Compruebe si la string es palíndromo después de eliminar todos los duplicados consecutivos

Dada una string str , la tarea es eliminar todos los duplicados consecutivos de la string str y verificar si la string final es palíndromo o no. Escriba “Sí” si es un palindrómico, de lo contrario escriba “No” .
Ejemplos: 
 

Entrada: str = “abbcbbbaaa” 
Salida: Sí 
Explicación: 
Al eliminar todos los caracteres duplicados consecutivos, la string se convierte en “abcba”, que es un palíndromo.
Entrada: str = “aaabbbaaccc” 
Salida: No 
Explicación: 
Al eliminar todos los caracteres duplicados consecutivos, la string se convierte en “abac”, que no es un palíndromo. 
 

Enfoque: La idea es crear una nueva string a partir de la string dada y verificar si la nueva string es palindrómica o no. A continuación se muestran los pasos:
 

• Inicialice la nueva string newStr = “” .
• Repita todos los caracteres de la string dada uno por uno y si el carácter actual es diferente del carácter anterior, agregue este carácter a la nueva string newStr .
• De lo contrario, busque el siguiente carácter.
• Compruebe si la string final formada es palíndromo o no. Escriba “Sí” si es un palindrómico, de lo contrario escriba “No” .

A continuación se muestra la implementación del enfoque anterior:
 

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if a string
// is palindrome or not
bool isPalindrome(string str)
{
    // Length of the string
    int len = str.length();
 
    // Check if its a palindrome
    for (int i = 0; i < len; i++) {
 
        // If the palindromic
        // condition is not met
        if (str[i] != str[len - i - 1])
            return false;
    }
 
    // Return true as str is palindromic
    return true;
}
 
// Function to check if string str is
// palindromic after removing every
// consecutive characters from the str
bool isCompressablePalindrome(string str)
{
    // Length of the string str
    int len = str.length();
 
    // Create an empty
    // compressed string
    string compressed = "";
 
    // The first character will
    // always be included in
    // the final string
    compressed.push_back(str[0]);
 
    // Check all the characters
    // of the string
    for (int i = 1; i < len; i++) {
 
        // If the current character
        // is not same as its previous
        // one, then insert it in the
        // final string
        if (str[i] != str[i - 1])
            compressed.push_back(str[i]);
    }
 
    // Check if the compressed
    // string is a palindrome
    return isPalindrome(compressed);
}
 
// Driver Code
int main()
{
 
    // Given string
    string str = "abbcbbbaaa";
 
    // Function call
    if (isCompressablePalindrome(str))
        cout << "Yes\n";
    else
        cout << "No\n";
 
    return 0;
}

// Java program for the above approach
import java.util.*;
class GFG{
 
// Function to check if a String
// is palindrome or not
static boolean isPalindrome(String str)
{
    // Length of the String
    int len = str.length();
 
    // Check if its a palindrome
    for (int i = 0; i < len; i++)
    {
 
        // If the palindromic
        // condition is not met
        if (str.charAt(i) != str.charAt(len - i - 1))
            return false;
    }
 
    // Return true as str is palindromic
    return true;
}
 
// Function to check if String str is
// palindromic after removing every
// consecutive characters from the str
static boolean isCompressablePalindrome(String str)
{
    // Length of the String str
    int len = str.length();
 
    // Create an empty
    // compressed String
    String compressed = "";
 
    // The first character will
    // always be included in
    // the final String
    compressed = String.valueOf(str.charAt(0));
 
    // Check all the characters
    // of the String
    for (int i = 1; i < len; i++)
    {
 
        // If the current character
        // is not same as its previous
        // one, then insert it in the
        // final String
        if (str.charAt(i) != str.charAt(i - 1))
            compressed += str.charAt(i);
    }
 
    // Check if the compressed
    // String is a palindrome
    return isPalindrome(compressed);
}
 
// Driver Code
public static void main(String[] args)
{
 
    // Given String
    String str = "abbcbbbaaa";
 
    // Function call
    if (isCompressablePalindrome(str))
        System.out.print("Yes\n");
    else
        System.out.print("No\n");
}
}
 
