
                    st.Pop(); 
                } 
  
                // If operators not found  
                if (flag == true) 
                { 
                    return true; 
                } 
            }  
            else
            { 
                st.Push(ch); // push open parenthesis '(',  
            }         // operators and operands to stack  
        } 
        return false; 
    } 
  
    // Function to check redundant brackets  
    static void findRedundant(String str) 
    { 
        bool ans = checkRedundancy(str); 
        if (ans == true)  
        { 
            Console.WriteLine("Yes"); 
        }  
        else
        { 
            Console.WriteLine("No"); 
        } 
    } 
  
    // Driver code  
    public static void Main(String[] args)  
    { 
        String str = "((a+b))"; 
        findRedundant(str); 
  
        str = "(a+(b)/c)"; 
        findRedundant(str); 
  
        str = "(a+b*(c-d))"; 
        findRedundant(str); 
    } 
} 
  
/* This code contributed by PrinciRaj1992 */

Output

Yes
Yes
No

Time complexity: O(n)
Auxiliary space: O(n)

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

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