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.
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.
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.
Thanks for learning with the DigitalOcean Community. Check out our offerings for compute, storage, networking, and managed databases.
While we believe that this content benefits our community, we have not yet thoroughly reviewed it. If you have any suggestions for improvements, please let us know by clicking the “report an issue“ button at the bottom of the tutorial.
Hi, Thank you for this post, but your code doesn’t work for cases like: input: abb output should be bb, but your code gives a input: bb output should be bb, but your code gives b I couldn’t figure out why? Do you think it’s possible to make your code work for these two cases?
- Peihong
Hi, I think there is two bugs in the code above. 1. when checking the mid with size 2, the original code does not check the two characters in the mid. So left should be initialized to mid and right should be initialized to mid + 1. 2. the logic of expanding left and right is not correct. In the code above, when the left and right characters are not the same, it does not end the while loop, instead it just jump over them and keep going. It should change to: while (left >= 0 && right longestRight - longestLeft){ longestLeft = left; longestRight = right; } left–; right++; }
- Zhenxiang Liang
for input 1223213 i am getting answer as 122321. Am i missing something?
- abhijit
for input 1223213 i am getting answer as 122321. Am i missing something? please reply to me at gaikwad.abhijit@gmail.com
- abhijit
sir plz tell me a code of java. Q.write a java program to add two numbers without using addition operator and utility function. plz reply on 12cs85@quest.edu.pk
- NAND LAL
public static int add(int a, int b){ if(b == 0) return a; int sum = a ^ b; //SUM of two integer is A XOR B int carry = (a & b) << 1; //CARRY of two integer is A AND B return add(sum, carry); }
- Anuj K
Your program is very perfect and it is useful to me
- hthahseen
package Practice; import java.util.HashSet; public class Permutationwithplaindrome { /** * @param args */ public static void main(String[] args) { String str=“abcaaa”; HashSet hset = new HashSet(); for(int i=0;i<str.length();i++){ for(int j=i+1;j0){ for(int i=0,j=str.length()-1;i<=j;i++,j–){ if(str.charAt(i)==str.charAt(j)){ continue; }else return false; } return true; } return false; } }
- sandeep kumar
private static boolean isPalindromeString(String str) { if (str == null) return false; int length = str.length(); System.out.println(length / 2); for (int i = 0; i < length / 2; i++) { if (str.charAt(i) != str.charAt(length - i - 1)) return false; } return true; }
- jhon
jhon… you code perfect works and it’s simple. :)
- Anand
it just finds whether string is palindrome or not
- madhushree
your code just searching the longest palindrome ,i mean ur code just search a single no
- manish kumar singh
how can u use return in main method cos in main method hv void nd u cnt return ny-thin using void
- manish kumar singh
This is just wrong, you are finding for pairs of letters and if you find some pair, even if the others don’t match in between that pair of letters, you accept it as a palindrome
- Carlos
Thanks for all the comments, I have found out the bug in the code and update it to work fine in all the cases it was failing before.
- Pankaj
Hi, Please can you explain a bit more in detail how this has been done… Regards, Rishi Chopra.
- Rishi Chopra
He is finding palindromes centering around all characters in the String 1 by 1. Just keeps the longest Palindrome centered at a character and then moves on the next character. I hope I was able to help a bit.
- Akshay
Thanks brother but it has taken some time to understand. Eventhough we are not writing 2nd if loop we are getting answer for both even and odd strings.
- PRASADARAO
Q.The sum of two number without add operator and Any other utility utility fun and the answer given by Anuj k is helpful
- hthahseen
Why is 1 palindrome in ur program? You strongly believe 1 is palin???
- Ellephy
a number is said to be a palindrome if that number and its reverse are equal. so every single digit number is palindrome.
