// C++ program to find duplicate parenthesis in a // balanced expression #include    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; } 

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 ');  }  } } 

# 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 

// 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 

// JavaScript program to find duplicate parentheses in a balanced expression function findDuplicateParenthesis(s) {  let stack = [];  // Iterate through the given expression  for (let ch of s) {    // If current character is a closing parenthesis ')'  if (ch === ')') {  let top = stack.pop();    // Count the number of elements  // inside the parentheses  let elementsInside = 0;  while (top !== '(') {  elementsInside++;  top = stack.pop();  }    // If there's nothing or only one element   // inside it's redundant  if (elementsInside < 1) {  return true;  }  }   // Push open parenthesis '(' operators and operands to stack  else {  stack.push(ch);  }  }  // No duplicates found  return false; } // Driver code let str = '(((a+(b))+(c+d)))'; if (findDuplicateParenthesis(str)) {  console.log('Duplicate Found'); } else {  console.log('No Duplicates Found'); } // This code is contributed by rag2127