Counting frequencies of array elements

Given an array which may contain duplicates, print all elements and their frequencies.

Examples:

Input :  arr[] = {10, 20, 20, 10, 10, 20, 5, 20}
Output : 10 3
         20 4
         5  1

Input : arr[] = {10, 20, 20}
Output : 10 1
         20 2 

A simple solution is to run two loops. For every item count number of times, it occurs. To avoid duplicate printing, keep track of processed items. 

Implementation:

C++

// CPP program to count frequencies of array items
#include <bits/stdc++.h>
using namespace std;

void countFreq(int arr[], int n)
{
    // Mark all array elements as not visited
    vector<bool> visited(n, false);

    // Traverse through array elements and
    // count frequencies
    for (int i = 0; i < n; i++) {

        // Skip this element if already processed
        if (visited[i] == true)
            continue;

        // Count frequency
        int count = 1;
        for (int j = i + 1; j < n; j++) {
            if (arr[i] == arr[j]) {
                visited[j] = true;
                count++;
            }
        }
        cout << arr[i] << " " << count << endl;
    }
}

int main()
{
    int arr[] = { 10, 20, 20, 10, 10, 20, 5, 20 };
    int n = sizeof(arr) / sizeof(arr[0]);
    countFreq(arr, n);
    return 0;
}

Java

// Java program to count frequencies of array items
import java.util.Arrays;

class GFG
{
public static void countFreq(int arr[], int n)
{
    boolean visited[] = new boolean[n];

    Arrays.fill(visited, false);

    // Traverse through array elements and
    // count frequencies
    for (int i = 0; i < n; i++) {

        // Skip this element if already processed
        if (visited[i] == true)
            continue;

        // Count frequency
        int count = 1;
        for (int j = i + 1; j < n; j++) {
            if (arr[i] == arr[j]) {
                visited[j] = true;
                count++;
            }
        }
        System.out.println(arr[i] + " " + count);
    }
}

// Driver code
public static void main(String []args)
{
    int arr[] = new int[]{ 10, 20, 20, 10, 10, 20, 5, 20 };
    int n = arr.length;
    countFreq(arr, n);
}
}

// This code contributed by Adarsh_Verma.

Python3

# Python 3 program to count frequencies
# of array items
def countFreq(arr, n):

    # Mark all array elements as not visited
    visited = [False for i in range(n)]

    # Traverse through array elements
    # and count frequencies
    for i in range(n):

        # Skip this element if already
        # processed
        if (visited[i] == True):
            continue

        # Count frequency
        count = 1
        for j in range(i + 1, n, 1):
            if (arr[i] == arr[j]):
                visited[j] = True
                count += 1

        print(arr[i], count)

# Driver Code
if __name__ == '__main__':
    arr = [10, 20, 20, 10, 10, 20, 5, 20]
    n = len(arr)
    countFreq(arr, n)

# This code is contributed by
# Shashank_Sharma

C#

// C# program to count frequencies of array items
using System;

class GFG
{
    public static void countFreq(int []arr, int n)
    {
        bool []visited = new bool[n];

        // Traverse through array elements and
        // count frequencies
        for (int i = 0; i < n; i++)
        {

            // Skip this element if already processed
            if (visited[i] == true)
                continue;

            // Count frequency
            int count = 1;
            for (int j = i + 1; j < n; j++)
            {
                if (arr[i] == arr[j])
                {
                    visited[j] = true;
                    count++;
                }
            }
            Console.WriteLine(arr[i] + " " + count);
        }
    }

    // Driver code
    public static void Main(String []args)
    {
        int []arr = new int[]{ 10, 20, 20, 10, 10, 20, 5, 20 };
        int n = arr.Length;
        countFreq(arr, n);
    }
}

// This code has been contributed by 29AjayKumar

Javascript

<script>

// JavaScript program to count frequencies of array items

function countFreq(arr, n)
{
    let visited = Array.from({length: n}, (_, i) => false);

    // Traverse through array elements and
    // count frequencies
    for (let i = 0; i < n; i++) {

        // Skip this element if already processed
        if (visited[i] == true)
            continue;

        // Count frequency
        let count = 1;
        for (let j = i + 1; j < n; j++) {
            if (arr[i] == arr[j]) {
                visited[j] = true;
                count++;
            }
        }
           document.write(arr[i] + " " + count + "<br/>");
    }
}

// Driver Code

    let arr = [ 10, 20, 20, 10, 10, 20, 5, 20 ];
    let n = arr.length;
    countFreq(arr, n);

</script>

Output

10 3
20 4
5 1

Complexity Analysis:

• Time Complexity : O(n²) 
• Auxiliary Space : O(n)

An efficient solution is to use hashing.

Implementation:

C++

// CPP program to count frequencies of array items
#include <bits/stdc++.h>
using namespace std;

void countFreq(int arr[], int n)
{
    unordered_map<int, int> mp;

    // Traverse through array elements and
    // count frequencies
    for (int i = 0; i < n; i++)
        mp[arr[i]]++;

    // Traverse through map and print frequencies
    for (auto x : mp)
        cout << x.first << " " << x.second << endl;
}

int main()
{
    int arr[] = { 10, 20, 20, 10, 10, 20, 5, 20 };
    int n = sizeof(arr) / sizeof(arr[0]);
    countFreq(arr, n);
    return 0;
}

Java

// Java program to count frequencies of array items
import java.util.*;

class GFG
{

    static void countFreq(int arr[], int n)
    {
        Map<Integer, Integer> mp = new HashMap<>();

        // Traverse through array elements and
        // count frequencies
        for (int i = 0; i < n; i++)
        {
            if (mp.containsKey(arr[i]))
            {
                mp.put(arr[i], mp.get(arr[i]) + 1);
            }
            else
            {
                mp.put(arr[i], 1);
            }
        }
        // Traverse through map and print frequencies
        for (Map.Entry<Integer, Integer> entry : mp.entrySet())
        {
            System.out.println(entry.getKey() + " " + entry.getValue());
        }
    }

    // Driver code
    public static void main(String args[])
    {
        int arr[] = {10, 20, 20, 10, 10, 20, 5, 20};
        int n = arr.length;
        countFreq(arr, n);
    }
}

// This code contributed by Rajput-Ji

Python3

# Python3 program to count frequencies
# of array items
def countFreq(arr, n):

    mp = dict()

    # Traverse through array elements
    # and count frequencies
    for i in range(n):
        if arr[i] in mp.keys():
            mp[arr[i]] += 1
        else:
            mp[arr[i]] = 1

    # Traverse through map and print
    # frequencies
    for x in mp:
        print(x, " ", mp[x])

# Driver code
arr = [10, 20, 20, 10, 10, 20, 5, 20 ]
n = len(arr)
countFreq(arr, n)

# This code is contributed by
# Mohit kumar 29

C#

// C# implementation of the approach
using System;
using System.Collections.Generic;

class GFG
{

    static void countFreq(int []arr, int n)
    {
        Dictionary<int, int> mp = new Dictionary<int,int>();

        // Traverse through array elements and
        // count frequencies
        for (int i = 0; i < n; i++)
        {
            if (mp.ContainsKey(arr[i]))
            {
                var val = mp[arr[i]];
                mp.Remove(arr[i]);
                mp.Add(arr[i], val + 1);
            }
            else
            {
                mp.Add(arr[i], 1);
            }
        }

        // Traverse through map and print frequencies
        foreach(KeyValuePair<int, int> entry in mp)
        {
            Console.WriteLine(entry.Key + " " + entry.Value);
        }
    }

    // Driver code
    public static void Main(String []args)
    {
        int []arr = {10, 20, 20, 10, 10, 20, 5, 20};
        int n = arr.Length;
        countFreq(arr, n);
    }
}

/* This code contributed by PrinciRaj1992 */

Javascript

<script>


// JavaScript program to count
// frequencies of array items



function countFreq(arr, n)
{
    const map ={}

    // Traverse through array elements and
    // count frequencies
    for (let i = 0; i < n; i++)
    {
       if(map[arr[i]){
          map[arr[i]]+=1
       }
        else{
           map[arr[i]] =1
         }
    }
   console.log(map)
}
var arr = [10, 20, 20, 10, 10, 20, 5, 20];
var n = arr.length;
countFreq(arr, n);

</script>

Output

5 1
20 4
10 3

Complexity Analysis:

• Time Complexity : O(n) 
• Auxiliary Space : O(n)

In the above efficient solution, how to print elements in same order as they appear in input? 

Implementation:

C++

// CPP program to count frequencies of array items
#include <bits/stdc++.h>
using namespace std;

void countFreq(int arr[], int n)
{
    unordered_map<int, int> mp;

    // Traverse through array elements and
    // count frequencies
    for (int i = 0; i < n; i++)
        mp[arr[i]]++;

    // To print elements according to first
    // occurrence, traverse array one more time
    // print frequencies of elements and mark
    // frequencies as -1 so that same element
    // is not printed multiple times.
    for (int i = 0; i < n; i++) {
      if (mp[arr[i]] != -1)
      {
          cout << arr[i] << " " << mp[arr[i]] << endl;
          mp[arr[i]] = -1;
      }
    }
}

int main()
{
    int arr[] = { 10, 20, 20, 10, 10, 20, 5, 20 };
    int n = sizeof(arr) / sizeof(arr[0]);
    countFreq(arr, n);
    return 0;
}

Java

// Java program to count frequencies of array items
import java.util.*;

class GFG
{

    static void countFreq(int arr[], int n)
    {
        Map<Integer, Integer> mp = new HashMap<>();

        // Traverse through array elements and
        // count frequencies
        for (int i = 0; i < n; i++)
        {
            mp.put(arr[i], mp.get(arr[i]) == null ? 1 : mp.get(arr[i]) + 1);
        }

        // To print elements according to first
        // occurrence, traverse array one more time
        // print frequencies of elements and mark
        // frequencies as -1 so that same element
        // is not printed multiple times.
        for (int i = 0; i < n; i++)
        {
            if (mp.get(arr[i]) != -1)
            {
                System.out.println(arr[i] + " " + mp.get(arr[i]));
                mp.put(arr[i], -1);
            }
        }
    }

    // Driver code
    public static void main(String[] args)
    {
        int arr[] = {10, 20, 20, 10, 10, 20, 5, 20};
        int n = arr.length;
        countFreq(arr, n);
    }
}

// This code contributed by Rajput-Ji

Python3

# Python3 program to count frequencies of array items
def countFreq(arr, n):

    mp = {}

    # Traverse through array elements and
    # count frequencies
    for i in range(n):
        if arr[i] not in mp:
            mp[arr[i]] = 0
        mp[arr[i]] += 1

    # To print elements according to first
    # occurrence, traverse array one more time
    # print frequencies of elements and mark
    # frequencies as -1 so that same element
    # is not printed multiple times.
    for i in range(n):
        if (mp[arr[i]] != -1):
            print(arr[i],mp[arr[i]])
        mp[arr[i]] = -1

# Driver code

arr = [10, 20, 20, 10, 10, 20, 5, 20]
n = len(arr)
countFreq(arr, n)

# This code is contributed by shubhamsingh10

C#

// C# program to count frequencies of array items
using System;
using System.Collections.Generic;

class GFG
{

    static void countFreq(int []arr, int n)
    {
        Dictionary<int,int> mp = new Dictionary<int,int>();

        // Traverse through array elements and
        // count frequencies

        for (int i = 0 ; i < n; i++)
        {
            if(mp.ContainsKey(arr[i]))
            {
                var val = mp[arr[i]];
                mp.Remove(arr[i]);
                mp.Add(arr[i], val + 1);
            }
            else
            {
                mp.Add(arr[i], 1);
            }
        }

        // To print elements according to first
        // occurrence, traverse array one more time
        // print frequencies of elements and mark
        // frequencies as -1 so that same element
        // is not printed multiple times.
        for (int i = 0; i < n; i++)
        {
            if (mp.ContainsKey(arr[i]) && mp[arr[i]] != -1)
            {
                Console.WriteLine(arr[i] + " " + mp[arr[i]]);
                mp.Remove(arr[i]);
                mp.Add(arr[i], -1);
            }
        }
    }

    // Driver code
    public static void Main(String[] args)
    {
        int []arr = {10, 20, 20, 10, 10, 20, 5, 20};
        int n = arr.Length;
        countFreq(arr, n);
    }
}

// This code is contributed by Princi Singh

Javascript

<script>


// Javascript program to count frequencies of array items

function countFreq(arr, n)
{
    var mp = new Map();

    // Traverse through array elements and
    // count frequencies
    for (var i = 0; i < n; i++)
    {
        if(mp.has(arr[i]))
            mp.set(arr[i], mp.get(arr[i])+1)
        else
            mp.set(arr[i], 1)
    }

    // To print elements according to first
    // occurrence, traverse array one more time
    // print frequencies of elements and mark
    // frequencies as -1 so that same element
    // is not printed multiple times.
    for (var i = 0; i < n; i++) {
      if (mp.get(arr[i]) != -1)
      {
          document.write( arr[i] + " " + mp.get(arr[i]) + "<br>");
          mp.set(arr[i], -1);
      }
    }
}

var arr = [10, 20, 20, 10, 10, 20, 5, 20];
var n = arr.length;
countFreq(arr, n);

// This code is contributed by rrrtnx.
</script>

Output

10 3
20 4
5 1

Complexity Analysis:

• Time Complexity : O(n) 
• Auxiliary Space : O(n)

This problem can be solved in Java using Hashmap. Below is the program. 

Implementation:

C++

// C++ program to count frequencies of
// integers in array using Hashmap
#include <bits/stdc++.h>
using namespace std;

void frequencyNumber(int arr[],int size)
{

  // Creating a HashMap containing integer
  // as a key and occurrences as a value
  unordered_map<int,int>freqMap;

  for (int i=0;i<size;i++) {
    freqMap[arr[i]]++;
  }

  // Printing the freqMap
  for (auto it : freqMap) {
    cout<<it.first<<" "<<it.second<<endl;
  }
}

int main()
{
  int arr[] = {10, 20, 20, 10, 10, 20, 5, 20};
  int size = sizeof(arr)/sizeof(arr[0]);
  frequencyNumber(arr,size);
}

// This code is contributed by shinjanpatra.

Java

// Java program to count frequencies of
// integers in array using Hashmap
import java.io.*;
import java.util.*;
class OccurenceOfNumberInArray {
    static void frequencyNumber(int arr[], int size)
    {
        // Creating a HashMap containing integer
        // as a key and occurrences as a value
        HashMap<Integer, Integer> freqMap
            = new HashMap<Integer, Integer>();

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

                // If number is present in freqMap,
                // incrementing it's count by 1
                freqMap.put(arr[i], freqMap.get(arr[i]) + 1);
            }
            else {

                // If integer is not present in freqMap,
                // putting this integer to freqMap with 1 as it's value
                freqMap.put(arr[i], 1);
            }
        }

        // Printing the freqMap
        for (Map.Entry entry : freqMap.entrySet()) {
            System.out.println(entry.getKey() + " " + entry.getValue());
        }
    }

    // Driver Code
    public static void main(String[] args)
    {
        int arr[] = {10, 20, 20, 10, 10, 20, 5, 20};
        int size = arr.length;
        frequencyNumber(arr,size);
    }
}

Python3

# Python program to count frequencies of
# integers in array using Hashmap

def frequencyNumber(arr,size):
    # Creating a HashMap containing integer
        # as a key and occurrences as a value
        freqMap = {}

        for i in range(size):
            if (arr[i] in freqMap):

                # If number is present in freqMap,
                # incrementing it's count by 1
                freqMap[arr[i]] = freqMap[arr[i]] + 1
            else:

                # If integer is not present in freqMap,
                # putting this integer to freqMap with 1 as it's value
                freqMap[arr[i]] = 1

        # Printing the freqMap
        for key, value in freqMap.items():
            print(f"{key} {value}")

# Driver Code
arr = [10, 20, 20, 10, 10, 20, 5, 20]
size = len(arr)
frequencyNumber(arr,size)

# This code is contributed by shinjanpatra

C#

// C# program to count frequencies of
// integers in array using Hashmap
using System;
using System.Collections.Generic;
class GFG
{
  static void frequencyNumber(int []arr,int size)
  {

    // Creating a Dictionary containing integer
    // as a key and occurrences as a value
    Dictionary<int, int> freqMap = new Dictionary<int,int>();

    for(int i = 0; i < size; i++){
      if (freqMap.ContainsKey(arr[i]))
      {
        var val = freqMap[arr[i]];
        freqMap.Remove(arr[i]);
        freqMap.Add(arr[i], val + 1);
      }
      else
      {
        freqMap.Add(arr[i], 1);
      }
    }

    // Printing the freqMap
    foreach(KeyValuePair<int, int> entry in freqMap)
    {
      Console.WriteLine(entry.Key + " " + entry.Value);
    }
  }

  public static void Main(String []args)
  {
    int []arr = {10, 20, 20, 10, 10, 20, 5, 20};
    int size = arr.Length;
    frequencyNumber(arr,size);
  }
}
// This code is contributed by Taranpreet

Javascript

<script>
// Javascript program to count frequencies of
// integers in array using Hashmap

function frequencyNumber(arr,size)
{
    // Creating a HashMap containing integer
        // as a key and occurrences as a value
        let freqMap
            = new Map();

        for (let i=0;i<size;i++) {
            if (freqMap.has(arr[i])) {

                // If number is present in freqMap,
                // incrementing it's count by 1
                freqMap.set(arr[i], freqMap.get(arr[i]) + 1);
            }
            else {

                // If integer is not present in freqMap,
                // putting this integer to freqMap with 1 as it's value
                freqMap.set(arr[i], 1);
            }
        }

        // Printing the freqMap
        for (let [key, value] of freqMap.entries()) {
            document.write(key + " " + value+"<br>");
        }
}

// Driver Code
let arr=[10, 20, 20, 10, 10, 20, 5, 20];
let size = arr.length;
frequencyNumber(arr,size);

// This code is contributed by patel2127
</script>

Output

5 1
20 4
10 3

Complexity Analysis:

• Time Complexity: O(n) since using a single loop to track frequency
• Auxiliary Space: O(n) for hashmap.

