Check if a string contains a palindromic sub-string of even length
Last Updated :
16 Oct, 2023
S is string containing only lowercase English alphabets. We need to find if there exists at least one palindromic sub-string whose length is even.
Examples:
Input : aassss
Output : YES
Input : gfg
Output : NO
Approach:
Approach to solve this problem is to check all even-length substrings of the given string and check if any of them is a palindrome. This can be done using nested loops. The outer loop will iterate over all possible starting indices of even-length substrings, while the inner loop will iterate over all possible lengths of such substrings. At each iteration, we can check if the current substring is a palindrome.
- Define a function named "isPalindrome" that takes a string as input and returns a boolean indicating whether the given string is a palindrome or not.
- Define another function named "hasEvenLengthPalindrome" that takes a string as input and returns a boolean indicating whether there exists at least one even-length palindromic substring in the given string or not.
- Traverse the string from the start and fix a starting position "i".
- Traverse the string from the fixed starting position and fix a length of the substring "len" such that len is even and (i+len) <= n.
- Check if the substring starting from position i and of length len is a palindrome or not by calling the isPalindrome function. If yes, return true.
- If the loop completes successfully, return false.
C++
#include <iostream>
#include <string>
using namespace std;
// function to check if a given string is a palindrome
bool isPalindrome(string s)
{
int n = s.length();
for (int i = 0; i < n / 2; i++) {
if (s[i] != s[n - i - 1]) {
return false;
}
}
return true;
}
// function to check if there exists at least one even-length palindromic substring
bool hasEvenLengthPalindrome(string s)
{
int n = s.length();
for (int i = 0; i < n; i++) {
for (int len = 2; i + len <= n; len += 2) {
if (isPalindrome(s.substr(i, len))) {
return true;
}
}
}
return false;
}
int main()
{
string s = "xzyyz";
if (hasEvenLengthPalindrome(s)) {
cout << "YES" << endl;
} else {
cout << "NO" << endl;
}
return 0;
}
public class Main {
// function to check if a given string is a palindrome
public static boolean isPalindrome(String s) {
int n = s.length();
for (int i = 0; i < n / 2; i++) {
if (s.charAt(i) != s.charAt(n - i - 1)) {
return false;
}
}
return true;
}
// function to check if there exists at least one even-length palindromic substring
public static boolean hasEvenLengthPalindrome(String s) {
int n = s.length();
for (int i = 0; i < n; i++) {
for (int len = 2; i + len <= n; len += 2) {
if (isPalindrome(s.substring(i, i + len))) {
return true;
}
}
}
return false;
}
public static void main(String[] args) {
String s = "xzyyz";
if (hasEvenLengthPalindrome(s)) {
System.out.println("YES");
} else {
System.out.println("NO");
}
}
}
// This code is contributed by shivamgupta0987654321
# Function to check if a given string is a palindrome
def isPalindrome(s):
n = len(s)
for i in range(n // 2):
if s[i] != s[n - i - 1]:
return False
return True
# Function to check if there exists at least one even-length palindromic substring
def hasEvenLengthPalindrome(s):
n = len(s)
for i in range(n):
for length in range(2, n - i + 1, 2):
if isPalindrome(s[i:i + length]):
return True
return False
if __name__ == "__main__":
s = "xzyyz"
if hasEvenLengthPalindrome(s):
print("YES")
else:
print("NO")
using System;
public class GFG
{
// function to check if a given string is a palindrome
public static bool IsPalindrome(string s)
{
int n = s.Length;
for (int i = 0; i < n / 2; i++)
{
if (s[i] != s[n - i - 1])
{
return false;
}
}
return true;
}
// function to check if there exists at least one even-length palindromic substring
public static bool HasEvenLengthPalindrome(string s)
{
int n = s.Length;
for (int i = 0; i < n; i++)
{
for (int len = 2; i + len <= n; len += 2)
{
if (IsPalindrome(s.Substring(i, len)))
{
return true;
}
}
}
return false;
}
//Driver Code
public static void Main()
{
string s = "xzyyz";
if (HasEvenLengthPalindrome(s))
{
Console.WriteLine("YES");
}
else
{
Console.WriteLine("NO");
}
}
}
// function to check if a given string is a palindrome
function isPalindrome(s) {
const n = s.length;
for (let i = 0; i < Math.floor(n / 2); i++) {
if (s[i] !== s[n - i - 1]) {
return false;
}
}
return true;
}
// function to check if there exists at least one even-length palindromic substring
function hasEvenLengthPalindrome(s) {
const n = s.length;
for (let i = 0; i < n; i++) {
for (let len = 2; i + len <= n; len += 2) {
if (isPalindrome(s.substr(i, len))) {
return true;
}
}
}
return false;
}
// Driver code
const s = "xzyyz";
if (hasEvenLengthPalindrome(s)) {
console.log("YES");
} else {
console.log("NO");
}
Time Complexity: O(N^3), where N is the length of the string s. This is because we are using nested loops to generate all possible substrings of even length and then checking if each substring is a palindrome. The isPalindrome function has a time complexity of O(N/2) = O(N) as we are iterating over half of the string.
