Dada una string numérica S formada únicamente por los caracteres ‘1’, ‘2’ y ‘3’ , la tarea es reemplazar los caracteres con un corchete abierto ( ‘(‘ ) o un corchete cerrado ( ‘)’ ) de modo que el la string recién formada se convierte en una secuencia de paréntesis balanceada .
Nota: Todas las ocurrencias de un carácter deben ser reemplazadas por los mismos paréntesis.
Ejemplos:
Entrada: S = “1123”
Salida: Sí, (())
Explicación: Reemplazo de ocurrencias del carácter ‘1’ con ‘(‘, ‘2’ con ‘)’ y ‘3’ con ‘)’. Por lo tanto, la secuencia de paréntesis obtenida es “(())”, que está balanceada.Entrada: S = “1121”
Salida : No
Enfoque: El problema dado se puede resolver con base en las siguientes observaciones:
- Para una secuencia de corchetes balanceada , es necesario que el primer y último carácter sean corchetes abiertos y cerrados respectivamente. Por lo tanto, el primero y el último carácter deben ser diferentes.
- Si el primero y el último carácter de una string son iguales, entonces es imposible obtener una secuencia de paréntesis balanceada.
- Si el primer y el último carácter de una string son diferentes, se reemplazan por corchetes abiertos y cerrados respectivamente. El tercer carácter se reemplaza por corchetes abiertos o cerrados .
- Verifique las dos formas de reemplazo uno por uno para el tercer carácter restante.
- Si ambos reemplazos del tercer carácter restante no pueden hacer una secuencia de paréntesis balanceada, entonces es imposible hacer una secuencia de paréntesis balanceada.
Siga los pasos a continuación para resolver el problema dado:
- Compruebe si el primer y el último carácter de la string S son iguales o no. Si se encuentra que es cierto, escriba “No” y regrese.
- Inicialice dos variables, digamos cntforOpen y cntforClose , para almacenar el recuento de paréntesis abiertos y cerrados.
- Iterar sobre los caracteres de la string y realizar las siguientes operaciones:
- Si el carácter actual es el mismo que el primer carácter de la string, incremente cntforOpen.
- Si el carácter actual es el mismo que el último carácter de la string, disminuya cntforOpen.
- Para el tercer carácter restante, incremente cntforOpen , es decir, reemplace ese carácter con ‘(‘ .
- Si en algún momento se encuentra que cntforOpen es negativo, entonces no se puede obtener una secuencia de paréntesis balanceada.
- De manera similar, verifique usando la variable cntforClose , es decir, reemplazando el tercer carácter con ‘)’ .
- Si ninguno de los dos métodos anteriores genera una secuencia de paréntesis equilibrada, imprima «No» . De lo contrario, escriba «Sí».
