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 2
         20 1

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++

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// 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;
}

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Java

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// 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.

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Python3

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# 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

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C#

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// 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

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Output:

10 3
20 4
5 1

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

An efficient solution is to use hashing.

C++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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 */

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Output:

5 1
10 3
20 4

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++

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// 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;
}

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Java

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// 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

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C#

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// 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

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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.

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// Java prorgam 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);
    }
}

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