C program to find the frequency of each element in the array

June 16, 2023

Frequency of Element in C

Here, in this page we will discuss the program to find the frequency of element in C programming language. We are given with an array and need to print the frequency of each given element.

Example

Input : arr[6] = [1, 2, 2, 3, 1, 2]
Output : 1 occurs 2 times
2 occurs 3 times
3 occurs 1 times

Method Discussed :

Problem Statement – Write a Program in C to count the Frequency of each element of an Array.

Method 1: Using Extra Space
Method 2: Naive approach without extra space.
Method 3: Using the Sorting Technique.

Method 1:

In this method, we will count the frequency of each element using two for loops.

To check the status of visited elements create a array of size n.
Run a loop from index 0 to n and check if (visited[i]==1) then skip that element.
Otherwise create a variable count = 1 to keep the count of frequency.
Run a loop from index i+1 to n
Check if(arr[i]==arr[j]), then increment the count by 1 and set visited[j]=1.
After complete iteration of for loop print element along with value of count.

Time and Space Complexity :

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

Sort first half in ascending order and second half in descending

Sort the elements of an array

Sorting elements of an array by frequency

Finding the Longest Palindrome in an Array

Counting Distinct Elements in an Array

Code in C

Run

#include<stdio.h>

// Main function to run the program
int main() 
{ 
    int arr[] = {10, 30, 10, 20, 10, 20, 30, 10}; 
    int n = sizeof(arr)/sizeof(arr[0]); 

    int visited[n];
 
    for(int i=0; i<n; i++){

       if(visited[i]==0){
          int count = 1;
          for(int j=i+1; j<n; j++){
             if(arr[i]==arr[j]){
                count++;
                visited[j]=1;
             }
          }

          printf("%d occurs %d times\n", arr[i], count);
       }
   }

   return 0; 
}

Output

10 occurs 4 times
30 occurs 2 times
20 occurs 2 times

Method 2 :

In this method we will use the naive way to find the frequency of elements in the given integer array without using any extra space.

Method 2 : Code in C

Run

#include<stdio.h>

void countFrequency(int *arr, int size){

    for (int i = 0; i < size; i++){
        int flag = 0;
        int count = 0;

        // Counting of any element has to be delayed to its last occurrence
        for (int j = i+1; j < size; j++){
            if (arr[i] == arr[j]){
                flag = 1;
                break;
            }
        }

        // The continue keyword is used to end the current iteration 
        // in a for loop (or a while loop), and continues to the next iteration
        if (flag == 1)
            continue;
            
        for(int j = 0;j<=i;j++){
            if(arr[i]==arr[j])
                count +=1;
        }
        
        printf("%d : %d\n", arr[i], count);
    }
}

int main()
{
    int arr[] = {5, 8, 5, 7, 8, 10};
    int size = sizeof(arr)/sizeof(arr[0]);
    
    countFrequency(arr, size);
    
    return 0;
}

Output

Method 3 :

In this method we will sort the array then, count the frequency of the elements.

Time and Space Complexity :

Time Complexity : O(nlogn)
Space Complexity : O(1)

Method 3 : Code in C

Run

#include<stdio.h>
 
void countDistinct(int arr[], int n)
{
    //Sorting of the array
    for(int i=0; i<n; i++){
        for(int j=i+1; j<n; j++){ if(arr[i]>arr[j]){
                int temp = arr[i];
                arr[i] = arr[j];
                arr[j] = temp;
            }
        }
    }
 
    // Traverse the sorted array
    for (int i = 0; i < n; i++){
        int count = 1;

        // Move the index ahead whenever
        // you encounter duplicates
        while (i < n - 1 && arr[i] == arr[i + 1]){
            i++;
            count++;
        }
        
        printf("%d : %d\n", arr[i], count);
    }
 
}
 
// Driver program to test above function
int main()
{
    int arr[] = {5, 8, 5, 7, 8, 10};
    int n = sizeof(arr) / sizeof(arr[0]);
    countDistinct(arr, n);
    return 0;
}

Output

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3 comments on “C program to find the frequency of each element in the array”

Gyanendra
import java.util.*;
public class Main
{
public static void main (String[]args)
{

int[] arr= {10,20,30,20,50,50,40};

// Hashamap(arr);
// boolean_array(arr);

}

public static void boolean_array(int[] arr){
boolean[] bool=new boolean[arr.length];
Arrays.fill(bool,false);

for(int i=0;i<arr.length;i++){
int count=1;

if(bool[i]!=true){
bool[i]=true;

for(int j=i+1;j<arr.length;j++){
if(arr[i]==arr[j]){
count++;
bool[j]=true;
}
}
System.out.println(arr[i]+" "+count);
}

// System.out.println(arr[i]+" "+count);
}
}

public static void Hashamap(int arr[]){
HashMap map=new HashMap();

for(int i=0;i<arr.length;i++){

if(map.containsKey(arr[i])){
map.put(arr[i],map.get(arr[i])+1);
}
else{
map.put(arr[i],1);
}
}

for(Map.Entry entry:map.entrySet()){
System.out.println(entry.getKey()+” “+entry.getValue());
}

}
}

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Isshita Ghosh_CSE_023
In method 1, when I am giving (10,20,30,20,50,50,40) The output is :
20 occurs 2 times
50 occurs 2 times

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Gyanendra
import java.util.*;
public class Main
{
public static void main (String[]args)
{

int[] arr= {10,20,30,20,50,50,40};

HashMap map=new HashMap();

for(int i=0;i<arr.length;i++){

if(map.containsKey(arr[i])){
map.put(arr[i],map.get(arr[i])+1);
}
else{
map.put(arr[i],1);
}
}

for(Map.Entry entry:map.entrySet()){
System.out.println(entry.getKey()+” “+entry.getValue());
}
}
}

0

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