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 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 |
6
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.