Find elements which are present in first array and not in second

Given two arrays, the task is that we find numbers which are present in first array, but not present in the second array.
Examples :

Input : a[] = {1, 2, 3, 4, 5, 10};
    b[] = {2, 3, 1, 0, 5};
Output : 4 10    
4 and 10 are present in first array, but
not in second array.

Input : a[] = {4, 3, 5, 9, 11};
        b[] = {4, 9, 3, 11, 10};
Output : 5  

Method 1 (Simple)
A Naive Approach is to use two loops and check element which not present in second array.

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ simple program to 
// find elements which are 
// not present in second array
#include<bits/stdc++.h>
using namespace std;
  
// Function for finding 
// elements which are there 
// in a[]  but not in b[].
void findMissing(int a[], int b[], 
                 int n, int m)
{
    for (int i = 0; i < n; i++)
    {
        int j;
        for (j = 0; j < m; j++)
            if (a[i] == b[j])
                break;
  
        if (j == m)
            cout << a[i] << " ";
    }
}
  
// Driver code
int main()
{
    int a[] = { 1, 2, 6, 3, 4, 5 };
    int b[] = { 2, 4, 3, 1, 0 };
    int n = sizeof(a) / sizeof(a[0]);
    int m = sizeof(b) / sizeof(b[1]);
    findMissing(a, b, n, m);
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java simple program to 
// find elements which are 
// not present in second array
class GFG 
{
      
    // Function for finding elements 
    // which are there in a[] but not
    // in b[].
    static void findMissing(int a[], int b[], 
                            int n, int m)
    {
        for (int i = 0; i < n; i++)
        {
            int j;
              
            for (j = 0; j < m; j++)
                if (a[i] == b[j])
                    break;
  
            if (j == m)
                System.out.print(a[i] + " ");
        }
    }
  
    // Driver Code
    public static void main(String[] args)
    {
        int a[] = { 1, 2, 6, 3, 4, 5 };
        int b[] = { 2, 4, 3, 1, 0 };
          
        int n = a.length;
        int m = b.length;
          
        findMissing(a, b, n, m);
    }
}
  
// This code is contributed
// by Anant Agarwal.

chevron_right


Python 3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python 3 simple program to find elements 
# which are not present in second array
  
# Function for finding elements which 
# are there in a[] but not in b[].
def findMissing(a, b, n, m):
  
    for i in range(n):
        for j in range(m):
            if (a[i] == b[j]):
                break
  
        if (j == m - 1):
            print(a[i], end = " ")
  
# Driver code
if __name__ == "__main__":
      
    a = [ 1, 2, 6, 3, 4, 5 ]
    b = [ 2, 4, 3, 1, 0 ]
    n = len(a)
    m = len(b)
    findMissing(a, b, n, m)
  
# This code is contributed 
# by ChitraNayal

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# simple program to find elements
// which are not present in second array
using System;
  
class GFG {
      
    // Function for finding elements 
    // which are there in a[] but not
    // in b[].
    static void findMissing(int []a, int []b, 
                            int n, int m)
    {
        for (int i = 0; i < n; i++)
        {
            int j;
              
            for (j = 0; j < m; j++)
                if (a[i] == b[j])
                    break;
  
            if (j == m)
                Console.Write(a[i] + " ");
        }
    }
  
    // Driver code
    public static void Main()
    {
        int []a = {1, 2, 6, 3, 4, 5};
        int []b = {2, 4, 3, 1, 0};
          
        int n = a.Length;
        int m = b.Length;
          
        findMissing(a, b, n, m);
    }
}
  
// This code is contributed by vt_m.

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP simple program to find 
// elements which are not 
// present in second array
  
// Function for finding 
// elements which are there
// in a[] but not in b[].
function findMissing( $a, $b, $n, $m)
{
    for ( $i = 0; $i < $n; $i++)
    {
        $j;
        for ($j = 0; $j < $m; $j++)
            if ($a[$i] == $b[$j])
                break;
  
        if ($j == $m)
            echo $a[$i] , " ";
    }
}
  
// Driver code
$a = array( 1, 2, 6, 3, 4, 5 );
$b = array( 2, 4, 3, 1, 0 );
$n = count($a);
$m = count($b);
findMissing($a, $b, $n, $m);
  
// This code is contributed by anuj_67.
?> 

chevron_right



Output :

6 5

 

Method 2 (Use Hashing)
In this method, we store all elements of second array in a hash table (unordered_set). One by one check all elements of first array and print all those elements which are not present in the hash table.

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ efficient program to 
// find elements which are not
// present in second array
#include<bits/stdc++.h>
using namespace std;
  
// Function for finding 
// elements which are there 
// in a[] but not in b[].
void findMissing(int a[], int b[], 
                 int n, int m)
{
    // Store all elements of 
    // second array in a hash table
    unordered_set <int> s;
    for (int i = 0; i < m; i++)
        s.insert(b[i]);
  
    // Print all elements of 
    // first array that are not
    // present in hash table
    for (int i = 0; i < n; i++)
        if (s.find(a[i]) == s.end())
            cout << a[i] << " ";
}
  
// Driver code
int main()
{
    int a[] = { 1, 2, 6, 3, 4, 5 };
    int b[] = { 2, 4, 3, 1, 0 };
    int n = sizeof(a) / sizeof(a[0]);
    int m = sizeof(b) / sizeof(b[1]);
    findMissing(a, b, n, m);
    return 0;

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java efficient program to find elements 
// which are not present in second array 
import java.util.HashSet;
import java.util.Set;
  
public class GfG{
  
    // Function for finding elements which  
    // are there in a[] but not in b[]. 
    static void findMissing(int a[], int b[],  
                     int n, int m) 
    
        // Store all elements of  
        // second array in a hash table 
        HashSet<Integer> s = new HashSet<>(); 
        for (int i = 0; i < m; i++) 
            s.add(b[i]); 
        
        // Print all elements of first array 
        // that are not present in hash table 
        for (int i = 0; i < n; i++) 
            if (!s.contains(a[i]))
                System.out.print(a[i] + " "); 
    
  
     public static void main(String []args){
          
        int a[] = { 1, 2, 6, 3, 4, 5 }; 
        int b[] = { 2, 4, 3, 1, 0 }; 
        int n = a.length; 
        int m = b.length; 
        findMissing(a, b, n, m);
     }
}
    
// This code is contributed by Rituraj Jain 

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 efficient program to find elements 
# which are not present in second array
  
# Function for finding elements which 
# are there in a[] but not in b[].
def findMissing(a, b, n, m):
      
    # Store all elements of second 
    # array in a hash table
    s = dict()
    for i in range(m):
        s[b[i]] = 1
  
    # Print all elements of first array 
    # that are not present in hash table
    for i in range(n):
        if a[i] not in s.keys():
            print(a[i], end = " ")
  
# Driver code
a = [ 1, 2, 6, 3, 4, 5 ]
b = [ 2, 4, 3, 1, 0 ]
n = len(a)
m = len(b)
findMissing(a, b, n, m)
  
# This code is contributed by mohit kumar

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# efficient program to find elements 
// which are not present in second array 
using System; 
using System.Collections.Generic;
  
class GfG
{
  
    // Function for finding elements which 
    // are there in a[] but not in b[]. 
    static void findMissing(int []a, int []b, 
                    int n, int m) 
    
        // Store all elements of 
        // second array in a hash table 
        HashSet<int> s = new HashSet<int>(); 
        for (int i = 0; i < m; i++) 
            s.Add(b[i]); 
          
        // Print all elements of first array 
        // that are not present in hash table 
        for (int i = 0; i < n; i++) 
            if (!s.Contains(a[i]))
                Console.Write(a[i] + " "); 
    
  
    // Driver code
    public static void Main(String []args)
    {
        int []a = { 1, 2, 6, 3, 4, 5 }; 
        int []b = { 2, 4, 3, 1, 0 }; 
        int n = a.Length; 
        int m = b.Length; 
        findMissing(a, b, n, m);
    }
}
  
/* This code contributed by PrinciRaj1992 */

chevron_right



Output :

6 5

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

This article is contributed by DANISH_RAZA . If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



My Personal Notes arrow_drop_up