Answer :
Answer:
The solution (Java programming Language) is given in the explanation section
Pay attention to comments for explanation
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\n", arr[i], freqArray[i]);
}
}
}
}