Number of balanced parenthesis substrings

Difficulty Level : Hard
Last Updated : 24 Apr, 2020

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

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

Examples :

Input : str = “()()()”
Output : 6
(), (), (), ()(), ()(), ()()()

Input : str = “(())()”
Output : 4
(), (()), (), (())()

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

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 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 

Output:

Time complexity : O(N)

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