Another Efficient Solution (Space optimization): we can find frequency of array elements using Binary search function . First we will sort the array for binary search . Our frequency of element will be ‘(last occ – first occ)+1’ of a element in a array .                  

Below is the implementation of the above approach: 

C++

// C++ implementation of the above approach

#include <bits/stdc++.h>
using namespace std;

//Function to find frequency of elements in the array
void countFreq(int arr[], int n)
{
    sort(arr,arr+n);//sort array for binary search

    for(int i = 0 ; i < n ;i++)
    {
      //index of first and last occ of arr[i]
      int first_index = lower_bound(arr,arr+n,arr[i])- arr;
      int last_index = upper_bound(arr,arr+n,arr[i])- arr-1;
      i=last_index;

      int fre=last_index-first_index+1;//finding frequency
      cout << arr[i] <<" "<<fre <<endl;//printing frequency
    }
}

// Drive code
int main()
{
    int arr[] = { 10, 20, 20, 10, 10, 20, 5, 20 };
    int n = sizeof(arr) / sizeof(arr[0]);

    // Function call
    countFreq(arr, n);
    return 0;

}

// This Code is contributed by nikhilsainiofficial546

Java

// Java program to count frequencies of
// integers in array using Hashmap
import java.io.*;
import java.util.*;
class OccurenceOfNumberInArray {
    public static void countFreq(int[] arr, int n) {
    Arrays.sort(arr); //sort array for binary search

    for (int i = 0; i < n; i++) {
        //index of first and last occ of arr[i]
        int first_index = Arrays.binarySearch(arr, arr[i]);
        int last_index = Arrays.binarySearch(arr, arr[i]);

        while (first_index > 0 && arr[first_index - 1] == arr[i]) {
            first_index--;
        }

        while (last_index < n - 1 && arr[last_index + 1] == arr[i]) {
            last_index++;
        }

        i = last_index;

        int fre = last_index - first_index + 1;//finding frequency
        System.out.println(arr[i] + " " + fre);//printing frequency
    }
}
    // Driver Code
    public static void main(String[] args)
    {
        int arr[] = {10, 20, 20, 10, 10, 20, 5, 20};
        int size = arr.length;
        countFreq(arr,size);
    }
}

Python3

from bisect import bisect_left, bisect_right

# Function to find frequency of elements in the array
def countFreq(arr, n):
    arr.sort()  # sort array for binary search

    i = 0
    while i < n:
        # index of first and last occ of arr[i]
        first_index = bisect_left(arr, arr[i])
        last_index = bisect_right(arr, arr[i]) - 1
        i = last_index + 1

        fre = last_index - first_index + 1  # finding frequency
        print(arr[i - 1], fre)  # printing frequency

# Driver code
arr = [10, 20, 20, 10, 10, 20, 5, 20]
n = len(arr)
countFreq(arr, n)

C#

using System;

public class GFG {

    // Function to find frequency of elements in the array
    public static void countFreq(int[] arr, int n) {
        Array.Sort(arr); // sort array for binary search

        for (int i = 0; i < n; i++) {
            // index of first and last occurrence of arr[i]
            int first_index = Array.IndexOf(arr, arr[i]);
            int last_index = Array.LastIndexOf(arr, arr[i]);

            int fre = last_index - first_index + 1; // finding frequency
            Console.WriteLine(arr[i] + " " + fre); // printing frequency
            i = last_index;
        }
    }

    // Driver code
    static public void Main () {
        int[] arr = { 10, 20, 20, 10, 10, 20, 5, 20 };
        int n = arr.Length;

        // Function call
        countFreq(arr, n);
    }
}

Javascript

// Javascript implementation of the above approach

//Function to find frequency of elements in the array
function countFreq(arr,n) {
    arr.sort((a, b) => a - b); //sort array for binary search
    let i = 0;
    while (i < n) {
        //index of first and last occ of arr[i]
        const first_index = arr.indexOf(arr[i]);
        const last_index = arr.lastIndexOf(arr[i]);
        i = last_index;

        const fre = last_index - first_index + 1; //finding frequency
        console.log(arr[i], fre); //printing frequency
        i++;
    }
}

// Driver code
const arr = [ 10, 20, 20, 10, 10, 20, 5, 20 ];
countFreq(arr,arr.length);

Output

5 1
10 3
20 4

Time Complexity: O(n*log₂n) , where O(log₂n) time for binary search function .
Auxiliary Space: O(1)