Count distinct elements in an array

Given an unsorted array, count all distinct elements in it.

Examples:

Input :   arr[] = {10, 20, 20, 10, 30, 10}
Output : 3
There are three distinct elements 10, 20 and 30.

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

A simple solution is to run two loops. For every element, check if it has appeared before. If yes, increment count of distinct elements.

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to count distinct elements
// in a given array
#include <iostream>
using namespace std;
  
int countDistinct(int arr[], int n)
{
    int res = 1;
  
    // Pick all elements one by one
    for (int i = 1; i < n; i++) {
        int j = 0;
        for (j = 0; j < i; j++)
            if (arr[i] == arr[j])
                break;
  
        // If not printed earlier, then print it
        if (i == j)
            res++;
    }
    return res;
}
  
// Driver program to test above function
int main()
{
    int arr[] = { 12, 10, 9, 45, 2, 10, 10, 45 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << countDistinct(arr, n);
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to count distinct 
// elements in a given array
class GFG
{
      
static int countDistinct(int arr[], int n)
{
    int res = 1;
  
    // Pick all elements one by one
    for (int i = 1; i < n; i++) 
    {
        int j = 0;
        for (j = 0; j < i; j++)
            if (arr[i] == arr[j])
                break;
  
        // If not printed earlier, 
        // then print it
        if (i == j)
            res++;
    }
    return res;
}
  
// Driver code
public static void main(String[] args)
{
    int arr[] = { 12, 10, 9, 45, 2, 10, 10, 45 };
    int n = arr.length;
    System.out.println(countDistinct(arr, n));
}
}
  
// This code is contributed by Code_Mech.

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 program to count distinct 
# elements in a given array
def countDistinct(arr, n):
  
    res = 1
  
    # Pick all elements one by one
    for i in range(1, n):
        j = 0
        for j in range(i):
            if (arr[i] == arr[j]):
                break
  
        # If not printed earlier, then prit
        if (i == j + 1):
            res += 1
      
    return res
  
# Driver Code
arr = [12, 10, 9, 45, 2, 10, 10, 45]
n = len(arr)
print(countDistinct(arr, n))
  
# This code is contributed by Mohit Kumar

chevron_right


Output:

5

Time Complexity of above solution is O(n2). We can Use Sorting to solve the problem in O(nLogn) time. The idea is simple, first sort the array so that all occurrences of every element become consecutive. Once the occurrences become consecutive, we can traverse the sorted array and count distinct elements in O(n) time. Following is the implementation of the idea.

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to count all distinct elements
// in a given array
#include <algorithm>
#include <iostream>
using namespace std;
  
int countDistinct(int arr[], int n)
{
    // First sort the array so that all
    // occurrences become consecutive
    sort(arr, arr + n);
  
    // Traverse the sorted array
    int res = 0;
    for (int i = 0; i < n; i++) {
  
        // Move the index ahead while
        // there are duplicates
        while (i < n - 1 && arr[i] == arr[i + 1])
            i++;
  
        res++;
    }
  
    return res;
}
  
// Driver program to test above function
int main()
{
    int arr[] = { 6, 10, 5, 4, 9, 120, 4, 6, 10 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << countDistinct(arr, n);
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to count all distinct elements
// in a given array
import java.util.Arrays;
  
class GFG 
{
  
    static int countDistinct(int arr[], int n) 
    {
        // First sort the array so that all
        // occurrences become consecutive
        Arrays.sort(arr);
  
        // Traverse the sorted array
        int res = 0;
        for (int i = 0; i < n; i++) 
        {
  
            // Move the index ahead while
            // there are duplicates
            while (i < n - 1 && 
                    arr[i] == arr[i + 1])
            {
                i++;
            }
            res++;
        }
        return res;
    }
  
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = {6, 10, 5, 4, 9, 120, 4, 6, 10};
        int n = arr.length;
        System.out.println(countDistinct(arr, n));
    }
  
// This code is contributed by 29AjayKumar

chevron_right


Output:

6

We can Use Hashing to solve this in O(n) time on average. The idea is to traverse the given array from left to right and keep track of visited elements in a hash set , as a set consists of only unique elements.
Following is the implementation of the idea.

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

/* CPP program to print all distinct elements 
   of a given array */
#include <bits/stdc++.h>
using namespace std;
  
// This function prints all distinct elements
int countDistinct(int arr[], int n)
{
    // Creates an empty hashset
    unordered_set<int> s;
  
    // Traverse the input array
    int res = 0;
    for (int i = 0; i < n; i++) {
  
        // If not present, then put it in
        // hashtable and increment result
        if (s.find(arr[i]) == s.end()) {
            s.insert(arr[i]);
            res++;
        }
    }
  
    return res;
}
  
// Driver program to test above function
int main()
{
    int arr[] = { 6, 10, 5, 4, 9, 120, 4, 6, 10 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << countDistinct(arr, n);
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java Program to count
// Unique elements in Array
import java.util.*;
  
class GFG
{
      
    // This method returns count 
    // of Unique elements 
    public static int countDistinct(int arr[],int n)
    {
  
        HashSet<Integer> hs = new HashSet<Integer>();
  
        for(int i = 0; i < n; i++)
        
            // add all the elements to the HashSet 
            hs.add(arr[i]);
        }
          
        // return the size of hashset as 
        // it consists of all Unique elements 
        return hs.size();     
    }
  
    // Driver code
    public static void main(String[] args) 
    
        int arr[] = new int[]{6, 10, 5, 4, 9,
                                120, 4, 6, 10};
        System.out.println(countDistinct(arr, 
                                arr.length));
    }
}
  
// This code is contributed by Adarsh_Verma

chevron_right


Output:

6


My Personal Notes arrow_drop_up