- Vivek Agrawal
Modified version of above code no need to call intermediatepalindrome method two times . package practice.palindrome; import java.util.ArrayList; import java.util.List; 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 List intermediatePalindrome(String s, int left, int right) { List list=new ArrayList(); int secondleft=left,secondright=right+1; if (left > right) return null; while (((left >= 0 && right < s.length())&&s.charAt(left)==s.charAt(right))|| ( (secondright~0)&&s.charAt(secondleft)==s.charAt(secondright))) { if(s.charAt(left)==s.charAt(right)) { left–; right++; } if(s.charAt(secondleft)==s.charAt(secondright)) { secondleft–; secondright++; } } list.add(s.substring(left + 1, right)); list.add(s.substring(secondleft + 1, secondright)); return list; } // 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 List palindromelist = intermediatePalindrome(s, i, i); for(String palindrome:palindromelist) { if (palindrome.length() > longest.length()) { longest = palindrome; } } } return longest; } }~
- Vivek Agrawal
class palindrome { public static void main(String args[]) { string original,reverse=" "; System.out.println(“Enter the String”); Scanner ash = new Scanner(System.in); original =ash.nextInt(); original =ash.nextLine(); for(int i=length-1;i>=0;i–) reverse=reverse+original.ChartAt(i); if(original.equals(reverse)) System.out.println(“enter the palindrme”); else System.out.println(“not palindrome”); } }
- Ashwini Ghanghav
Can we solve it the below way: I am getting the output correct. If not this way can you please help me what I did wrong, it will help me to code in a systematic way in future, Thanks. /* * To change this license header, choose License Headers in Project Properties. * To change this template file, choose Tools | Templates * and open the template in the editor. */ package javaapplication15; import java.io.*; import java.util.HashSet; import java.util.*; import java.util.Set; import java.util.logging.Filter; import java.util.logging.LogRecord; /** * * @author Vishal */ public class JavaApplication15 { public static HashSet set = new HashSet(); /** * @param args the command line arguments */ public static void main(String[] args) throws Exception { ArrayList al = new ArrayList(); al.add(“ab2ba”); al.add(“abc2cba”); al.add(“abd2dba”); al.add(“wab2baw”); al.add(“lzyadb2bdayzl”); al.add(“ooab2baoo”); for (int i = 0; i < al.size(); i++) { plaindrome(al.get(i)); } } public static void plaindrome(String str) { System.out.println(str); boolean flag = true; int mid = str.length()/2; System.out.println(mid); for (int i = 1; i <= mid; i++) //*** needs to be equalto { Character char1 = str.charAt(mid+i); System.out.println(“char1=”+char1); Character char2 = str.charAt(mid-i); System.out.println(“char2=”+char2); if(!char1.equals(char2)) { flag = false; } } if(flag == true) { System.out.println(“It is a plaindrome”); set.add(str); } else { System.out.println(“It is Not a plaindrome”); } longestPlaindrome(); } public static void longestPlaindrome() { Iterator itr = set.iterator(); String temp =“”; while(itr.hasNext()) { for (int i = 0; i temp.length()) { temp = s1; } } } System.out.println(“The longest plaindrome is =”+temp); } } Output: run: ab2ba 2 char1=b char2=b char1=a char2=a It is a plaindrome ab2ba abc2cba 3 char1=c char2=c char1=b char2=b char1=a char2=a It is a plaindrome abc2cba abd2dba 3 char1=d char2=d char1=b char2=b char1=a char2=a It is a plaindrome abc2cba wab2baw 3 char1=b char2=b char1=a char2=a char1=w char2=w It is a plaindrome abc2cba lzyadb2bdayzl 6 char1=b char2=b char1=d char2=d char1=a char2=a char1=y char2=y char1=z char2=z char1=l char2=l It is a plaindrome lzyadb2bdayzl ooab2baoo 4 char1=b char2=b char1=a char2=a char1=o char2=o char1=o char2=o It is a plaindrome The longest plaindrome is =lzyadb2bdayzl BUILD SUCCESSFUL (total time: 0 seconds)
- Vishal
Your isPalindrome() method will work wrong if the length of the string is even.
- karthikeya
1223213 for this string palindrome combinations can be 22 , 232, 1 etc. But y u we r getting output only 232…could u pls fix this error i really want to know how to solve this program.
- prema
static String check(String s) { String longestPalindrome=“”; for (int i = 0; i ~=i; j–) { if (s.charAt(i) == s.charAt(j)) { String ss=s.substring(i, j+1); if(checkPalindrome(ss)) longestPalindrome=(longestPalindrome.length()<ss.length())?ss:longestPalindrome; } } } return longestPalindrome; } static boolean checkPalindrome(String s) { for (int i = 0, j = s.length() - 1; i < (s.length() / 2); i++, j–) { if (s.charAt(i) != s.charAt(j)) return false; } return true; }~
- shravankumar
/* ---------- run ---------- Longest Palindrome is = AAJKJAA Output completed (0 sec consumed) - Normal Termination*/
- ANKIT SINGH
MR Ankit, your code is not working …
- VJ
class LongestPal { public static String rev(String s) { String s1 = “”; for (int i = 0; i < s.length(); i++) s1 = s.charAt(i) + s1; return s1; } public static void main(String[] args) { String s = “madamlevelnitin”; String s1 = “”; String s2 = “”; String s3 = “”; for(int i = 0; i < s.length(); i++) { for(int j = i; j < s.length(); j++) { s1 = s1 + s.charAt(j); s2 = rev(s1); s3 = s1.equals(s2) && (s3.length() <= s2.length()) ? (s3.length() 0) ? s2 : s3)) : s3; } s1 = “”; } System.out.println("Longest Pallindrome = " + (s3.length() > 1 ? s3 : “Not found”)); } } /* ---------- run ---------- Longest Pallindrome = nitin Output completed (0 sec consumed) - Normal Termination */
- Ankit Singh
thanks for the code. Really helful.
- mohit kumar
There’s a possible Null reference exception. intermediatePalindrome will return null if (left > right). But longestPalindromeString doesn’t check if a null is returned from intermediatePalindrome before calling palindrome.length(). Admittedly I don’t see how this could occur in the given context, but it’s good form to always check your return values.
- Corey
public class LongstPalndrome { public static void main(String[] args) { String s=“abbsdsdsd”; int strlen=s.length(); outerloop: for(int i=0;i<strlen;i++) { for(int j=0;j<strlen;j++){ for (int k=0;k<=j;k++) { String l=s.substring(i+k, strlen-j+k); if(l.length()<2){ System.out.println(“No palindrome found”); break outerloop; } if(l.equals(new StringBuffer(l).reverse().toString())) { { System.out.println("The longest palindrome is “+l+” with length: "+l.length()); break outerloop; } } } } } } } /*Works.Time complexity sucks.Space is good i guess.*/
- Ranadeep Poddar
Thanks for this perfect code. It helped.
- Anushka
Just tried with my own logic. Sharing here. private static String longestPalindromeString(String string) { String p = “”; for (int i = 0; i < string.length(); i++) { String str = “”; for (int j = i; j < string.length(); j++) { str += string.charAt(j); String revStr = new StringBuilder(str).reverse().toString(); if (string.contains(revStr) && p.length() < revStr.length()) { p = revStr; } if (p.length() == string.length()) { break; } } } return p; }
- Srikanth K
Try for input abacdfgdcaba It is returning wrong answer dcaba
- Shaju
this code did not pass input(contraint 10,00000) please give the efficent code
- Dinesh
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
String st = “abbccbba”; String out = “”; for (int i = st.length() - 1; i >= 0; i–) { out += st.charAt(i); } System.out.println(out + " is " + (st.equals(out) ? “Palindrom.” : “not Palindrom.”));
- krissn
String st = “abbccbba”; System.out.println(st + " is " + (st.equals(new StringBuilder(st).reverse().toString()) ? “Palindrom.” : “not Palindrom.”));
- krissn
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
Hi Panksj, How to find the start index of the longest palindrome in your code?
- JJ
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
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
change int indexofnd = checkString.indexOf(fnd); to int indexofnd = i;
- Satyadip Paul
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 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
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
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