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++

// 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;
}

Java

// 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

C#

// 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.


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.

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Improved By : vt_m