Dada una string str , la tarea es verificar si la string dada es Palindrome par-impar o no.
Una string de palíndromo par-impar se define como una string cuyos caracteres en índices pares forman un palíndromo mientras que los caracteres en índices impares también forman un palíndromo por separado.
Ejemplos:
Entrada: str=”abzzab”
Salida: SI
Explicación:
String formada por caracteres en índices impares: bzb , que es un Palíndromo.
String formada por caracteres en índices pares: aza , que es un palíndromo.
Por lo tanto, la string dada es un palíndromo par-impar.
Entrada: str=”daccad”
Salida: NO
Enfoque: para resolver el problema, cree una nueva string agregando los caracteres impares indexados de la string dada y verifique si las strings formadas son palindrómicas o no. Del mismo modo, verifique los caracteres indexados uniformes . Si ambas strings son palindrómicas, imprima «Sí» . De lo contrario, escriba “No” .
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to check if the string
// str is palindromic or not
bool isPalindrome(string str)
{
// Pointers to iterate the
// string from both ends
int l = 0;
int h = str.size() - 1;
while (h > l) {
// If characters are found
// to be distinct
if (str[l++] != str[h--]) {
return false;
}
}
// Return true if the
// string is palindromic
return true;
}
// Function to generate string
// from characters at odd indices
string makeOddString(string str)
{
string odd = "";
for (int i = 1; i < str.size();
i += 2) {
odd += str[i];
}
return odd;
}
// Function to generate string
// from characters at even indices
string makeevenString(string str)
{
string even = "";
for (int i = 0; i < str.size();
i += 2) {
even += str[i];
}
return even;
}
// Functions to checks if string
// is Even-Odd Palindrome or not
void checkevenOddPalindrome(string str)
{
// Generate odd indexed string
string odd = makeOddString(str);
// Generate even indexed string
string even = makeevenString(str);
// Check for Palindrome
if (isPalindrome(odd)
&& isPalindrome(even))
cout << "Yes" << endl;
else
cout << "No" << endl;
}
// Driver Code
int main()
{
string str = "abzzab";
checkevenOddPalindrome(str);
return 0;
}
Java
// Java program implementation
// of the approach
import java.util.*;
import java.io.*;
class GFG{
// Function to check if the string
// str is palindromic or not
static boolean isPalindrome(String str)
{
// Pointers to iterate the
// string from both ends
int l = 0;
int h = str.length() - 1;
while (h > l)
{
// If characters are found
// to be distinct
if (str.charAt(l++) !=
str.charAt(h--))
return false;
}
// Return true if the
// string is palindromic
return true;
}
// Function to generate string
// from characters at odd indices
static String makeOddString(String str)
{
String odd = "";
for(int i = 1; i < str.length(); i += 2)
{
odd += str.charAt(i);
}
return odd;
}
// Function to generate string
// from characters at even indices
static String makeevenString(String str)
{
String even = "";
for(int i = 0; i < str.length(); i += 2)
{
even += str.charAt(i);
}
return even;
}
// Functions to checks if string
// is Even-Odd Palindrome or not
static void checkevenOddPalindrome(String str)
{
// Generate odd indexed string
String odd = makeOddString(str);
// Generate even indexed string
String even = makeevenString(str);
// Check for Palindrome
if (isPalindrome(odd) && isPalindrome(even))
System.out.println("Yes");
else
System.out.println("No");
}
// Driver code
public static void main(String[] args)
{
String str = "abzzab";
checkevenOddPalindrome(str);
}
}
// This code is contributed by sanjoy_62
Python3
# Python3 program to implement
# the above approach
# Function to check if the string
# str is palindromic or not
def isPalindrome(Str):
# Pointers to iterate the
# string from both ends
l = 0
h = len(Str) - 1
while (h > l):
# If characters are found
# to be distinct
if (Str[l] != Str[h]):
return False
l += 1
h -= 1
# Return true if the
# string is palindromic
return True
# Function to generate string
# from characters at odd indices
def makeOddString(Str):
odd = ""
for i in range(1, len(Str), 2):
odd += Str[i]
return odd
# Function to generate string
# from characters at even indices
def makeevenString(Str):
even = ""
for i in range(0, len(Str), 2):
even += Str[i]
return even
# Functions to checks if string
# is Even-Odd Palindrome or not
def checkevenOddPalindrome(Str):
# Generate odd indexed string
odd = makeOddString(Str)
# Generate even indexed string
even = makeevenString(Str)
# Check for Palindrome
if (isPalindrome(odd) and
isPalindrome(even)):
print("Yes")
else:
print("No")
# Driver code
Str = "abzzab"
checkevenOddPalindrome(Str)
# This code is contributed by himanshu77
C#
// C# program implementation
// of the approach
using System;
class GFG{
// Function to check if the string
// str is palindromic or not
static bool isPalindrome(string str)
{
// Pointers to iterate the
// string from both ends
int l = 0;
int h = str.Length - 1;
while (h > l)
{
// If characters are found
// to be distinct
if (str[l++] != str[h--])
return false;
}
// Return true if the
// string is palindromic
return true;
}
// Function to generate string
// from characters at odd indices
static string makeOddString(string str)
{
string odd = "";
for(int i = 1; i < str.Length; i += 2)
{
odd += str[i];
}
return odd;
}
// Function to generate string
// from characters at even indices
static string makeevenString(string str)
{
string even = "";
for(int i = 0; i < str.Length; i += 2)
{
even += str[i];
}
return even;
}
// Functions to checks if string
// is Even-Odd Palindrome or not
static void checkevenOddPalindrome(string str)
{
// Generate odd indexed string
string odd = makeOddString(str);
// Generate even indexed string
string even = makeevenString(str);
// Check for Palindrome
if (isPalindrome(odd) && isPalindrome(even))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
// Driver code
public static void Main()
{
string str = "abzzab";
checkevenOddPalindrome(str);
}
}
// This code is contributed by sanjoy_62
Javascript
<script>
// Javascript program implementation
// of the approach
// Function to check if the string
// str is palindromic or not
function isPalindrome(str)
{
// Pointers to iterate the
// string from both ends
var l = 0;
var h = str.length - 1;
while (h > l)
{
// If characters are found
// to be distinct
if (str.charAt(l++) !=
str.charAt(h--))
return false;
}
// Return true if the
// string is palindromic
return true;
}
// Function to generate string
// from characters at odd indices
function makeOddString(str)
{
var odd = "";
for(var i = 1; i < str.length; i += 2)
{
odd += str.charAt(i);
}
return odd;
}
// Function to generate string
// from characters at even indices
function makeevenString(str)
{
var even = "";
for(var i = 0; i < str.length; i += 2)
{
even += str.charAt(i);
}
return even;
}
// Functions to checks if string
// is Even-Odd Palindrome or not
function checkevenOddPalindrome(str)
{
// Generate odd indexed string
var odd = makeOddString(str);
// Generate even indexed string
var even = makeevenString(str);
// Check for Palindrome
if (isPalindrome(odd) && isPalindrome(even))
document.write("Yes");
else
document.write("No");
}
// Driver code
var str = "abzzab";
checkevenOddPalindrome(str);
// This code is contributed by Khushboogoyal499
</script>
Producción:
Yes
Complejidad de tiempo: O(n)
Espacio Auxiliar: O(n)
Publicación traducida automáticamente
Artículo escrito por grand_master y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA