Length of Longest Balanced Subsequence

Given a string S, find the length of longest balanced subsequence in it. A balanced string is defined as:-

  • A Null string is a balanced string.
  • If X and Y are balanced strings, then (X)Y and XY are balanced strings.

Examples :

Input : S = "()())"
Output : 4

()() is the longest balanced subsequence 
of length 4.

Input : s = "()(((((()"
Output : 4



A brute force approach is to find all subsequence of the given string S and check for all possible subsequence if it form a balanced sequence, if yes, compare it with maximum value.

The better approach is to use Dynamic Programming.

Longest Balananced Subsequence (LBS), can be recursively defined as below.

LBS of substring str[i..j] : 
If str[i] == str[j]
    LBS(str, i, j) = LBS(str, i + 1, j - 1) + 2
Else
    LBS(str, i, j) = max(LBS(str, i, k) +
                         LBS(str, k + 1, j))
                     Where i <= k < j

Declare a 2D matrix dp[][], where our state dp[i][j] will denote the length of longest balanced subsequence from index i to j. We will compute this state in order of increasing j – i. For a particular state dp[i][j], we will try to match the jth symbol with kth symbol, that can be done only if S[k] is ‘(‘ and S[j] is ‘)’, we will take the max of 2 + dp[i][k – 1] + dp[k + 1][j – 1] for all such possible k and also max(dp[i + 1][j], dp[i][j – 1]) and put the value in dp[i][j]. In this way we can fill all the dp states. dp[0][length of string – 1] (considering 0 indexing) will be our answer.

Below is the implementation of this approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to find length of 
// the longest balanced subsequence
#include <bits/stdc++.h>
using namespace std;
  
int maxLength(char s[], int n)
{
    int dp[n][n];
    memset(dp, 0, sizeof(dp));
  
    // Considering all balanced 
    // substrings of length 2
    for (int i = 0; i < n - 1; i++)
        if (s[i] == '(' && s[i + 1] == ')')
            dp[i][i + 1] = 2;
  
    // Considering all other substrings
    for (int l = 2; l < n; l++) 
    {
        for (int i = 0, j = l; j < n; i++, j++) 
        {
            if (s[i] == '(' && s[j] == ')')
                dp[i][j] = 2 + dp[i + 1][j - 1];
  
            for (int k = i; k < j; k++) 
                dp[i][j] = max(dp[i][j],
                              dp[i][k] + 
                          dp[k + 1][j]);         
        }
    }
  
    return dp[0][n - 1];
}
  
// Driver Code
int main()
{
    char s[] = "()(((((()";
    int n = strlen(s);
    cout << maxLength(s, n) << endl;
    return 0;
}

chevron_right






                        filter_none









Java














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















// Java program to find length of the 


// longest balanced subsequence. 


import java.io.*; 


  


class GFG  


{ 


static int maxLength(String s, int n) 


{ 


    int dp[][] = new int[n][n]; 


      


  


    // Considering all balanced substrings  


    // of length 2 


    for (int i = 0; i < n - 1; i++) 


        if (s.charAt(i) == '(' &&   


                s.charAt(i + 1)    == ')') 


            dp[i][i + 1] = 2; 


  


    // Considering all other substrings 


    for (int l = 2; l < n; l++)  


    { 


        for (int i = 0, j = l; j < n; i++, j++)  


        { 


            if (s.charAt(i) == '(' &&  


                      s.charAt(j) == ')') 


                dp[i][j] = 2 + dp[i + 1][j - 1]; 


  


            for (int k = i; k < j; k++)  


                dp[i][j] = Math.max(dp[i][j], 


                        dp[i][k] + dp[k + 1][j]);      


                              


        } 


    } 


  


    return dp[0][n - 1]; 


} 


  


// Driver Code 


public static void main(String[] args) 


{ 


    String s = "()(((((()"; 


    int n = s.length() ; 


    System.out.println(maxLength(s, n)); 


} 


} 


// This code is contributed by Prerna Saini 



















                        chevron_right







                        filter_none









Python3














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















# Python3 program to find length of  


# the longest balanced subsequence  


  


def maxLength(s, n): 


      


    dp = [[0 for i in range(n)] 


             for i in range(n)] 


               


    # Considering all balanced  


    # substrings of length 2  


    for i in range(n - 1):  


        if (s[i] == '(' and s[i + 1] == ')'):  


            dp[i][i + 1] = 2


  


    # Considering all other substrings  


    for l in range(2, n): 


        i = -1


        for j in range(l, n): 


            i += 1


            if (s[i] == '(' and s[j] == ')'):  


                dp[i][j] = 2 + dp[i + 1][j - 1]  


            for k in range(i, j):  


                dp[i][j] = max(dp[i][j], dp[i][k] +


                                         dp[k + 1][j]) 


    return dp[0][n - 1]  


  


# Driver Code  


s= "()(((((()"


n = len(s) 


print(maxLength(s, n))  


  


# This code is contributed  


# by sahishelangia 



















                        chevron_right







                        filter_none









C#














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















// C# program to find length of the 


// longest balanced subsequence. 


using System; 


  


class GFG  


{ 


  


static int maxLength(String s, int n) 


{ 


    int [,]dp = new int[n, n]; 


      


  


    // Considering all balanced substrings  


    // of length 2 


    for (int i = 0; i < n - 1; i++) 


        if (s[i] == '(' && s[i + 1] == ')') 


            dp[i, i + 1] = 2; 


  


    // Considering all other substrings 


    for (int l = 2; l < n; l++)  


    { 


        for (int i = 0, j = l; j < n; i++, j++)  


        { 


            if (s[i] == '(' && s[j] == ')') 


                dp[i, j] = 2 + dp[i + 1, j - 1]; 


  


            for (int k = i; k < j; k++)  


                dp[i, j] = Math.Max(dp[i, j], 


                        dp[i, k] + dp[k + 1, j]);      


        } 


    } 


  


    return dp[0, n - 1]; 


} 


  


    // Driver Code 


    public static void Main() 


    { 


        string s = "()(((((()"; 


        int n = s.Length ; 


        Console.WriteLine(maxLength(s, n)); 


    } 


} 


  


// This code is contributed by vt_m. 



















                        chevron_right







                        filter_none









PHP









Output:


4



Time Complexity : O(n2)

Auxiliary Space : O(n2)


This article is contributed by Anuj Chauhan. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.


Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



          
          
          
          

            

    

        My Personal Notes
        arrow_drop_up
    

    

        
        

            
            
        

    


Recommended Posts: