Find if an expression has duplicate parenthesis or not

Given a balanced expression, find if it contains duplicate parenthesis or not. A set of parenthesis are duplicate if the same subexpression is surrounded by multiple parenthesis.

Examples:

Below expressions have duplicate parenthesis - 
((a+b)+((c+d)))
The subexpression "c+d" is surrounded by two
pairs of brackets.

(((a+(b)))+(c+d))
The subexpression "a+(b)" is surrounded by two 
pairs of brackets.

(((a+(b))+c+d))
The whole expression is surrounded by two 
pairs of brackets.

Below expressions don't have any duplicate parenthesis -
((a+b)+(c+d)) 
No subsexpression is surrounded by duplicate
brackets.

((a+(b))+(c+d))
No subsexpression is surrounded by duplicate
brackets.

It may be assumed that the given expression is valid and there are not any white spaces present.



The idea is to use stack. 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, push it to the top of the stack. If the character is close parenthesis ‘)’, then pop characters from the stack till matching open parenthesis ‘(‘ is found and a counter is used, whose value is incremented for every character encountered till the opening parenthesis ‘(‘ is found. If the number of characters encountered between the opening and closing parenthesis pair, which is equal to the value of the counter, is less than 1, then a pair of duplicate parenthesis is found else there is no occurrence of redundant parenthesis pairs. For example, (((a+b))+c) has duplicate brackets around “a+b”. When the second “)” after a+b is encountered, the stack contains “((“. Since the top of stack is a opening bracket, it can be concluded that there are duplicate brackets.

Below is the implementation of above idea :

C++

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// C++ program to find duplicate parenthesis in a
// balanced expression
#include <iostream>
#include <stack>
using namespace std;
  
// Function to find duplicate parenthesis in a
// balanced expression
bool findDuplicateparenthesis(string str)
{
    // create a stack of characters
    stack<char> Stack;
  
    // Iterate through the given expression
    for (char ch : str)
    {
        // if current character is close parenthesis ')'
        if (ch == ')')
        {
            // pop character from the stack
            char top = Stack.top();
            Stack.pop();
  
            // stores the number of characters between a 
            // closing and opening parenthesis
            // if this count is less than or equal to 1
            // then the brackets are redundant else not
            int elementsInside = 0;
            while (top != '(')
            {
                elementsInside++;
                top = Stack.top();
                Stack.pop();
            }
            if(elementsInside < 1) {
                return 1;
            }
        }
  
        // push open parenthesis '(', operators and
        // operands to stack
        else
            Stack.push(ch);
    }
  
    // No duplicates found
    return false;
}
  
  
// Driver code
int main()
{
    // input balanced expression
    string str = "(((a+(b))+(c+d)))";
  
    if (findDuplicateparenthesis(str))
        cout << "Duplicate Found ";
    else
        cout << "No Duplicates Found ";
  
    return 0;
}

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Java

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import java.util.Stack;
  
// Java program to find duplicate parenthesis in a 
// balanced expression 
public class GFG {
  
// Function to find duplicate parenthesis in a 
// balanced expression 
    static boolean findDuplicateparenthesis(String s) {
        // create a stack of characters 
        Stack<Character> Stack = new Stack<>();
  
        // Iterate through the given expression 
        char[] str = s.toCharArray();
        for (char ch : str) {
            // if current character is close parenthesis ')' 
            if (ch == ')') {
                // pop character from the stack 
                char top = Stack.peek();
                Stack.pop();
  
                // stores the number of characters between a 
                // closing and opening parenthesis 
                // if this count is less than or equal to 1 
                // then the brackets are redundant else not 
                int elementsInside = 0;
                while (top != '(') {
                    elementsInside++;
                    top = Stack.peek();
                    Stack.pop();
                }
                if (elementsInside < 1) {
                    return true;
                }
            } // push open parenthesis '(', operators and 
            // operands to stack 
            else {
                Stack.push(ch);
            }
        }
  
        // No duplicates found 
        return false;
    }
  
// Driver code 
public static void main(String[] args) {
  
        // input balanced expression 
        String str = "(((a+(b))+(c+d)))";
  
        if (findDuplicateparenthesis(str)) {
            System.out.println("Duplicate Found ");
        } else {
            System.out.println("No Duplicates Found ");
        }
  
    }
}

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Python3

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# Python3 program to find duplicate 
# parenthesis in a balanced expression 
  
# Function to find duplicate parenthesis 
# in a balanced expression 
def findDuplicateparenthesis(string): 
  
    # create a stack of characters 
    Stack = [] 
  
    # Iterate through the given expression 
    for ch in string:
      
        # if current character is 
        # close parenthesis ')' 
        if ch == ')'
          
            # pop character from the stack 
            top = Stack.pop() 
  
            # stores the number of characters between 
            # a closing and opening parenthesis 
            # if this count is less than or equal to 1 
            # then the brackets are redundant else not 
            elementsInside = 0
            while top != '('
              
                elementsInside += 1
                top = Stack.pop() 
              
            if elementsInside < 1
                return True
  
        # push open parenthesis '(', operators 
        # and operands to stack 
        else:
            Stack.append(ch) 
      
    # No duplicates found 
    return False
  
# Driver Code
if __name__ == "__main__"
  
    # input balanced expression 
    string = "(((a+(b))+(c+d)))"
  
    if findDuplicateparenthesis(string) == True
        print("Duplicate Found"
    else:
        print("No Duplicates Found"
  
# This code is contributed by Rituraj Jain

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C#

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// C# program to find duplicate parenthesis 
// in a balanced expression 
using System;
using System.Collections.Generic;
  
class GFG 
{
  
// Function to find duplicate parenthesis  
// in a balanced expression 
static Boolean findDuplicateparenthesis(String s) 
{
    // create a stack of characters 
    Stack<char> Stack = new Stack<char>();
  
    // Iterate through the given expression 
    char[] str = s.ToCharArray();
    foreach (char ch in str) 
    {
        // if current character is 
        // close parenthesis ')' 
        if (ch == ')'
        {
            // pop character from the stack 
            char top = Stack.Peek();
            Stack.Pop();
  
            // stores the number of characters between
            // a closing and opening parenthesis 
            // if this count is less than or equal to 1 
            // then the brackets are redundant else not 
            int elementsInside = 0;
            while (top != '('
            {
                elementsInside++;
                top = Stack.Peek();
                Stack.Pop();
            }
            if (elementsInside < 1) 
            {
                return true;
            }
        }  
          
        // push open parenthesis '(', 
        // operators and operands to stack 
        else 
        {
            Stack.Push(ch);
        }
    }
  
    // No duplicates found 
    return false;
}
  
// Driver code 
public static void Main(String[] args)
{
  
    // input balanced expression 
    String str = "(((a+(b))+(c+d)))";
  
    if (findDuplicateparenthesis(str))
    {
        Console.WriteLine("Duplicate Found ");
    
    else 
    {
        Console.WriteLine("No Duplicates Found ");
    }
}
}
  
// This code is contributed by 29AjayKumar

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Output:

Duplicate Found

Time complexity of above solution is O(n).
Auxiliary space used by the program is O(n).

This article is contributed by Aditya Goel. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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