isAnagramOfAStringPalindrome

LOGIC:
Store the character count in an array of length 26. Loop through all the counts, to find if the odd occurence is greater
than 1. Ideally if palindrome exists in then, the total odd occurences would be 0 or 1. Find this, using your own
examples. Work with your own examples.



import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
        Scanner myScan = new Scanner(System.in);
        String inputString = myScan.nextLine();
       
        String ans="";
        boolean got = checkPalindrome(inputString);
        if(got==true)
            ans="YES";
        else
            ans="NO";
        // Assign ans a value of YES or NO, depending on whether or not inputString satisfies the required condition
        System.out.println(ans);
        myScan.close();
    }
     public static boolean checkPalindrome(String input)
{
int [] count = new int[26];
for( int i = 0; i < input.length(); i++ )
{
char ch = input.charAt(i);
count[ch-'a']++;
}
int oddOccur = 0;
for( int cnt:count )
{
if( oddOccur > 1)
return false;
if( cnt%2 == 1 )
oddOccur++;
}
return true;
}
}