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Number of balanced parenthesis substrings
• Difficulty Level : Hard
• Last Updated : 06 May, 2021

Given a balanced parenthesis string which consists of ‘(‘ and ‘)‘. The task is to find the number of balanced parenthesis substrings in the given string

Examples :

Input : str = “()()()”
Output :
(), (), (), ()(), ()(), ()()()
Input : str = “(())()”
Output :
(), (()), (), (())()

Approach :
Let us assume that whenever we encounter with opening bracket the depth increases by one and with a closing bracket the depth decreases by one. Whenever we encounter the closing bracket increase our required answer by one and then increment our required answer by the already formed balanced substrings at this depth.
Below is the implementation of the above approach :

## C++

 `// CPP program to find number of``// balanced parenthesis sub strings``#include ``using` `namespace` `std;` `// Function to find number of``// balanced parenthesis sub strings``int` `Balanced_Substring(string str, ``int` `n)``{``    ``// To store required answer``    ``int` `ans = 0;` `    ``// Vector to stores the number of``    ``// balanced brackets at each depth.``    ``vector<``int``> arr(n / 2 + 1, 0);` `    ``// d strores checks the depth of our sequence``    ``// For example level of () is 1``    ``// and that of (()) is 2.``    ``int` `d = 0;``    ``for` `(``int` `i = 0; i < n; i++) {``        ``// If open bracket``        ``// increase depth``        ``if` `(str[i] == ``'('``)``            ``d++;` `        ``// If closing bracket``        ``else` `{``            ``if` `(d == 1) {``                ``for` `(``int` `j = 2; j <= n / 2 + 1 && arr[j] != 0; j++)``                    ``arr[j] = 0;``            ``}``            ``++ans;``            ``ans += arr[d];``            ``arr[d]++;``            ``d--;``        ``}``    ``}` `    ``// Return the required answer``    ``return` `ans;``}` `// Driver code``int` `main()``{``    ``string str = ``"()()()"``;` `    ``int` `n = str.size();` `    ``// Function call``    ``cout << Balanced_Substring(str, n);` `    ``return` `0;``}`

## Java

 `// Java program to find number of``// balanced parenthesis sub strings``class` `GFG {` `    ``// Function to find number of``    ``// balanced parenthesis sub strings``    ``public` `static` `int` `Balanced_Substring(String str,``                                         ``int` `n)``    ``{` `        ``// To store required answer``        ``int` `ans = ``0``;` `        ``// Vector to stores the number of``        ``// balanced brackets at each depth.``        ``int``[] arr = ``new` `int``[n / ``2` `+ ``1``];` `        ``// d strores checks the depth of our sequence``        ``// For example level of () is 1``        ``// and that of (()) is 2.``        ``int` `d = ``0``;``        ``for` `(``int` `i = ``0``; i < n; i++) {` `            ``// If open bracket``            ``// increase depth``            ``if` `(str.charAt(i) == ``'('``)``                ``d++;` `            ``// If closing bracket``            ``else` `{``                ``if` `(d == ``1``) {``                    ``for` `(``int` `j = ``2``; j <= n / ``2` `+ ``1` `&& arr[j] != ``0``; j++)``                        ``arr[j] = ``0``;``                ``}``                ``++ans;``                ``ans += arr[d];``                ``arr[d]++;``                ``d--;``            ``}``        ``}` `        ``// Return the required answer``        ``return` `ans;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``String str = ``"()()()"``;``        ``int` `n = str.length();` `        ``// Function call``        ``System.out.println(Balanced_Substring(str, n));``    ``}``}` `// This code is contributed by``// sanjeev2552`

## Python3

 `# Python3 program to find number of``# balanced parenthesis sub strings` `# Function to find number of``# balanced parenthesis sub strings``def` `Balanced_Substring(s, n):` `    ``# To store required answer``    ``ans ``=` `0``;` `    ``# Vector to stores the number of``    ``# balanced brackets at each depth.``    ``arr ``=` `[``0``] ``*` `(``int``(n ``/` `2``) ``+` `1``);` `    ``# d strores checks the depth of our sequence``    ``# For example level of () is 1``    ``# and that of (()) is 2.``    ``d ``=` `0``;``    ``for` `i ``in` `range``(n):` `        ``# If open bracket``        ``# increase depth``        ``if` `(s[i] ``=``=` `'('``):``            ``d ``+``=` `1``;` `        ``# If closing bracket``        ``else``:``            ``if` `(d ``=``=` `1``):``                ``j ``=` `2``                ``while` `(j <``=` `n``/``/``2` `+` `1` `and` `arr[j] !``=` `0``):``                    ``arr[j] ``=` `0``            ``ans ``+``=` `1``;``            ``ans ``+``=` `arr[d];``            ``arr[d] ``+``=` `1``;``            ``d ``-``=` `1``;` `    ``# Return the required answer``    ``return` `ans;` `# Driver code``s ``=` `"()()()"``;``n ``=` `len``(s);` `# Function call``print``(Balanced_Substring(s, n));` `# This code contributed by Rajput-Ji`

## C#

 `// C# program to find number of``// balanced parenthesis sub strings``using` `System;` `class` `GFG {` `    ``// Function to find number of``    ``// balanced parenthesis sub strings``    ``public` `static` `int` `Balanced_Substring(String str,``                                         ``int` `n)``    ``{` `        ``// To store required answer``        ``int` `ans = 0;` `        ``// Vector to stores the number of``        ``// balanced brackets at each depth.``        ``int``[] arr = ``new` `int``[n / 2 + 1];` `        ``// d strores checks the depth of our sequence``        ``// For example level of () is 1``        ``// and that of (()) is 2.``        ``int` `d = 0;``        ``for` `(``int` `i = 0; i < n; i++) {` `            ``// If open bracket``            ``// increase depth``            ``if` `(str[i] == ``'('``)``                ``d++;` `            ``// If closing bracket``            ``else` `{``                ``if` `(d == 1) {``                    ``for` `(``int` `j = 2; j <= n / 2 + 1 && arr[j] != 0; j++)``                        ``arr[j] = 0;``                ``}``                ``++ans;``                ``ans += arr[d];``                ``arr[d]++;``                ``d--;``            ``}``        ``}` `        ``// Return the required answer``        ``return` `ans;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``String str = ``"()()()"``;``        ``int` `n = str.Length;` `        ``// Function call``        ``Console.WriteLine(Balanced_Substring(str, n));``    ``}``}` `// This code is contributed by Princi Singh`

## Javascript

 ``
Output:
`6`

Time complexity : O(N)

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