Expression contains redundant bracket or not

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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.

Examples:

Input: str = “((a+b))”
Output: YES
Explanation: ((a+b)) can reduced to (a+b), this Redundant

Input: str = “(a+(b)/c)”
Output: YES
Explanation: (a+(b)/c) can reduced to (a+b/c) because b is surrounded by () which is redundant.

Checking Redundant Bracket using Stack

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

Follow the steps mentioned below to implement the approach:

We iterate through the given expression and for each character in the expression
- if the character is an 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 conditions will arise while popping.
- If immediate pop hits an open parenthesis ‘(‘, then we have found a duplicate parenthesis. For example, (((a+b))+c) has duplicate brackets around a+b. When we reach the second “)” after a+b, we have “((” in the stack. Since the top of the stack is an 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 contains unwanted () around a thus it is redundant.

Below is the implementation of the above approach:

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 (!st.empty() and 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);
    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);
 
    }
}

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) 
 
 
# 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);
    }
}
 
/* This code contributed by PrinciRaj1992 */

Javascript

<script>
 
/* JavaScript Program to check whether valid 
 expression is redundant or not*/
 
 
// Function to check redundant brackets in a
// balanced expression
function checkRedundancy(str)
{
    // create a stack of characters
    var st = [];
    var ans = false;
    // Iterate through the given expression
    str.split('').forEach(ch => {
         
 
        // if current character is close parenthesis ')'
        if (ch == ')') {
            var top = st[st.length-1];
            st.pop();
 
            // If immediate pop have open parenthesis '('
            // duplicate brackets found
            var flag = true;
 
            while (st.length!=0 && top != '(') {
 
                // Check for operators in expression
                if (top == '+' || top == '-' || 
                    top == '*' || top == '/')
                    flag = false;
 
                // Fetch top element of stack
                top = st[st.length-1];
                st.pop();
            }
 
            // If operators not found
            if (flag == true)
                ans = true;
        }
 
        else
            st.push(ch); // push open parenthesis '(',
                  // operators and operands to stack
    });
    return ans;
}
 
// Function to check redundant brackets
function findRedundant(str)
{
    var ans = checkRedundancy(str);
    if (ans == true)
        document.write( "Yes<br>");
    else
        document.write( "No<br>");
}
 
// Driver code
 
var str = "((a+b))";
findRedundant(str);
 
 
 
</script>

Output

Yes

Time Complexity: O(N)
Auxiliary Space: O(N)

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