Number of balanced parenthesis substrings

Difficulty Level : Hard
Last Updated : 26 Jul, 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 : 6
(), (), (), ()(), ()(), ()()()
Input : str = “(())()”
Output : 4
(), (()), (), (())()

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 <bits/stdc++.h>
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 stores 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 stores 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 stores 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 stores 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

<script>
 
// Javascript program to find number of
// balanced parenthesis sub strings
 
// Function to find number of
// balanced parenthesis sub strings
function Balanced_Substring(str, n)
{
    // To store required answer
    let ans = 0;
 
    // Vector to stores the number of
    // balanced brackets at each depth.
    let arr = new Array(n / 2 + 1).fill(0);
 
    // d stores checks the depth of our sequence
    // For example level of () is 1
    // and that of (()) is 2.
    let d = 0;
    for (let i = 0; i < n; i++) {
        // If open bracket
        // increase depth
        if (str[i] == '(')
            d++;
 
        // If closing bracket
        else {
            if (d == 1) {
                for (let j = 2; j <= parseInt(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
    let str = "()()()";
 
    let n = str.length;
 
    // Function call
    document.write(Balanced_Substring(str, n));
 
</script>

Output:

Time complexity : O(N)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes