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.

((a+(b))+(c+d))
(b) and ((a+(b)) is surrounded by two
pairs of brackets but, it will not be counted as duplicate.

Below expressions don't have any duplicate parenthesis -
((a+b)+(c+d)) 
No subexpression 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++ program to find duplicate parenthesis in a // balanced expression #include <bits/stdc++.h> 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 

Duplicate Found 

Output: 

Duplicate Found

Time complexity of above solution is O(n). 

Auxiliary space used by the program is O(n).