Answer:
The solution (Java programming Language) is given in the explanation section
Pay attention to comments for explanation
Step-by-step explanation:
import java.util.Scanner;
public class num2 {
 public static void main(String[] args) {
 //Calling the CountFreq in the main method
 CountFreq();
 }
 //Defining the CountFreq Method
 public static void CountFreq(){
 Scanner in = new Scanner(System.in);
 System.out.println("Enter array size");
 int arrSize = in.nextInt();
 //Creating the array
 int [] arr = new int[arrSize];
 //Creating another array to hold number frequencies
 int [] freqArray = new int[arr.length];
 //For counting occurence of a the same number
 int count;
 //Receiving the elements of the array
 System.out.println("Enter array Elements: ");
 for(int i=0; i<arr.length; i++)
 {
 System.out.println("Enter the values");
 arr[i] = in.nextInt();
 System.out.printf("%d",arr[i]);
 //Intialize frequency to -1
 freqArray[i] = -1;
 }
 //Checking for occurence of numbers, Increasing the count variable
 //Avoiding duplicates
 for(int i=0; i<arr.length; i++)
 {
 count = 1;
 for(int j=i+1; j<arr.length; j++)
 {
 //Check for duplicate
 if(arr[i]==arr[j])
 {
 count++;
 // Avoid counting same element twice
 freqArray[j] = 0;
 }
 }
 //If frequency of current element is not counted 
 if(freqArray[i] != 0)
 {
 freqArray[i] = count;
 }
 }
 //Output of frequencies
 System.out.println("Frequency of all elements of array");
 for(int i=0; i<arr.length; i++)
 {
 if(freqArray[i] != 0)
 {
 System.out.printf("%d occurs %d times\\", arr[i], freqArray[i]);
 }
 }
 }
 }