GeeksforGeeks App
Open App
Browser
Continue

# Find the lexicographical next balanced bracket sequence

Given a balanced bracket sequence as a string str containing character ‘(‘ or ‘)’, the task is to find the next lexicographical order balanced sequence if possible else print -1.
Examples:

Input: str = “(())”
Output: ()()
Input: str = “((()))”
Output: (()())

Approach: First find the rightmost opening bracket which we can replace it by a closing bracket to get the lexicographically larger bracket string. The updated string might not be balanced, we can fill the remaining part of the string with the lexicographically minimal one: i.e. first with as much opening brackets as possible, and then fill up the remaining positions with closing brackets. In other words we try to leave a long as possible prefix unchanged, and the suffix gets replaced by the lexicographically minimal one.
To find this position, we can iterate over the character from right to left, and maintain the balance depth of open and closing brackets. When we meet an opening brackets, we will decrement depth, and when we meet a closing bracket, we increase it. If we are at some point meet an opening bracket, and the balance after processing this symbol is positive, then we have found the rightmost position that we can change. We change the symbol, compute the number of opening and closing brackets that we have to add to the right side, and arrange them in the lexicographically minimal way.
If we find no suitable position, then this sequence is already the maximal possible one, and there is no answer.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to find the lexicographically``// next balanced bracket``// expression if possible``string next_balanced_sequence(string& s)``{``    ``string next = ``"-1"``;``    ``int` `length = s.size();``    ``int` `depth = 0;``    ``for` `(``int` `i = length - 1; i >= 0; --i) {` `        ``// Decrement the depth for``        ``// every opening bracket``        ``if` `(s[i] == ``'('``)``            ``depth--;` `        ``// Increment for the``        ``// closing brackets``        ``else``            ``depth++;` `        ``// Last opening bracket``        ``if` `(s[i] == ``'('` `&& depth > 0) {``            ``depth--;``            ``int` `open = (length - i - 1 - depth) / 2;``            ``int` `close = length - i - 1 - open;` `            ``// Generate the required string``            ``next = s.substr(0, i) + ``')'``                   ``+ string(open, ``'('``)``                   ``+ string(close, ``')'``);``            ``break``;``        ``}``    ``}``    ``return` `next;``}` `// Driver code``int` `main()``{``    ``string s = ``"((()))"``;` `    ``cout << next_balanced_sequence(s);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``class` `Sol``{``    ` `// makes a string containing char d``// c number of times``static` `String string(``int` `c, ``char` `d)``{``    ``String s = ``""``;``    ``for``(``int` `i = ``0``; i < c; i++)``    ``s += d;``    ` `    ``return` `s;``}``    ` `// Function to find the lexicographically``// next balanced bracket``// expression if possible``static` `String next_balanced_sequence(String s)``{``    ``String next = ``"-1"``;``    ``int` `length = s.length();``    ``int` `depth = ``0``;``    ``for` `(``int` `i = length - ``1``; i >= ``0``; --i)``    ``{` `        ``// Decrement the depth for``        ``// every opening bracket``        ``if` `(s.charAt(i) == ``'('``)``            ``depth--;` `        ``// Increment for the``        ``// closing brackets``        ``else``            ``depth++;` `        ``// Last opening bracket``        ``if` `(s.charAt(i) == ``'('` `&& depth > ``0``)``        ``{``            ``depth--;``            ``int` `open = (length - i - ``1` `- depth) / ``2``;``            ``int` `close = length - i - ``1` `- open;` `            ``// Generate the required String``            ``next = s.substring(``0``, i) + ``')'``                ``+ string(open, ``'('``)``                ``+ string(close, ``')'``);``            ``break``;``        ``}``    ``}``    ``return` `next;``}` `// Driver code``public` `static` `void` `main(String args[])``{``    ``String s = ``"((()))"``;` `    ``System.out.println(next_balanced_sequence(s));``}``}` `// This code is contributed by Arnab Kundu`

## Python3

 `# Python3 implementation of the approach` `# Function to find the lexicographically``# next balanced bracket``# expression if possible``def` `next_balanced_sequence(s) :` `    ``next` `=` `"-1"``;``    ``length ``=` `len``(s);``    ``depth ``=` `0``;``    ` `    ``for` `i ``in` `range``(length ``-` `1``, ``-``1``, ``-``1``) :``        ` `        ``# Decrement the depth for``        ``# every opening bracket``        ``if` `(s[i] ``=``=` `'('``) :``            ``depth ``-``=` `1``;` `        ``# Increment for the``        ``# closing brackets``        ``else` `:``            ``depth ``+``=` `1``;` `        ``# Last opening bracket``        ``if` `(s[i] ``=``=` `'('` `and` `depth > ``0``) :``            ` `            ``depth ``-``=` `1``;``            ``open` `=` `(length ``-` `i ``-` `1` `-` `depth) ``/``/` `2``;``            ``close ``=` `length ``-` `i ``-` `1` `-` `open``;` `            ``# Generate the required string``            ``next` `=` `s[``0` `: i] ``+` `')'` `+` `open` `*` `'('` `+` `close``*` `')'``;``            ``break``;``            ` `    ``return` `next``;`  `# Driver code``if` `__name__ ``=``=` `"__main__"` `:` `    ``s ``=` `"((()))"``;` `    ``print``(next_balanced_sequence(s));` `    ``# This code is contributed by AnkitRai01`

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG``{``    ` `// makes a string containing char d``// c number of times``static` `String strings(``int` `c, ``char` `d)``{``    ``String s = ``""``;``    ``for``(``int` `i = 0; i < c; i++)``    ``s += d;``    ` `    ``return` `s;``}``    ` `// Function to find the lexicographically``// next balanced bracket``// expression if possible``static` `String next_balanced_sequence(String s)``{``    ``String next = ``"-1"``;``    ``int` `length = s.Length;``    ``int` `depth = 0;``    ``for` `(``int` `i = length - 1; i >= 0; --i)``    ``{` `        ``// Decrement the depth for``        ``// every opening bracket``        ``if` `(s[i] == ``'('``)``            ``depth--;` `        ``// Increment for the``        ``// closing brackets``        ``else``            ``depth++;` `        ``// Last opening bracket``        ``if` `(s[i] == ``'('` `&& depth > 0)``        ``{``            ``depth--;``            ``int` `open = (length - i - 1 - depth) / 2;``            ``int` `close = length - i - 1 - open;` `            ``// Generate the required String``            ``next = s.Substring(0, i) + ``')'` `+``                        ``strings(open, ``'('``) +``                        ``strings(close, ``')'``);``            ``break``;``        ``}``    ``}``    ``return` `next;``}` `// Driver code``public` `static` `void` `Main(String []args)``{``    ``String s = ``"((()))"``;` `    ``Console.WriteLine(next_balanced_sequence(s));``}``}` `// This code is contributed by Princi Singh`

## Javascript

 ``

Output:

`(()())`

Time Complexity: O(n)
Auxiliary Space: O(1)

My Personal Notes arrow_drop_up