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include <iostream>
using namespace std;
// Function to check if the given
// string can be converted to a
// balanced bracket sequence or not
void balBracketSequence(string str)
{
int n = str.size();
// Check if the first and
// last characters are equal
if (str[0]
== str[n - 1])
{
cout << "No" << endl;
}
else {
// Initialize two variables to store
// the count of open and closed brackets
int cntForOpen = 0, cntForClose = 0;
int check = 1;
for (int i = 0; i < n; i++) {
// If the current character is
// same as the first character
if (str[i] == str[0])
cntForOpen++;
// If the current character is
// same as the last character
else if (str[i] == str[n - 1])
cntForOpen--;
else
cntForOpen++;
// If count of open brackets
// becomes less than 0
if (cntForOpen < 0) {
check = 0;
break;
}
}
if (check && cntForOpen == 0) {
cout << "Yes, ";
// Print the new string
for (int i = 0; i < n; i++) {
if (str[i] == str[n - 1])
cout << ')';
else
cout << '(';
}
return;
}
else {
for (int i = 0; i < n; i++) {
// If the current character is
// same as the first character
if (str[i] == str[0])
cntForClose++;
else
cntForClose--;
// If bracket sequence
// is not balanced
if (cntForClose
< 0) {
check = 0;
break;
}
}
// Check for unbalanced
// bracket sequence
if (check
&& cntForClose
== 0) {
cout << "Yes, ";
// Print the sequence
for (int i = 0; i < n;
i++) {
if (str[i] == str[0])
cout << '(';
else
cout << ')';
}
return;
}
}
cout << "No";
}
}
// Driver Code
int main()
{
// Given Input
string str = "123122";
// Function Call
balBracketSequence(str);
return 0;
}
Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to check if the given
// string can be converted to a
// balanced bracket sequence or not
static void balBracketSequence(String str)
{
int n = str.length();
// Check if the first and
// last characters are equal
if (str.charAt(0)
== str.charAt(n - 1))
{
System.out.println("No");
}
else {
// Initialize two variables to store
// the count of open and closed brackets
int cntForOpen = 0, cntForClose = 0;
int check = 1;
for (int i = 0; i < n; i++) {
// If the current character is
// same as the first character
if (str.charAt(i) == str.charAt(0))
cntForOpen++;
// If the current character is
// same as the last character
else if (str.charAt(i) == str.charAt(n - 1))
cntForOpen -= 1;
else
cntForOpen += 1;
// If count of open brackets
// becomes less than 0
if (cntForOpen < 0) {
check = 0;
break;
}
}
if (check != 0 && cntForOpen == 0) {
System.out.print("Yes, ");
// Print the new string
for (int i = 0; i < n; i++) {
if (str.charAt(i) == str.charAt(n - 1))
System.out.print(')');
else
System.out.print('(');
}
return;
}
else {
for (int i = 0; i < n; i++) {
// If the current character is
// same as the first character
if (str.charAt(i) == str.charAt(0))
cntForClose++;
else
cntForClose--;
// If bracket sequence
// is not balanced
if (cntForClose
< 0) {
check = 0;
break;
}
}
// Check for unbalanced
// bracket sequence
if (check != 0
&& cntForClose
== 0) {
System.out.print("Yes, ");
// Print the sequence
for (int i = 0; i < n;
i++) {
if (str.charAt(i) == str.charAt(0))
System.out.print('(');
else
System.out.print(')');
}
return;
}
}
System.out.print("No");
}
}
// Driver Code
public static void main(String args[])
{
// Given Input
String str = "123122";
// Function Call
balBracketSequence(str);
}
}
// This code is contributed by ipg2016107.
Python3
# Python program for the above approach;
# Function to check if the given
# string can be converted to a
# balanced bracket sequence or not
def balBracketSequence(str):
n = len(str)
# Check if the first and
# last characters are equal
if (str[0] == str[n - 1]):
print("No", end="")
else:
# Initialize two variables to store
# the count of open and closed brackets
cntForOpen = 0
cntForClose = 0
check = 1
for i in range(n):
# If the current character is
# same as the first character
if (str[i] == str[0]):
cntForOpen += 1
# If the current character is
# same as the last character
elif str[i] == str[n - 1] :
cntForOpen -= 1
else:
cntForOpen += 1
# If count of open brackets
# becomes less than 0
if (cntForOpen < 0):
check = 0
break
if (check and cntForOpen == 0):
print("Yes, ", end="")
