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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 

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 */

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++

// 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 prelements according to first 
    # occurrence, traverse array one more time 
    # prfrequencies 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 

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.

// 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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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

C++

Java

Python3

C#

C++

Java

Python3

C#

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