Length of longest balanced parentheses prefix
Given a string of open bracket ‘(‘ and closed bracket ‘)’. The task is to find the length of longest balanced prefix.
Examples:
Input : S = "((()())())(("
Output : 10
From index 0 to index 9, they are forming
a balanced parentheses prefix.
Input : S = "()(())((()"
Output : 6
The idea is take value of open bracket ‘(‘ as 1 and value of close bracket ‘)’ as -1. Now start finding the prefix sum of the given string. The farthest index, say maxi, where the value of sum is 0 is the index upto which longest balanced prefix exists. So the answer would be maxi + 1.
Below is the implementation of this approach:
C++
// CPP Program to find length of longest balanced// parentheses prefix.#include <bits/stdc++.h>using namespace std;// Return the length of longest balanced parentheses// prefix.int maxbalancedprefix(char str[], int n){ int sum = 0; int maxi = 0; // Traversing the string. for (int i = 0; i < n; i++) { // If open bracket add 1 to sum. if (str[i] == '(') sum += 1; // If closed bracket subtract 1 // from sum else sum -= 1; // if first bracket is closing bracket // then this condition would help if (sum < 0) break; // If sum is 0, store the index // value. if (sum == 0) maxi = i + 1; } return maxi;}// Driven Programint main(){ char str[] = "((()())())(("; int n = strlen(str); cout << maxbalancedprefix(str, n) << endl; return 0;} |
Java
// Java Program to find length of longest// balanced parentheses prefix.import java.io.*;class GFG { // Return the length of longest // balanced parentheses prefix. static int maxbalancedprefix(String str, int n) { int sum = 0; int maxi = 0; // Traversing the string. for (int i = 0; i < n; i++) { // If open bracket add 1 to sum. if (str.charAt(i) == '(') sum += 1; // If closed bracket subtract 1 // from sum else sum -= 1; // if first bracket is closing bracket // then this condition would help if (sum < 0) break; // If sum is 0, store the index // value. if (sum == 0) maxi = i + 1; } return maxi; } // Driven Program public static void main(String[] args) { String str = "((()())())(("; int n = str.length(); System.out.println(maxbalancedprefix(str, n)); }}// This code is contributed by vt_m |
Python3
# Python3 code to find length of# longest balanced parentheses prefix.# Function to return the length of# longest balanced parentheses prefix.def maxbalancedprefix (str, n): _sum = 0 maxi = 0 # Traversing the string. for i in range(n): # If open bracket add 1 to sum. if str[i] == '(': _sum += 1 # If closed bracket subtract 1 # from sum else: _sum -= 1 # if first bracket is closing bracket # then this condition would help if _sum < 0: break # If sum is 0, store the # index value. if _sum == 0: maxi = i + 1 return maxi # Driver Codestr = '((()())())(('n = len(str)print(maxbalancedprefix (str, n))# This code is contributed by "Abhishek Sharma 44" |
C#
// C# Program to find length of longest// balanced parentheses prefix.using System;class GFG { // Return the length of longest // balanced parentheses prefix. static int maxbalancedprefix(string str, int n) { int sum = 0; int maxi = 0; // Traversing the string. for (int i = 0; i < n; i++) { // If open bracket add 1 to sum. if (str[i] == '(') sum += 1; // If closed bracket subtract 1 // from sum else sum -= 1; // if first bracket is closing bracket // then this condition would help if (sum < 0) break; // If sum is 0, store the index // value. if (sum == 0) maxi = i + 1; } return maxi; } // Driven Program public static void Main() { string str = "((()())())(("; int n = str.Length; Console.WriteLine(maxbalancedprefix(str, n)); }}// This code is contributed by vt_m |
PHP
<?php// PHP Program to find length// of longest balanced// parentheses prefix.// Return the length of longest// balanced parentheses prefix.function maxbalancedprefix($str, $n){ $sum = 0; $maxi = 0; // Traversing the string. for ($i = 0; $i <$n; $i++) { // If open bracket add 1 to sum. if ($str[$i] == '(') $sum += 1; // If closed bracket subtract 1 // from sum else $sum -= 1; if ($sum < 0) break; // If sum is 0, store the index // value. if ($sum == 0) $maxi = $i+1; } return $maxi;}// Driver Code$str = array('(', '(', '(', ')', '(', ')', ')', '(', ')', ')', '(', '(');$n = count($str);echo maxbalancedprefix($str, $n);// This code is contributed by anuj_67..?> |
Javascript
<script>// Javascript Program to find// length of longest balanced// parentheses prefix.// Return the length of longest// balanced parentheses// prefix.function maxbalancedprefix( str, n){ var sum = 0; var maxi = 0; // Traversing the string. for (var i = 0; i < n; i++) { // If open bracket add 1 to sum. if (str[i] == '(') sum += 1; // If closed bracket subtract 1 // from sum else sum -= 1; // if first bracket is closing bracket // then this condition would help if (sum < 0) break; // If sum is 0, store the index // value. if (sum == 0) maxi = i + 1; } return maxi;}// Driven Programvar str = "((()())())((";var n = str.length;document.write( maxbalancedprefix(str, n));</script> |
Output:
10
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