Longest palindrome substring in a string is a very common java interview question. To find out the longest palindrome in String, first of all, we need to identify the logic to do it.
Longest Palindrome Substring in a String Algorithm
The key point here is that from the mid of any palindrome string if we go to the right and left by 1 place, it’s always the same character. For example 12321, here mid is 3 and if we keep moving one position on both sides, we get 2 and then 1. We will use the same logic in our java program to find out the longest palindrome. However, if the palindrome length is even, the mid-size is also even. So we need to make sure in our program that this is also checked. For example, 12333321, here mid is 33 and if we keep moving one position in both sides, we get 3, 2 and 1.
Longest Palindrome Substring in a String Java Program
In our java program, we will iterate over the input string with mid as 1st place and check the right and left character. We will have two global variables to save the start and the end position for palindrome. We also need to check if there is already a longer palindrome found since there can we multiple palindromes in the given string. Here is the final program that works fine for all the cases.
package com.journaldev.util;
public class LongestPalindromeFinder {
public static void main(String[] args) {
System.out.println(longestPalindromeString("1234"));
System.out.println(longestPalindromeString("12321"));
System.out.println(longestPalindromeString("9912321456"));
System.out.println(longestPalindromeString("9912333321456"));
System.out.println(longestPalindromeString("12145445499"));
System.out.println(longestPalindromeString("1223213"));
System.out.println(longestPalindromeString("abb"));
}
static public String intermediatePalindrome(String s, int left, int right) {
if (left > right) return null;
while (left >= 0 && right < s.length()
&& s.charAt(left) == s.charAt(right)) {
left--;
right++;
}
return s.substring(left + 1, right);
}
// O(n^2)
public static String longestPalindromeString(String s) {
if (s == null) return null;
String longest = s.substring(0, 1);
for (int i = 0; i < s.length() - 1; i++) {
//odd cases like 121
String palindrome = intermediatePalindrome(s, i, i);
if (palindrome.length() > longest.length()) {
longest = palindrome;
}
//even cases like 1221
palindrome = intermediatePalindrome(s, i, i + 1);
if (palindrome.length() > longest.length()) {
longest = palindrome;
}
}
return longest;
}
}
Below image shows the output of the above longest palindrome java program. 
We can improve the above code by moving the palindrome and longest lengths check into a different function. However, I have left that part for you. :) Please let me know if there are any other better implementations or if it fails in any case.
You can download the complete example code from our GitHub Repository.
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public static boolean isPalindrome(String strToCheck){ int mid=-1; if (strToCheck.length()<2 && !strToCheck.isEmpty()){ return true; } else if (!strToCheck.isEmpty()){ mid = strToCheck.length() / 2; boolean isEven = strToCheck.length()%2==0; if (strToCheck.substring(0,mid).equals(new StringBuffer(strToCheck.substring((isEven?mid:mid+1), strToCheck.length())).reverse().toString())){ return true; } } return false; }
- Chetan R
public static boolean isPalindrome(String strToCheck){ int mid = strToCheck.length() / 2; boolean isEven = strToCheck.length()%2==0; if (strToCheck.substring(0,mid).equals(new StringBuffer(strToCheck.substring((isEven?mid:mid+1), strToCheck.length())).reverse().toString())){ return true; } return false; }
- Chetan R
package practisesessions; public class PracticeSession { public static boolean isEmpty(String s) { return s==null || s.length()==0; } public static String reverseString(String input) { String reverse=“”; for(int count=input.length()-1;count>=0;count–) { reverse +=input.charAt(count); } return reverse; } public static String lcp(String s, String t){ int n = Math.min(s.length(),t.length()); for(int i = 0; i < n; i++){ if(s.charAt(i) != t.charAt(i)){ return s.substring(0,i); } } return s.substring(0,n); } private static boolean longestPanlindrome(String substring, String reverse) { if(reverse.indexOf(substring)!=-1) { return true; }else { return false; } } public static void main(String…strings) { String input = “adbmadamthdmaadaamssszzzzzzzzzzsum”; String reverse = reverseString(input); String longestPalindrome=“”; if(!isEmpty(input)) { for(int i=0;i<input.length();i++) { for(int j=i+1;j<=input.length();j++) { if(longestPanlindrome(input.substring(i, j),reverse)) { if(longestPalindrome.length() < (input.substring(i, j)).length()) { longestPalindrome =input.substring(i, j); } }else { break; } } } } System.out.println("Longest Palindrome is "+longestPalindrome); } }