Space Complexity: O(1) as we are not using any extra data structure to store the substrings.
Notice that a palindrome of even length must contain two same alphabets in the middle. So we just need to check for this condition. If we find two consecutive same alphabets in the string then we output "YES" otherwise "NO".
Below is the implementation:
C++
// C++ program to check if there is a substring
// palindrome of even length.
#include <bits/stdc++.h>
using namespace std;
// function to check if two consecutive same
// characters are present
bool check(string s)
{
for (int i = 0; i < s.length() - 1; i++)
if (s[i] == s[i + 1])
return true;
return false;
}
int main()
{
string s = "xzyyz";
if (check(s))
cout << "YES" << endl;
else
cout << "NO" << endl;
return 0;
}
// Java program to check if there is a substring
// palindrome of even length.
class GFG {
// function to check if two consecutive same
// characters are present
static boolean check(String s)
{
for (int i = 0; i < s.length() - 1; i++)
if (s.charAt(i) == s.charAt(i+1))
return true;
return false;
}
// Driver Code
public static void main(String[] args) {
String s = "xzyyz";
if (check(s))
System.out.println("YES");
else
System.out.println("NO");
}
}
# Python 3 program to check if there is
# a substring palindrome of even length.
# function to check if two consecutive
# same characters are present
def check(s):
for i in range (0, len(s)):
if (s[i] == s[i + 1]):
return True
return False
# Driver Code
s = "xzyyz"
if(check(s)):
print("YES")
else:
print("NO")
# This code is contributed
# by iAyushRAJ
// C# program to check if there is a substring
// palindrome of even length.
using System;
public class GFG {
// function to check if two consecutive same
// characters are present
static bool check(String s)
{
for (int i = 0; i < s.Length - 1; i++)
if (s[i] == s[i+1])
return true;
return false;
}
// Driver Code
public static void Main() {
String s = "xzyyz";
if (check(s))
Console.WriteLine("YES");
else
Console.WriteLine("NO");
}
}
<script>
// Javascript program to check if there is
// a substring palindrome of even length.
// Function to check if two consecutive same
// characters are present
function check(s)
{
for(let i = 0; i < s.length - 1; i++)
if (s[i] == s[i + 1])
return true;
return false;
}
// Driver code
let s = "xzyyz";
if (check(s))
document.write("YES");
else
document.write("NO");
// This code is contributed by suresh07
</script>
<?php
// PHP program to check if there is a
// substring palindrome of even length.
// function to check if two consecutive
// same characters are present
function check($s)
{
for ($i = 0; $i < strlen($s) - 1; $i++)
if ($s[$i] == $s[$i + 1])
return true;
return false;
}
// Driver Code
$s = "xzyyz";
if (check($s))
echo "YES","\n";
else
echo "NO" ,"\n";
// This code is contributed by ajit
?>
Time complexity: O(N) where N is the length of the given string.
Auxiliary space: O(1), as constant extra space is used.
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