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  • Difficulty Level : Easy
  • Last Updated : 21 Jun, 2022

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. 

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

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

An efficient solution is to use hashing.

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)
{
    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)
    }
  
    var keys = [];
    mp.forEach((value, key) => {
        keys.push(key);
    });
    keys.sort((a,b)=> a-b);
      
    // Traverse through map and print frequencies
    keys.forEach((key) => {
          
        document.write(key + " " + mp.get(key)+ "<br>");
    });
}
  
var arr = [10, 20, 20, 10, 10, 20, 5, 20];
var n = arr.length;
countFreq(arr, n);
  
</script>

Output

5 1
10 3
20 4

Complete Interview Preparation - GFG

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

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

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

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

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

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
10 3
20 4

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


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