// This code is contributed by amal kumar choubey

# Python3 program for the above approach
 
# Function to check if a string
# is palindrome or not
def isPalindrome(Str):
 
    # Length of the string
    Len = len(Str)
 
    # Check if its a palindrome
    for i in range(Len):
 
        # If the palindromic
        # condition is not met
        if (Str[i] != Str[Len - i - 1]):
            return False
 
    # Return true as Str is palindromic
    return True
 
# Function to check if string str is
# palindromic after removing every
# consecutive characters from the str
def isCompressablePalindrome(Str):
 
    # Length of the string str
    Len = len(Str)
 
    # Create an empty compressed string
    compressed = ""
 
    # The first character will always
    # be included in final string
    compressed += Str[0]
 
    # Check all the characters
    # of the string
    for i in range(1, Len):
 
        # If the current character
        # is not same as its previous
        # one, then insert it in
        # the final string
        if (Str[i] != Str[i - 1]):
            compressed += Str[i]
 
    # Check if the compressed
    # string is a palindrome
    return isPalindrome(compressed)
 
# Driver Code
if __name__ == '__main__':
     
    # Given string
    Str = "abbcbbbaaa"
 
    # Function call
    if (isCompressablePalindrome(Str)):
        print("Yes")
    else:
        print("No")
 
# This code is contributed by himanshu77

// C# program for the above approach
using System;
 
class GFG{
 
// Function to check if a String
// is palindrome or not
static bool isPalindrome(String str)
{
     
    // Length of the String
    int len = str.Length;
 
    // Check if its a palindrome
    for(int i = 0; i < len; i++)
    {
        
       // If the palindromic
       // condition is not met
       if (str[i] != str[len - i - 1])
           return false;
    }
 
    // Return true as str is palindromic
    return true;
}
 
// Function to check if String str is
// palindromic after removing every
// consecutive characters from the str
static bool isCompressablePalindrome(String str)
{
     
    // Length of the String str
    int len = str.Length;
 
    // Create an empty
    // compressed String
    String compressed = "";
 
    // The first character will
    // always be included in
    // the readonly String
    compressed = String.Join("", str[0]);
 
    // Check all the characters
    // of the String
    for(int i = 1; i < len; i++)
    {
        
       // If the current character
       // is not same as its previous
       // one, then insert it in the
       // readonly String
       if (str[i] != str[i - 1])
           compressed += str[i];
    }
 
    // Check if the compressed
    // String is a palindrome
    return isPalindrome(compressed);
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given String
    String str = "abbcbbbaaa";
 
    // Function call
    if (isCompressablePalindrome(str))
        Console.Write("Yes\n");
    else
        Console.Write("No\n");
}
}
 
// This code is contributed by amal kumar choubey

<script>
    // Javascript program for the above approach
     
    // Function to check if a String
    // is palindrome or not
    function isPalindrome(str)
    {
 
        // Length of the String
        let len = str.length;
 
        // Check if its a palindrome
        for(let i = 0; i < len; i++)
        {
 
           // If the palindromic
           // condition is not met
           if (str[i] != str[len - i - 1])
               return false;
        }
 
        // Return true as str is palindromic
        return true;
    }
 
    // Function to check if String str is
    // palindromic after removing every
    // consecutive characters from the str
    function isCompressablePalindrome(str)
    {
 
        // Length of the String str
        let len = str.length;
 
        // Create an empty
        // compressed String
        let compressed = "";
 
        // The first character will
        // always be included in
        // the readonly String
        compressed = str[0];
 
        // Check all the characters
        // of the String
        for(let i = 1; i < len; i++)
        {
 
           // If the current character
           // is not same as its previous
           // one, then insert it in the
           // readonly String
           if (str[i] != str[i - 1])
               compressed += str[i];
        }
 
        // Check if the compressed
        // String is a palindrome
        return isPalindrome(compressed);
    }
     
    // Given String
    let str = "abbcbbbaaa";
   
    // Function call
    if (isCompressablePalindrome(str))
        document.write("Yes" + "</br>");
    else
        document.write("No");
 
// This code is contributed by divyeshrabadiya07.
</script>

Yes

Complejidad de tiempo: O(N) , donde N es la longitud de la string.