Dada una string de corchetes abiertos ‘(‘ y corchetes cerrados ‘)’. La tarea es encontrar la longitud del prefijo balanceado más largo.
Ejemplos:
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 Program
int 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 Code
str = '((()())())(('
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 Program
var str = "((()())())((";
var n = str.length;
document.write( maxbalancedprefix(str, n));
</script>