Related Articles

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

• Difficulty Level : Easy
• Last Updated : 11 Aug, 2021

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  ```

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

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``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);``    ``int` `m = ``sizeof``(b) / ``sizeof``(b);``    ``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

 ``

## Javascript

 ``

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``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);``    ``int` `m = ``sizeof``(b) / ``sizeof``(b);``    ``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 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 */`

## Javascript

 ``

Output :

`6 5`

Time complexity : O(n+m)
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 write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.