# Print the new string
for i in range(n):
if (str[i] == str[n - 1]):
print(')', end="")
else:
print('(', end="")
return
else:
for i in range(n):
# If the current character is
# same as the first character
if (str[i] == str[0]):
cntForClose += 1
else:
cntForClose -= 1
# If bracket sequence
# is not balanced
if (cntForClose < 0):
check = 0
break
# Check for unbalanced
# bracket sequence
if (check and cntForClose == 0):
print("Yes, ", end="")
# Print the sequence
for i in range(n):
if (str[i] == str[0]):
print('(', end="")
else:
print(')', end="")
return
print("NO", end="")
# Driver Code
# Given Input
str = "123122"
# Function Call
balBracketSequence(str)
# This code is contributed by gfgking
C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to check if the given
// string can be converted to a
// balanced bracket sequence or not
static void balBracketSequence(string str)
{
int n = str.Length;
// Check if the first and
// last characters are equal
if (str[0] == str[n - 1])
{
Console.Write("No");
}
else
{
// Initialize two variables to store
// the count of open and closed brackets
int cntForOpen = 0, cntForClose = 0;
int check = 1;
for(int i = 0; i < n; i++)
{
// If the current character is
// same as the first character
if (str[i] == str[0])
cntForOpen++;
// If the current character is
// same as the last character
else if (str[i] == str[n - 1])
cntForOpen--;
else
cntForOpen++;
// If count of open brackets
// becomes less than 0
if (cntForOpen < 0)
{
check = 0;
break;
}
}
if (check != 0 && cntForOpen == 0)
{
Console.Write("Yes, ");
// Print the new string
for(int i = 0; i < n; i++)
{
if (str[i] == str[n - 1])
Console.Write(')');
else
Console.Write('(');
}
return;
}
else
{
for(int i = 0; i < n; i++)
{
// If the current character is
// same as the first character
if (str[i] == str[0])
cntForClose++;
else
cntForClose--;
// If bracket sequence
// is not balanced
if (cntForClose < 0)
{
check = 0;
break;
}
}
// Check for unbalanced
// bracket sequence
if (check != 0 && cntForClose == 0)
{
Console.Write("Yes, ");
// Print the sequence
for(int i = 0; i < n; i++)
{
if (str[i] == str[0])
Console.Write('(');
else
Console.Write(')');
}
return;
}
}
Console.Write("No");
}
}
// Driver Code
public static void Main()
{
// Given Input
string str = "123122";
// Function Call
balBracketSequence(str);
}
}
// This code is contributed by sanjoy_62
Javascript
<script>
// JavaScript program for the above approach;
// Function to check if the given
// string can be converted to a
// balanced bracket sequence or not
function balBracketSequence(str)
{
let n = str.length;
// Check if the first and
// last characters are equal
if (str[0]
== str[n - 1])
{
document.write( "No");
}
else {
// Initialize two variables to store
// the count of open and closed brackets
let cntForOpen = 0, cntForClose = 0;
let check = 1;
for (let i = 0; i < n; i++) {
// If the current character is
// same as the first character
if (str[i] == str[0])
cntForOpen++;
// If the current character is
// same as the last character
else if (str[i] == str[n - 1])
cntForOpen--;
else
cntForOpen++;
// If count of open brackets
// becomes less than 0
if (cntForOpen < 0) {
check = 0;
break;
}
}
if (check && cntForOpen == 0) {
document.write("Yes, ");
// Print the new string
for (let i = 0; i < n; i++) {
if (str[i] == str[n - 1])
document.write(')');
else
document.write('(');
}
return;
}
else {
for (let i = 0; i < n; i++) {
// If the current character is
// same as the first character
if (str[i] == str[0])
cntForClose++;
else
cntForClose--;
// If bracket sequence
// is not balanced
if (cntForClose
< 0) {
check = 0;
break;
}
}
// Check for unbalanced
// bracket sequence
if (check
&& cntForClose
== 0) {
document.write("Yes, ");
// Print the sequence
for (let i = 0; i < n;
i++) {
if (str[i] == str[0])
document.write('(');
else
document.write(')');
}
return;
}
}
document.write("NO") ;
}
}
// Driver Code
// Given Input
let str = "123122";
// Function Call
balBracketSequence(str);
// This code is contributed by Potta Lokesh
</script>
Producción:
Yes,()(())
Complejidad temporal: O(N)
Espacio auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por ramashishkushwaha819 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA