can any solve this question? only two test case passed..

Raja loves Strings. In Strings he specially loves palindromes.Palindromes
are strings that read the same when read forward or backwards. Here
Palindromes considered are only of even length(maybe 0). His Teacher
gave me him a long string consisting of only digits as a gift on his
birthday. Now Raja wants to know The longest subarray whose elements
(i.e digits) can be rearranged to form a palindrome of even length. Raja is
unable to figure out the solution to the problem so he asks for your help.

The function contains one argument - A String S consisting of digits (0-9).
First and only line of input consists of S (1 <= |S| <= 100000).
You must return a single integer denoting the longest subarray whose
elements (digits) can be rearranged to form a palindrome of even
length.If no such subarray can be found return 0.
Example -
Sample Test Case 1




Explanation : No subarray can be rearranged to form even length of palindrome .Hence 0.

Explanation : Here the longest subarray is 345354 which can be rearranged to form 345543 which is a palindrome of even length 6.Hence and is 6.

Sample Test Case 2-





The only thing you have to do is to store the number of times a digit has occcured . Initialize ans=0 and keep on adding count of those digits which has occured even number of times .

and first we have to find longest sub array with non contigious string element in string of even no.then we have to make the palindrome …

This is my code i dont know where it is wrong… can u identify it


// A utility function to get max of two integers
int max (int x, int y) { return (x > y)? x : y; }

// Returns the length of the longest palindromic subsequence in seq
int lps(char *seq, int i, int j)
// Base Case 1: If there is only 1 character
if (i == j)
return 1;

// Base Case 2: If there are only 2 characters and both are same
if (seq[i] == seq[j] && i + 1 == j)
return 2;

// If the first and last characters match
if (seq[i] == seq[j])
return lps (seq, i+1, j-1) + 2;

// If the first and last characters do not match
return max( lps(seq, i, j-1), lps(seq, i+1, j) );

int main()
char seq[] = “12345354987”;//12345
int n = strlen(seq);
printf (“The length is %d”, lps(seq, 0, n-1));
return 0;

Can you please elaborate it more?

Just counting the number of even digits is not enough it would result in the rearrangement of sub sequences and sub array .For sub array I believe you need go for sliding window algorithm and manage the state at each window point.

Initial version


Corrected Version


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@athulpai Unfortunately your solution is giving wrong answer for many of the test cases. See this

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Thanks for pointing it out . I made corrections to my code .Initially max_len has to be initialised to the maximum even consecutive element.
Here is the edited version of the code

still your approach is wrong.

Check out for this:



Please don’t try to mislead others by posting wrong solution at here.

Below is my answer in Java

import java.util.HashMap;
import java.util.Map;
import java.util.Map.Entry;

public class Problem2 {

	public static void main(String arg[])
		String number = "12345354987";//"123225411482222211111";
		Map<Character, Integer> map = new HashMap<>();
		char[] array = null;
		int len = number.length();
		boolean palindrom = false;
		int maxPalindrom = 0;
		for(int j=2; j < len; j +=2)
			for(int i=0; i <= len - j; i++)
				array = number.substring(i, i+j).toCharArray();
				palindrom = true;
				for(char c : array)
					map.put(c, map.containsKey(c) ? (map.get(c) + 1) : 1);
				for (Entry<Character, Integer> entry : map.entrySet())
					if(entry.getValue() % 2 != 0){
						palindrom = false;
					maxPalindrom = array.length;