Check if all the palindromic sub-strings are of odd length

Given a string ‘s’ check if all of its palindromic sub-strings are of odd length or not. If yes then print “YES” or “NO” otherwise.

Examples:

Input: str = “geeksforgeeks” 
Output: NO 
Since, “ee” is a palindromic sub-string of even length.

Input: str = “madamimadam” 
Output: YES 

Brute Force Approach: Simply, iterate over each sub-string of ‘s’ and check if it is a palindrome. If it is a palindrome then it must of odd length.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include<bits//stdc++.h>
using namespace std;

// Function to check if
// the string is palindrome
bool checkPalindrome(string s)
{
    for (int i = 0; i < s.length(); i++)
    {
        if(s[i] != s[s.length() - i - 1])
            return false;
    }
    return true;
}

// Function that checks whether
// all the palindromic
// sub-strings are of odd length.
bool CheckOdd(string s)
{
int n = s.length();
for (int i = 0; i < n; i++)
{

    // Creating each substring
    string x = "";
    for (int j = i; j < n; j++)
    {
        x += s[j];

        // If the sub-string is
        // of even length and
        // is a palindrome then,
        // we return False
        if(x.length() % 2 == 0 &&
        checkPalindrome(x) == true)
            return false;
        }
    }

    return true;
}

// Driver code
int main()
{
    string s = "geeksforgeeks";
    if(CheckOdd(s))
        cout<<("YES");
    else
        cout<<("NO");
}
// This code is contributed by
// Sahil_shelangia

Java

// Java implementation of the approach
import java.util.*;
class GFG
{

// Function to check if
// the string is palindrome
static boolean checkPalindrome(String s)
{
    for (int i = 0; i < s.length(); i++)
    {
        if(s.charAt(i) != s.charAt(s.length() - i - 1))
            return false;
    }
    return true;
}

// Function that checks whether
// all the palindromic
// sub-strings are of odd length.
static boolean CheckOdd(String s)
{
int n = s.length();
for (int i = 0; i < n; i++)
{

    // Creating each substring
    String x = "";
    for (int j = i; j < n; j++)
    {
        x += s.charAt(j);

        // If the sub-string is
        // of even length and
        // is a palindrome then,
        // we return False
        if(x.length() % 2 == 0 &&
        checkPalindrome(x) == true)
            return false;
        }
    }

    return true;
}

// Driver code
public static void main(String args[])
{
    String s = "geeksforgeeks";
    if(CheckOdd(s))
        System.out.print("YES");
    else
        System.out.print("NO");
}
}

// This code is contributed
// by Arnab Kundu

Python

# Python implementation of the approach

# Function to check if
# the string is palindrome


def checkPalindrome(s):
    for i in range(len(s)):
        if(s[i] != s[len(s)-i-1]):
            return False
    return True

# Function that checks whether
# all the palindromic
# sub-strings are of odd length.


def CheckOdd(s):
    n = len(s)
    for i in range(n):

        # Creating each substring
        x = ""
        for j in range(i, n):
            x += s[j]
            # If the sub-string is
            # of even length and
            # is a palindrome then,
            # we return False
            if(len(x) % 2 == 0
                    and checkPalindrome(x) == True):
                return False
    return True


# Driver code
s = "geeksforgeeks"
if(CheckOdd(s)):
    print("YES")
else:
    print("NO")

C#

// C# implementation of the approach
using System;


public class GFG {

// Function to check if
// the string is palindrome
static bool checkPalindrome(String s)
{
    for (int i = 0; i < s.Length; i++)
    {
        if(s[i] != s[(s.Length - i - 1)])
            return false;
    }
    return true;
}

// Function that checks whether
// all the palindromic
// sub-strings are of odd length.
static bool CheckOdd(String s)
{
int n = s.Length;
for (int i = 0; i < n; i++)
{

    // Creating each substring
    String x = "";
    for (int j = i; j < n; j++)
    {
        x += s[j];

