# Find if an expression has duplicate parenthesis or not

Given an balanced expression, find if it contains duplicate parenthesis or not. A set of parenthesis are duplicate if 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. We 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, we push it to the stack. If the character is close parenthesis ‘)’, then pop characters from the stack till matching open parenthesis ‘(‘ is found. However if the immediate pop hits is open parenthesis ‘(‘, then we have found a duplicate parenthesis. For example, (((a+b))+c) has duplicate brackets around “a+b”. When we reach second “)” after a+b, we have “((” in the stack. Since the top of stack is a opening bracket, we conclude that there are duplicate brackets..

Below is C++ implementation of above idea :

`// 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();`
`            ``// if immediate pop is a open parenthesis '(',`
`            ``// we have found duplicate`
`            ``if` `(top == ``'('``)`
`                ``return` `true``;`
`            ``// else we continue popping characters from the`
`            ``// stack till open parenthesis '(' is encountered`
`            ``else`
`            ``{`
`                ``while` `(top != ``'('``)`
`                ``{`
`                    ``top = Stack.top();`
`                    ``Stack.pop();`
`                ``}`
`            ``}`
`        ``}`
`        ``// 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;`
`}`

Output:

```Duplicate Found
```

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

Disclaimer: This does not belong to TechCodeBit, its an article taken from the below
source and credits.
source and credits: http://www.geeksforgeeks.org
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rakesh