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++

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// C++ implementation of the approach
#include <bits/stdc++.h>
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;
}

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Java

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// 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

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Python3

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# 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

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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

Output:

(()())


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