        // If the sub-string is
        // of even length and
        // is a palindrome then,
        // we return False
        if(x.Length % 2 == 0 &&
        checkPalindrome(x) == true)
            return false;
        }
    }

    return true;
}

// Driver code
public static void Main()
{
    String s = "geeksforgeeks";
    if(CheckOdd(s))
        Console.Write("YES");
    else
        Console.Write("NO");
}
}

/* This code is contributed by 29AjayKumar*/

PHP

<?php
// PHP implementation of the approach

// Function to check if the string
// is palindrome
function checkPalindrome($s)
{
    for ($i = 0; $i < strlen($s); $i++)
    {
        if($s[$i] != $s[strlen($s) - $i - 1])
            return false;
    }
    return true;
}

// Function that checks whether all the
// palindromic sub-strings are of odd length.
function CheckOdd($s)
{
    $n = strlen($s);
    for ($i = 0; $i < $n; $i++)
    {

        // Creating each substring
        $x = "";
        for ($j = $i; $j < $n; $j++)
        {
            $x = $x.$s[$i];

            // If the sub-string is
            // of even length and
            // is a palindrome then,
            // we return False
            if(strlen($x) % 2 == 0 &&
            checkPalindrome($x) == true)
                return false;
        }
    }

    return true;
}

// Driver code
$s = "geeksforgeeks";
if(CheckOdd($s))
    echo "YES";
else
    echo "NO";

// This code is contributed by ita_c
?>

Javascript

<script>

// JavaScript implementation of the approach

// Function to check if
// the string is palindrome
function checkPalindrome(s)
{
    for (let i = 0; i < s.length; i++)
    {
        if(s[i] != s[s.length - i - 1])
            return false;
    }
    return true;
}

// Function that checks whether
// all the palindromic
// sub-strings are of odd length.
function CheckOdd(s)
{
    let n = s.length;
for (let i = 0; i < n; i++)
{

    // Creating each substring
    let x = "";
    for (let j = i; j < n; j++)
    {
        x += s[j];

        // If the sub-string is
        // of even length and
        // is a palindrome then,
        // we return False
        if(x.length % 2 == 0 &&
        checkPalindrome(x) == true)
            return false;
        }
    }

    return true;
}

// Driver code
let s = "geeksforgeeks";
if(CheckOdd(s))
    document.write("YES");
else
    document.write("NO");


// This code is contributed by avanitrachhadiya2155

</script>

Output

NO

Time Complexity: O(n³), where n is the length of the given string.
Auxiliary Space: O(n), where n is the length of the given string.

Efficient Approach: To check if all palindromic substrings of s have odd lengths, we can search for an even length palindromic substring of it. We know that every even length palindrome has at least two consecutive characters that are identical (e.g. cxxa, ee). Therefore, we can check two consecutive characters at a time to see if they are the same. If so, then s has an even length palindromic substring and hence output will be NO, and if we find no even length substring the answer will be YES. We can complete this checking after one string traversal.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include<bits//stdc++.h>
using namespace std;

// Function that checks whether s
// contains a even length palindromic
// sub-strings or not.
bool CheckEven(string s)
{
  for (int i = 1; i < s.size(); ++i) {
        if (s[i] == s[i - 1]) {
            return true;
        }
    }
    return false;
}

// Driver code
int main()
{
    string s = "geeksforgeeks";
    if(CheckEven(s)==false)
        cout<<("YES");
    else
        cout<<("NO");
}
// This code is contributed by
// Aditya Jaiswal

Java

// Java implementation of the approach
import java.util.*;
class GFG
{

// Function to check if
// the string is palindrome
static boolean checkPalindrome(String s)
{
    for (int i = 0; i < s.length(); i++)
    {
        if(s.charAt(i) != s.charAt(s.length() - i - 1))
            return false;
    }
    return true;
}

// Function that checks whether
// all the palindromic
// sub-strings are of odd length.
static boolean CheckOdd(String s)
{
int n = s.length();
for (int i = 0; i < n; i++)
{

    // Creating each substring
    String x = "";
    for (int j = i; j < n; j++)
    {
        x += s.charAt(j);

