Find the Frequency of Each Element in the Array in Java

We may count the occurrence of each element in the array of items. Maintaining one array to store the counts of each array element is one strategy for solving this issue. The frequency of each element is calculated by iterating through the array, and it is then stored in another array called fr.

Example

Input : arr [ ] = [1,2,4,3,3,2,1]

Output:

Element    Frequency

1. 2
2. 2
3. 2
4. 1

Method 1

By using Loops

Algorithm

• STEP 1: BEGIN
• Initiate array[] = [1, 2, 7, 3, 2, 2, 2, 1] .
• Create a frequency [ ] of array[] length in step three.
• SET vis = -1 in step four.
• for (i=0; iarr.length; i++)
  • SET count is 1.
• for (i=0;i<frequency.length;i++) 
• PRINT array [i] and frequenct [i] if (frequency [i]!=visited)

FrequencyOfElements.java

import java.util.*;
public class FrequencyOfElements {  
// Main section of the program where execution begins
    public static void main (String[] s) {  
        // creating object for the scanner class
        Scanner sc=new Scanner (System.in);
        System.out.println(" Enter array size ");
        // enter the size of the array
        int n=sc.nextInt();
        // array declaration 
        int array[] = new int [n];
        System.out.println(" Enter array elements ");
        // enter the array elements
        // array initialisation
        for(int i=0;i<n;i++)
        {
            array[i]=sc.nextInt();
        }
        // printing the array elements
        System.out.println(" The array elements are ");
        for(int i=0;i<n;i++)
        {
            System.out.print(array[i]+" ");
        }
        System.out.println();
        //Array frequency will store frequencies of an  element  
        int [] frequency = new int [array.length];  
        int vis = -1;  
        for(int j = 0; j < array.length; j++){  
            int count = 1;  
            // running the for loop from j+1 index to end of the array 
            for(int i = j+1; i < array.length; i++){ 
                // if 2 elements are equal then count cariable will be incremented
                if(array[i] == array[j]){  
                    count++;  
                    //To avoid counting same element again  
                    frequency[i] = vis;  
                }  
            } 
            // if the element is not visited then only count will be updated in the frequency array
            if(frequency[j] != vis)  
                frequency[j] = count;  
        }  
  
        //Displays the frequency of each element present in array  
        System.out.println(" - -- --- --- --- -----------");  
        System.out.println(" Elements | Frequency");  
        // printing the frequencies of array Elements with respect to their array elements
        for(int j = 0; j < frequency.length; j++){  
            if(frequency[j] != vis)  
                System.out.println("    " + array[j] + "    |    " + frequency[j]);  
        }  
    }}  

Output

Enter array size 
5
 Enter array elements 
1 2 1 3 3
The array elements are
1 2 1 3 3 
-----------------------
 Elements | Frequency
    1    |    2
    2    |    1
    3    |    2

Method 2

FrequencyOfElements2.java

By Using Hashing

import java.util.*;
class FrequencyOfElements2
{
static void Freqz (int arr[], int n)
{
Map<Integer, Integer> m1 = new HashMap < > () ;
// Go through the elements of an array and 
for (int i = 0; i < n; i++)
{
if (m1.containsKey(arr[i]))
{
m1.put(arr[i], m1.get(arr[i]) + 1);
}
else
{
m1.put(arr[i], 1);
}
}
// Traversing  through map and printing the  frequencies
for (Map.Entry<Integer, Integer> traversal : m1.entrySet())
{
System.out.println(traversal.getKey() + " " + traversal.getValue());
}
}
public static void main(String s[])
{
    Scanner sc = new Scanner(System.in);
    System.out.println(" Enter size of the array ");
    int n=sc.nextInt();
    int a[]=new int [n];
    System.out.println("Enter array elements");
    for(int i=0;i<n;i++)
    {
        a[i]= sc.nextInt();
    }
Freqz(a, n);
}
}

Output

Enter size of the array 
8
Enter array elements
12 12 32 44 44 10 10 67
32 1
67 1
10 2
12 2
44 2

By using HashMap

FrequencyOfElements3.java

import java.io.*;
import java.util.*;
class FrequencyOfElements3
 {
static void Freqz(int arr[], int size)
{
HashMap<Integer, Integer> h1 = new HashMap<Integer, Integer> () ;


for (int j=0;j<size;j++) {
if (h1.containsKey(arr[j])) {


h1.put(arr[j], h1.get(arr[j]) + 1);
}
else {


h1.put(arr[j], 1);
}
}
// Printing the freqMap
for (Map.Entry e : h1.entrySet()) {
System.out.println(e.getKey() + " " + e.getValue());
}
}


// Driver Code
public static void main(String s[])
{
    Scanner sc = new Scanner(System.in);
    System.out.println(" Enter size of the array ");
    int n=sc.nextInt();
    int a[]=new int [n];
    System.out.println("Enter array elements");
    for(int i=0;i<n;i++)
    {
        a[i]= sc.nextInt();
    }

Freqz(a, n);
}
}

Output:

Enter size of the array 
5
Enter array elements
10 20 10 30 20
20 2
10 2
30 1