# Find an equal point in a string of brackets

Given a string of brackets, the task is to find an index k which decides the number of opening brackets is equal to the number of closing brackets.
String must be consists of only opening and closing brackets i.e. ‘(‘ and ‘)’.

An equal point is an index such that the number of opening brackets before it is equal to the number of closing brackets from and after.

Examples:

```Input : str = "(())))("
Output:   4
After index 4, string splits into (())
and ))(. Number of opening brackets in the
first part is equal to number of closing
brackets in the second part.

Input : str = "))"
Output: 2
As after 2nd position i.e. )) and "empty"
string will be split into these two parts:
So, in this number of opening brackets i.e.
0 in the first part is equal to number of
closing brackets in the second part i.e.
also 0.
```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

1. Store the number of opening brackets appears in the string up to every index, it must start from starting index.
2. Similarly, Store the number of closing brackets appears in the string upto each and every index but it should be done from last index.
3. Check if any index has the same value of opening and closing brackets.

## C++

```// C++ program to find an index k which
// decides the number of opening brackets
// is equal to the number of closing brackets
#include<bits/stdc++.h>
using namespace std;

// Function to find an equal index
int findIndex(string str)
{
int len = str.length();
int open[len+1], close[len+1];
int index = -1;
memset(open, 0, sizeof (open));
memset(close, 0, sizeof (close));

open[0] = 0;
close[len] = 0;
if (str[0]=='(')
open[1] = 1;
if (str[len-1] == ')')
close[len-1] = 1;

// Store the number of opening brackets
// at each index
for (int i = 1; i < len; i++)
{
if ( str[i] == '(' )
open[i+1] = open[i] + 1;
else
open[i+1] = open[i];
}

// Store the number of closing brackets
// at each index
for (int i = len-2; i >= 0; i--)
{
if ( str[i] == ')' )
close[i] = close[i+1] + 1;
else
close[i] = close[i+1];
}

// check if there is no opening or closing
// brackets
if (open[len] == 0)
return len;
if (close[0] == 0)
return 0;

// check if there is any index at which
// both brackets are equal
for (int i=0; i<=len; i++)
if (open[i] == close[i])
index = i;

return index;
}

// Driver code
int main()
{
string str = "(()))(()()())))";
cout << findIndex(str);
return 0;
}
```

## Java

```// Java program to find an index k which
// decides the number of opening brackets
// is equal to the number of closing brackets

public class GFG
{
// Method to find an equal index
static int findIndex(String str)
{
int len = str.length();
int open[] = new int[len+1];
int	close[] = new int[len+1];
int index = -1;

open[0] = 0;
close[len] = 0;
if (str.charAt(0)=='(')
open[1] = 1;
if (str.charAt(len-1) == ')')
close[len-1] = 1;

// Store the number of opening brackets
// at each index
for (int i = 1; i < len; i++)
{
if ( str.charAt(i) == '(' )
open[i+1] = open[i] + 1;
else
open[i+1] = open[i];
}

// Store the number of closing brackets
// at each index
for (int i = len-2; i >= 0; i--)
{
if ( str.charAt(i) == ')' )
close[i] = close[i+1] + 1;
else
close[i] = close[i+1];
}

// check if there is no opening or closing
// brackets
if (open[len] == 0)
return len;
if (close[0] == 0)
return 0;

// check if there is any index at which
// both brackets are equal
for (int i=0; i<=len; i++)
if (open[i] == close[i])
index = i;

return index;
}

// Driver Method
public static void main(String[] args)
{
String str = "(()))(()()())))";
System.out.println(findIndex(str));
}
}
```

Output:

```9
```

Time Complexity : O(n)

This article is contributed by Sahil Chhabra (akku). 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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