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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++

// 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;
}

Java

// 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.

Python 3

# 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

C#

// 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.

PHP

<?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.
?> 


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++

// 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;

Java

// 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 

Python3

# 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

C#

// 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 */


Output :

6 5

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

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



This article is attributed to GeeksforGeeks.org

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