Java Program to Check Whether a String is a Palindrome
Last Updated :
12 Jul, 2025
A string in Java can be called a palindrome if we read it from forward or backward, it appears the same or in other words, we can say if we reverse a string and it is identical to the original string for example we have a string s = "jahaj " and when we reverse it s = "jahaj"(reversed) so they look identical so we can say that "jahaj" is a palindrome string. In this article, we will go through different approaches to check if a string is a palindrome in Java.
Example Input/Output:
Input: s = "level"
Output: True
Input: s = "Geeks"
Output: True
Input s = "G"
Output: True.
Brute Force Approach
The brute force or naive approach to check if a string is a palindrome is by reversing the string, and then we can compare it with the original. Here, to ensure the comparison is case-insensitive, the string is converted to lowercase before the check. For example, if we do not convert the string to lowercase so it won't work with the different cases, such as Level and level, it will give false, but it is a palindrome.
Algorithm:
- First, convert the string to lowercase to ensure a case-insensitive comparison.
- Reverse the string using a loop.
- Compare the original string with the reversed String. If both are equal, then it is a palindrome. Otherwise, it is not.
Example: Java program to check whether a string is a palindrome by comparing it with its reversed string.
Java
// Java Program to Check if a
// String is a Palindrome
import java.io.*;
public class Geeks
{
// Method to check if a string is a palindrome
public static boolean isPalindrome(String s) {
// Convert string to lowercase for
// case-insensitive comparison
s = s.toLowerCase();
// Reverse the string
String rev = "";
for (int i = s.length() - 1; i >= 0; i--) {
rev = rev + s.charAt(i);
}
// Compare the original string with
// the reversed string
return s.equals(rev);
}
public static void main(String[] args) {
// Input string
String s = "level";
// Check if the string is a palindrome
boolean res = isPalindrome(s);
// Print the result with enhanced output
if (res) {
System.out.println('"' + s + '"' + " is a palindrome.");
} else {
System.out.println('"' + s + '"' + " is not a palindrome.");
}
}
}
Output"level" is a palindrome.
Explanation: In the above example, It checks if the given string is palindrome by converting it to lowercase, reversing it, and then comparing the reversed string with the original.
- Time Complexity: O(n)
- Space Complexity: O(n)
Other Methods to Check a String Palindrome
1. Two Pointer Approach to Check if incrementa String Is Palindrome
This approach uses two pointers, one starting at the beginning "i" and the other at the end "j" of the string. By comparing characters at these pointers and moving them inward, we can determine if the string is a palindrome or not.
Algorithm:
- We initialize two pointers (i=0) from the start and (j = s.length -1) from the end.
- Compare the characters at index i and j, if both are not equal then return false( it is not a palindrome).
- If they match then increment i by 1 and decrement j by 1 and continue the process.
- If all the characters match and the loop stops at ( i = j), then it is a palindrome and returns true.
Example:
Java
// Java program to check whether a
// string is a Palindrome
// Using two pointing variables
public class Geeks
{
// Method to check if a string is a palindrome
public static boolean isPalindrome(String s) {
int i = 0, j = s.length() - 1;
// Compare characters while i < j
while (i < j) {
if (s.charAt(i) != s.charAt(j)) {
return false;
}
i++;
j--;
}
return true;
}
public static void main(String[] args) {
// Input strings
String s1 = "geeks";
String s2 = "Racecar";
// Convert strings to lowercase for
// case-insensitive comparison
s1 = s1.toLowerCase();
s2 = s2.toLowerCase();
// Check and print results for s1
if (isPalindrome(s1)) {
System.out.println("\"" + s1 + "\" is a palindrome.");
} else {
System.out.println("\"" + s1 + "\" is not a palindrome.");
}
// Check and print results for s2
if (isPalindrome(s2)) {
System.out.println("\"" + s2 + "\" is a palindrome.");
} else {
System.out.println("\"" + s2 + "\" is not a palindrome.");
}
}
}
Output"geeks" is not a palindrome.
"racecar" is a palindrome.
Explanation: In this example, the isPalindrome method checks for mismatched characters while i < j and returns false if found. The main method converts input strings to lowercase and prints whether they are palindrome.
- Time Complexity: O(n)
- Space Complexity: O(1)
2. Recursive Approach to Check if a String is Palindrome
Now for recursion we are using the same approach as we used in the two pointer, here we create two pointer I from start and j from the end in the recursion. and Increaement the i by one and decrement j by 1, if the characters at the index i not match the character present at index j then it is not a palidrome if it the base condition hit i=j and
Algorithm:
- We will take two pointers, "i" pointing to the start of the string and "j" pointing to the end of the string.
- Base Case: If i >= j, the recursion stops, as the string has been fully checked, and it is confirmed to be a palindrome.
- Character Comparison: Check if the characters at positions i and j are equal.
- If not, return false immediately (the string is not a palindrome).
- If they match, make a recursive call with the next set of indices (i + 1 and j - 1).
- The recursion continues until the base case is satisfied.
If all characters match until the pointers meet or cross, the string is a palindrome. Otherwise, it is not.
Example:
Java
// Java program to check whether a
// string is a Palindrome using recursion
public class Geeks
{
// Recursive method to check
// if a string is a palindrome
public static boolean isPalindrome(int i, int j, String s) {
// If pointers have crossed,
// it's a palindrome
if (i >= j) {
return true;
}
// If characters at i and j are not the same,
// return false
if (s.charAt(i) != s.charAt(j)) {
return false;
}
// Recursive call for the
//next pair of pointers
return isPalindrome(i + 1, j - 1, s);
}
// Overloaded method to simplify the call
public static boolean isPalindrome(String s) {
return isPalindrome(0, s.length() - 1, s);
}
public static void main(String[] args) {
// Input strings
String s1 = "geeks";
String s2 = "Racecar";
// Convert strings to lowercase for
// case-insensitive comparison
s1 = s1.toLowerCase();
s2 = s2.toLowerCase();
// Check and print results for s1
if (isPalindrome(s1)) {
System.out.println("\"" + s1 + "\" is a palindrome.");
} else {
System.out.println("\"" + s1 + "\" is not a palindrome.");
}
// Check and print results for s2
if (isPalindrome(s2)) {
System.out.println("\"" + s2 + "\" is a palindrome.");
} else {
System.out.println("\"" + s2 + "\" is not a palindrome.");
}
}
}
Output"geeks" is not a palindrome.
"racecar" is a palindrome.
- Time Complexity: O(n)
- Space Complexity: O(n)
3. StringBuilder Approach to Check a String is Palindrome
In this approach, we will use the StringBuilder class to reverse a string and compare it with the original string. Below are the steps:
- First we create an object of StringBuilder using the input string.
- Then we use the reverse() method of StringBuilder class to reverse the string.
- Now, convert the reversed string back to a String using the method toString().
- Compare the reversed string with the original string using the equals() method.
- If the equals method returns true then it is a palindrome otherwise it is a not a palindrome.
Example:
Java
// Java program to check if a
// string is a palindrome using StringBuilder
public class Geeks
{
public static void main(String[] args) {
// Input string
String s = "GeeksForGeeks";
// Create a StringBuilder object
//with the original string
StringBuilder s1 = new StringBuilder(s);
// Reverse the string
// using the reverse() method
s1.reverse();
// Compare the reversed string
// with the original string
if (s.equals(s1.toString())) {
System.out.println("\"" + s + "\" is a palindrome string.");
} else {
System.out.println("\"" + s + "\" is not a palindrome string.");
}
}
}
Output"GeeksForGeeks" is not a palindrome string.
- Time Complexity: O(n)
- Space Complexity: O(n)
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