- Suman Madyala
For a given question to indemnify the longest palindrome in a given set of palindromes. can you please give us more detail like if we don’t find the longest palindrome then would be ? Do we need to see for the next second longest palindrome in a given set and so on ? Please clarify this
- Mohammed
package JAVA_Concepts; public class StringPalindrome { public Boolean checkPalindrome(String str) { StringBuilder strPal = new StringBuilder(str); // System.out.println("Original String " + strPal); // StringBuilder strRev = strPal.reverse(); StringBuilder strRev = new StringBuilder(strPal).reverse(); // System.out.println("Reversed String " + strRev.toString()); // System.out.println("Original String " + strPal.toString()); if(str.equals(strRev.toString())) { // System.out.println(“Its a Palindrome”); return true; }else { // System.out.println(“Its NOT a Palindrome”); return false; } } public Boolean interimPalindrome(String str) { String checkString = str; int lenStr = checkString.length(); // System.out.println(“Length Of String .:” + lenStr); int MaxLen = 0; int startindex = 0; int endindex = 0; boolean isPalindrome = false; for(int i=0;i<lenStr;i++) { char fnd = checkString.charAt(i); int indexofnd = checkString.indexOf(fnd); int indexofNxtfnd = checkString.indexOf(fnd, indexofnd+1); // int lastIndexoffnd = checkString.lastIndexOf(fnd); // System.out.println("FirstIndex.: " + indexofnd); while(indexofNxtfnd != -1) { // System.out.println("SecondIndex.: " + indexofNxtfnd); if((checkPalindrome(str.substring(indexofnd, indexofNxtfnd+1)))) { isPalindrome = true; if(MaxLen < str.substring(indexofnd, indexofNxtfnd+1).length()) { MaxLen = str.substring(indexofnd, indexofNxtfnd+1).length(); startindex = indexofnd; endindex = indexofNxtfnd+1; } } indexofNxtfnd = checkString.indexOf(fnd, indexofNxtfnd+1); } } if(isPalindrome) { System.out.println(“Max length Palindrome.:” + MaxLen); System.out.println(“Palindrome.:” + str.substring(startindex, endindex)); return true; }else { System.out.println(“There is No Palindrome”); return false; } } public static void main(String[] args) { // TODO Auto-generated method stub StringPalindrome onj1 = new StringPalindrome(); // System.out.println(onj1.checkPalindrome(“1234”)); onj1.interimPalindrome(“1234”); onj1.interimPalindrome(“12321”); onj1.interimPalindrome(“9912321456”); onj1.interimPalindrome(“9912333321456”); onj1.interimPalindrome(“12145445499”); onj1.interimPalindrome(“1223213”); onj1.interimPalindrome(“abb”); } }
- Satyadip Paul
import java.util.*; import java.lang.*; import java.io.*; class longpal { public static void main (String[] args) throws java.lang.Exception { int n=0,i,j; String s,d=" "; Scanner in=new Scanner(System.in); s=in.nextLine(); for(i=0;i<s.length()-1;i++) { for(j=i+1;jn) { n=j-i; d=s.substring(i,j); } } } System.out.print(d); } }
- Vismaya
Hi Panksj, How to find the start index of the longest palindrome in your code?
- JJ
Please comment on my program and let me know if I missed anything package stringprograms; public class LongestPalindrome { public static void main(String[] args) { String name = “mohanabbaganesh”; String longestPalindrome = “”; for (int i = 0; i < name.length(); ++i) { String tempString = “”; tempString = tempString + name.charAt(i); for (int j = i + 1; j longestPalindrome.length()) longestPalindrome = tempString; } } } System.out.println("longestPalindrome – " + longestPalindrome); } }
- mohan
String st = “abbccbba”; System.out.println(st + " is " + (st.equals(new StringBuilder(st).reverse().toString()) ? “Palindrom.” : “not Palindrom.”));
- krissn
String st = “abbccbba”; String out = “”; for (int i = st.length() - 1; i >= 0; i–) { out += st.charAt(i); } System.out.println(out + " is Palindrom.");
- krissn