Check if a given string is Even-Odd Palindrome or not

Given a string str, the task is to check if the given string is Even-Odd Palindrome or not. 

An Even-Odd Palindrome string is defined to be a string whose characters at even indices form a Palindrome while the characters at odd indices also form a Palindrome separately. 

Examples:

Input: str=”abzzab” 
Output:YES 
Explanation:
String formed by characters at odd indices: bzb, which is a Palindrome. 
String formed by characters at even indices: aza, which is a Palindrome. 
Hence, the given string is an Even-Odd Palindrome.
Input: str=”daccad” 
Output: NO 

Approach: To solve the problem, create a new String by appending the Odd Indexed Characters of the given string and check if the strings formed are palindromic or not. Similarly, check for Even Indexed Characters. If both the strings are palindromic, then print “Yes”. Otherwise, print “No”.
Below is the implementation of the above approach: 

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>

Output:

Yes

Time Complexity: O(n)

Auxiliary Space: O(n)