Whether String is Palindrome or not

package com.array;

import java.util.HashMap;

import java.util.Map;

/* To find whether String is a Palindrome or not

 * Convert String to Map with char as key and count as value

 * Calculate if count is even or count is odd only once */

public class StringPalindrome {

     public static void main(String[] args) {

           String arrayStr[] = { "NitiN", "Paras", "tactactoto" };

           for(String str:arrayStr){

                System.out.println("String is "+ str);

                Map map = countEveryCharInMap(str);

                System.out.println("String conversion into Map " + map);

                palindrome(map);

           }

     }

     static Map<Character, Integer> countEveryCharInMap(String str) {

           Map<Character, Integer> map = new HashMap<>();

           for (int i = 0; i < str.toCharArray().length; i++) {

                char key = str.charAt(i);

                int count = 1;

                if (map.containsKey(key)) {

                     Integer value = map.get(key);

                     Integer newValue = ++value;

                     map.put(key, newValue);

                } else

                     map.put(key, count);

           }

           return map;

     }

     static void palindrome(Map<Character, Integer> map) {

           int oddCountflag = 0;

           for (Character c : map.keySet()) {

                if (map.get(c) % 2 != 0)

                     oddCountflag++;

           }

           if (oddCountflag <= 1)

                System.out.println("PALINDROME");

           else

                System.out.println("NOT A PALINDROME");

     }

}

Output:

String is NitiN

String conversion into Map {t=1, i=2, N=2}

PALINDROME

String is Paras

String conversion into Map {P=1, a=2, r=1, s=1}

NOT A PALINDROME

String is tactactoto

String conversion into Map {a=2, c=2, t=4, o=2}

PALINDROME