Find the most frequent element in an array using Java

import java.util.Arrays;
import java.util.HashMap;

public class Main {

    public static void main(String[] args) {
        int result;

        int[] arr1 = new int[]{1, 2, 3, 4, 5, 5, 6, 7, 8, 4, 6, 5};
        int[] arr2 = new int[]{1, 2, 3, 4, 5, 6, 7, 8, 9};

        result = findFrequentNumber(arr1);
        printResult(arr1, result);

        System.out.println();

        result = findFrequentNumber(arr2);
        printResult(arr2, result);

    }

    /**
     * print the result
     *
     * @param arr    : input array
     * @param result : result . if -1 , means all elements frequency is 1
     */
    private static void printResult(int[] arr, int result) {
        System.out.println(Arrays.toString(arr));
        if (result == -1) {
            System.out.println("All elements are same ");
        } else {
            System.out.println(("Most frequent number : " + result));
        }
    }

    /**
     * Find most frequent number in a array
     *
     * @param inputArr : input array
     * @return : most frequent element in inputArr
     */
    private static int findFrequentNumber(int[] inputArr) {

        //create a hashmap to store the count of each element . key is number and value is count for the number
        HashMap<Integer, Integer> numberMap = new HashMap<>();

        int result = -1; //result will hold the most frequent element for the given array
        int frequency = -1; //frequency is the count for the most frequent element

        int value;

        for (int i = 0; i < inputArr.length; i++) { //scan all elements of the array one by one

            value = -1;
            // set value as -1
            if (numberMap.containsKey(inputArr[i])) {
                value = numberMap.get(inputArr[i]); // if the element is in the map, get the count
            }
            if (value != -1) {
                // value is not -1 , that means the element is in the map. Increment the value and check if it is
                // greater than the maximum
                value += 1;
                if (value > frequency) {
                    //if the value is greater than frequency, it means it is the maximum value
                    // till now. store it
                    frequency = value;
                    result = inputArr[i];
                }

                numberMap.put(inputArr[i], value); // put the updated value in the map
            } else {
                //element is not in the map. put it with value or count as 1
                numberMap.put(inputArr[i], 1);
            }

        }

        if (frequency == 1)
            return -1;

        return result;
    }

}

[1, 2, 3, 4, 5, 5, 6, 7, 8, 4, 6, 5]
Most frequent number : 5

[1, 2, 3, 4, 5, 6, 7, 8, 9]
All elements are same