# 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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

```// CPP 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];
}

// Driven Program
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];
}

// Driven Program
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
```

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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