Verifique que la string dada sea extrañamente palíndromo o no – Part 1

Dada la string str , la tarea es verificar si los caracteres en los índices impares de str forman una string palíndromo o no. Si no, escriba «No» , de lo contrario, escriba «Sí» .
Ejemplos: 
 

Entrada: str = “osafdfgsg”, N = 9 
Salida: Sí 
Explicación: 
Los caracteres impares indexados son = { s, f, f, s }, 
por lo que creará una string palindrómica, “sffs”.
Entrada: str = «addwfefwkll», N = 11 
Salida: No 
Explicación: 
Los caracteres impares indexados son = {d, w, e, w, l}, 
por lo que no generará una string de palíndromo, «dwewl» 
 

Enfoque ingenuo: el enfoque ingenuo es crear una nueva string agregando caracteres impares indexados de la string dada. Luego, simplemente verifique si la cuerda formada es palindrómica o no. Si la string es palindrómica, imprima «Sí» , de lo contrario, imprima «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 the string str
// is palindromic or not
bool isPalindrome(string str)
{
 
    // Iterate the string str from left
    // and right pointers
    int l = 0;
    int h = str.size() - 1;
 
    // Keep comparing characters
    // while they are same
    while (h > l) {
 
        // If they are not same
        // then return false
        if (str[l++] != str[h--]) {
            return false;
        }
    }
 
    // Return true if the string is
    // palindromic
    return true;
}
 
// Function to make string using odd
// indices of string str
string makeOddString(string str)
{
    string odd = "";
    for (int i = 1; i < str.size();
         i += 2) {
        odd += str[i];
    }
 
    return odd;
}
 
// Functions checks if characters at
// odd index of the string forms
// palindrome or not
void checkOddlyPalindrome(string str)
{
 
    // Make odd indexed string
    string odd = makeOddString(str);
 
    // Check for Palindrome
    if (isPalindrome(odd))
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
}
 
// Driver Code
int main()
{
    // Given string
    string str = "ddwfefwde";
 
    // Function Call
    checkOddlyPalindrome(str);
    return 0;
}

// Java program for the above approach
class GFG{
     
// Function to check if the String str
// is palindromic or not
public static boolean isPalindrome(String str)
{
     
    // Iterate the String str from left
    // and right pointers
    int l = 0;
    int h = str.length() - 1;
 
    // Keep comparing characters
    // while they are same
    while (h > l)
    {
         
        // If they are not same
        // then return false
        if (str.charAt(l) != str.charAt(h))
        {
            return false;
        }
         
        l++;
        h--;
    }
 
    // Return true if the String is
    // palindromic
    return true;
}
 
// Function to make String using odd
// indices of String str
public static String makeOddString(String str)
{
    String odd = "";
     
    for(int i = 1; i < str.length(); i += 2)
    {
       odd += str.charAt(i);
    }
    return odd;
}
 
// Functions checks if characters at
// odd index of the String forms
// palindrome or not
public static void checkOddlyPalindrome(String str)
{
 
    // Make odd indexed String
    String odd = makeOddString(str);
 
    // Check for Palindrome
    if (isPalindrome(odd))
        System.out.println("Yes");
    else
        System.out.println("No");
}
 
// Driver Code
public static void main(String []args)
{
     
    // Given String
    String str = "ddwfefwde";
 
    // Function Call
    checkOddlyPalindrome(str);
}
}
 
// This code is contributed by grand_master

# Python3 program for the above approach
 
# Function to check if the string 
# str is palindromic or not
def isPalindrome(str):
 
    # Iterate the string str from
    # left and right pointers
    l = 0;
    h = len(str) - 1;
 
    # Keep comparing characters
    # while they are same
    while (h > l):
 
        # If they are not same
        # then return false
        if (str[l] != str[h]):
            return False;
             
        l += 1
        h -= 1
         
    # Return true if the string is
    # palindromic
    return True;
 
# Function to make string using odd
# indices of string str
def makeOddString(str):
 
    odd = "";
    for i in range(1, len(str), 2):
        odd += str[i];
     
    return odd;
 
# Functions checks if characters at
# odd index of the string forms
# palindrome or not
def checkOddlyPalindrome(str):
 
    # Make odd indexed string
    odd = makeOddString(str);
 
    # Check for Palindrome
    if (isPalindrome(odd)):
        print("Yes")
    else:
        print("No")
 
# Driver code
 
# Given string
str = "ddwfefwde";
 
# Function call
checkOddlyPalindrome(str);
 
# This code is contributed by grand_master

// C# program for the above approach
using System;
 
class GFG{
     
// Function to check if the String str
// is palindromic or not
static bool isPalindrome(string str)
{
     
    // Iterate the String str from left
    // and right pointers
    int l = 0;
    int h = str.Length - 1;
 
    // Keep comparing characters
    // while they are same
    while (h > l)
    {
         
        // If they are not same
        // then return false
        if (str[l] != str[h])
        {
            return false;
        }
         
        l++;
        h--;
    }
 
    // Return true if the String is
    // palindromic
    return true;
}
 
// Function to make String using odd
// indices of String str
static string makeOddString(string str)
{
    string odd = "";
     
    for(int i = 1; i < str.Length; i += 2)
    {
        odd += str[i];
    }
    return odd;
}
 
// Functions checks if characters at
// odd index of the String forms
// palindrome or not
static void checkOddlyPalindrome(string str)
{
 
    // Make odd indexed String
    string odd = makeOddString(str);
 
    // Check for Palindrome
    if (isPalindrome(odd))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
 
// Driver code
static void Main()
{
     
    // Given String
    string str = "ddwfefwde";
     
    // Function Call
    checkOddlyPalindrome(str);
}
}
 
// This code is contributed by divyeshrabadiya07

<script>
 
// Javascript program for the above approach
 
// Function to check if the string str
// is palindromic or not
function isPalindrome(str)
{
 
    // Iterate the string str from left
    // and right pointers
    var l = 0;
    var h = str.length - 1;
 
    // Keep comparing characters
    // while they are same
    while (h > l) {
 
        // If they are not same
        // then return false
        if (str[l++] != str[h--]) {
            return false;
        }
    }
 
    // Return true if the string is
    // palindromic
    return true;
}
 
// Function to make string using odd
// indices of string str
function makeOddString(str)
{
    var odd = "";
    for (var i = 1; i < str.length;
         i += 2) {
        odd += str[i];
    }
 
    return odd;
}
 
// Functions checks if characters at
// odd index of the string forms
// palindrome or not
function checkOddlyPalindrome(str)
{
 
    // Make odd indexed string
    var odd = makeOddString(str);
 
    // Check for Palindrome
    if (isPalindrome(odd))
        document.write( "Yes" );
    else
        document.write( "No" );
}
 
// Driver Code
 
// Given string
var str = "ddwfefwde";
 
// Function Call
checkOddlyPalindrome(str);
 
// This code is contributed by itsok.
</script>

Yes

Complejidad de tiempo: O(N) , N es la longitud de la string. 
Espacio Auxiliar: O(N/2)