How to check a string can be eligible to form a palindrome

import java.util.HashMap;
import java.util.Scanner;

public class CheckStringForPalindrome {

	public static void main(String[] args) {
		HashMap<Character, Integer> hmCharsCount = new HashMap<Character, Integer>();

		Scanner inpStr=new Scanner(System.in);
		System.out.println("Enter the input string is=");
		String palinStr=inpStr.nextLine();

		// Use the Hash Map collection to find the count each character occurrences
		for(int i=0;i<palinStr.length();i++) 
			if(!hmCharsCount.containsKey(palinStr.charAt(i)))
				hmCharsCount.put(palinStr.charAt(i), 1);
			else
				hmCharsCount.put(palinStr.charAt(i), hmCharsCount.get(palinStr.charAt(i))+1);

		System.out.println("The Hashmap of each character occurences are: "+hmCharsCount);

		Character[] charKeys=new Character[hmCharsCount.size()];
		int[] charCnts=new int[hmCharsCount.size()];
		int i=0;

		//Separate the Character keys and their respective counts into separate arrays
		for (HashMap.Entry<Character, Integer> entry: hmCharsCount.entrySet() ) {
			charKeys[i]=entry.getKey();
			charCnts[i++]=entry.getValue();
		}
		inpStr.close();
		for(i=0;i<hmCharsCount.size();i++)
			System.out.println("The respective Character key="+charKeys[i]+" and number of occurances is="+charCnts[i]);

		// Get the odd Character counts and even Character Counts
		int oddCntOfChars=0,evnCntOfChars=0;
		for (int k=0;k<charCnts.length;k++) {
			
			if (charCnts[k]%2 == 0)
				evnCntOfChars++;
			else
				oddCntOfChars++;
		}
		System.out.println("The Even count of Character keys are="+evnCntOfChars);
		System.out.println("The Odd count of Character keys are="+oddCntOfChars);

		/* There must be one and only one character key in the input string with odd number of times.
		Any number of character keys with even number of count are allowed. Hence the, below 
		condition check is the eligibility criteria to form the successful Palindrome */

		if (oddCntOfChars <= 1)
			System.out.println("The string "+palinStr+" can be rearranged as a Palindrome");
		else
			System.out.println("The string "+palinStr+" can not be rearranged as a Palindrome");		
	}
}