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.

Asked in : Amazon

- Store the number of opening brackets appears in the string up to every index, it must start from starting index.
- Similarly, Store the number of closing brackets appears in the string upto each and every index but it should be done from last index.
- 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)

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