# Find if an expression has duplicate parenthesis or not

• Difficulty Level : Medium
• Last Updated : 25 Jul, 2021

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.

Below expressions don't have any duplicate parenthesis -
((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++

 `// 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;` `}`

## Java

 `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 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

 `# 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#

 `// 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

 ``

Output:

`Duplicate Found`

Time complexity of above solution is O(n).

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