        // If the sub-string is
        // of even length and
        // is a palindrome then,
        // we return False
        if(x.length() % 2 == 0 &&
           checkPalindrome(x) == true)
            return false;
        }
    }

    return true;
}

// Driver code
public static void main(String args[])
{
    String s = "geeksforgeeks";
    if(CheckOdd(s))
        System.out.print("YES");
    else
        System.out.print("NO");
}
}

// This code is contributed
// by Arnab Kundu

Python3

# Python implementation of the approach

# Function to check if
# the string is palindrome
def checkPalindrome(s):
    for i in range(len(s)):
        if(s[i] != s[len(s)-i-1]):
            return False
    return True

# Function that checks whether
# all the palindromic
# sub-strings are of odd length.
def CheckOdd(s):
    n = len(s)
    for i in range(n):

        # Creating each substring
        x = ""
        for j in range(i, n):
            x += s[j]
            # If the sub-string is
            # of even length and
            # is a palindrome then,
            # we return False
            if(len(x)% 2 == 0
                  and checkPalindrome(x) == True):
                return False
    return True

# Driver code
s = "geeksforgeeks"
if(CheckOdd(s)):
    print("YES")
else:
    print("NO")

C#

// C# implementation of the approach
using System;


public class GFG {

// Function to check if
// the string is palindrome
static bool checkPalindrome(String s)
{
    for (int i = 0; i < s.Length; i++)
    {
        if(s[i] != s[(s.Length - i - 1)])
            return false;
    }
    return true;
}

// Function that checks whether
// all the palindromic
// sub-strings are of odd length.
static bool CheckOdd(String s)
{
int n = s.Length;
for (int i = 0; i < n; i++)
{

    // Creating each substring
    String x = "";
    for (int j = i; j < n; j++)
    {
        x += s[j];

        // If the sub-string is
        // of even length and
        // is a palindrome then,
        // we return False
        if(x.Length % 2 == 0 &&
           checkPalindrome(x) == true)
            return false;
        }
    }

    return true;
}

// Driver code
public static void Main()
{
    String s = "geeksforgeeks";
    if(CheckOdd(s))
        Console.Write("YES");
    else
        Console.Write("NO");
}
}

/* This code is contributed by 29AjayKumar*/

PHP

<?php
// PHP implementation of the approach

// Function to check if the string
// is palindrome
function checkPalindrome($s)
{
    for ($i = 0; $i < strlen($s); $i++)
    {
        if($s[$i] != $s[strlen($s) - $i - 1])
            return false;
    }
    return true;
}

// Function that checks whether all the
// palindromic sub-strings are of odd length.
function CheckOdd($s)
{
    $n = strlen($s);
    for ($i = 0; $i < $n; $i++)
    {

        // Creating each substring
        $x = "";
        for ($j = $i; $j < $n; $j++)
        {
            $x = $x.$s[$i];

            // If the sub-string is
            // of even length and
            // is a palindrome then,
            // we return False
            if(strlen($x) % 2 == 0 &&
            checkPalindrome($x) == true)
                return false;
        }
    }

    return true;
}

// Driver code
$s = "geeksforgeeks";
if(CheckOdd($s))
    echo "YES";
else
    echo "NO";

// This code is contributed by ita_c
?>

Javascript

<script>
// Javascript implementation of the approach

// Function to check if
// the string is palindrome
function  checkPalindrome(s)
{
    for (let i = 0; i < s.length; i++)
    {
        if(s[i] != s[(s.length - i - 1)])
            return false;
    }
    return true;
}

// Function that checks whether
// all the palindromic
// sub-strings are of odd length.
function CheckOdd(s)
{
    let n = s.length;
for (let i = 0; i < n; i++)
{

    // Creating each substring
    let x = "";
    for (let j = i; j < n; j++)
    {
        x += s[j];

        // If the sub-string is
        // of even length and
        // is a palindrome then,
        // we return False
        if(x.length % 2 == 0 &&
           checkPalindrome(x) == true)
            return false;
        }
    }

    return true;
}

// Driver code
let s = "geeksforgeeks";
if(CheckOdd(s))
    document.write("YES");
else
    document.write("NO");


// This code is contributed by rag2127
</script>

Output

NO

Time Complexity: O(N)
Auxiliary Space: O(1)