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.

Implementation:

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.
?>

Javascript

<script>

// Javascript 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 (let i = 0; i < n; i++)
        {
            let j;

            for (j = 0; j < m; j++)
                if (a[i] == b[j])
                    break;

            if (j == m)
                document.write(a[i] + " ");
        }
    }

    // Driver Code
    let a=[ 1, 2, 6, 3, 4, 5 ];
    let b=[2, 4, 3, 1, 0];
    let n = a.length;
    let m = b.length;
    findMissing(a, b, n, m);


    // This code is contributed by avanitrachhadiya2155

</script>

Output

6 5 

Time complexity: O(n*m) since using inner and outer loops
Auxiliary Space : O(1)

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.

Implementation:

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 */

Javascript

<script>
// Javascript 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[].
    function findMissing(a,b,n,m)
    {
        // Store all elements of
        // second array in a hash table
        let s = new Set();
        for (let i = 0; i < m; i++)
            s.add(b[i]);

        // Print all elements of first array
        // that are not present in hash table
        for (let i = 0; i < n; i++)
            if (!s.has(a[i]))
                document.write(a[i] + " ");
    }

    let a=[1, 2, 6, 3, 4, 5 ];
    let b=[2, 4, 3, 1, 0];
    let n = a.length;
    let m = b.length;
    findMissing(a, b, n, m);


// This code is contributed by patel2127
</script>

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. 

Approach 3: Recursion

Algorithm:

1. “findMissing” function takes four parameters, array “a” of size “n” and array “b” of size “m”. 
2. Base case : If n==0 ,  then there are no more elements left to check, so return from the function.
3. Recursive case : Check if the first element of array “a” is present in array “b”.  For this, use a for loop and iterate over all elements of array “b”.
4. If first element of array “a” is not in array “b”, print it.
5. Recursively call the “findMissing” function with the remaining elements of array “a” and array “b”. For this, increment the pointer of array “a” and decrease it’s size “n” by 1.
6. Call the recursive function “findMissing” in main() with array “a” and array “b”, and their respective sizes “n” and “m”.

Here’s the implementation:

C++

#include <iostream>
using namespace std;

void findMissing(int a[], int b[], int n, int m) {
    // Base case: if n is zero, then there are no more elements to check
    if (n == 0) {
        return;
    }

    // Recursive case: check if the first element of a[] is in b[]
    int i;
    for (i = 0; i < m; i++) {
        if (a[0] == b[i]) {
            break;
        }
    }

    // If the first element of a[] is not in b[], print it
    if (i == m) {
        cout << a[0] << " ";
    }

    // Recursively call findMissing with the remaining elements of a[] and b[]
    findMissing(a+1, b, n-1, m);
}

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);
    cout << endl;

    return 0;
}

// This code is contributed by Vaibhav Saroj

C

#include <stdio.h>

void findMissing(int a[], int b[], int n, int m) {
    // Base case: if n is zero, then there are no more elements to check
    if (n == 0) {
        return;
    }

    // Recursive case: check if the first element of a[] is in b[]
    int i;
    for (i = 0; i < m; i++) {
        if (a[0] == b[i]) {
            break;
        }
    }

    // If the first element of a[] is not in b[], print it
    if (i == m) {
        printf("%d ", a[0]);
    }

    // Recursively call findMissing with the remaining elements of a[] and b[]
    findMissing(a+1, b, n-1, m);
}

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[0]);

    findMissing(a, b, n, m);
    printf("\n");

    return 0;
}



// This code is contributed by Vaibhav Saroj

Java

import java.util.*;

class Main {
    public static void findMissing(int[] a, int[] b, int n, int m) {
        // Base case: if n is zero, then there are no more elements to check
        if (n == 0) {
            return;
        }

        // Recursive case: check if the first element of a[] is in b[]
        int i;
        for (i = 0; i < m; i++) {
            if (a[0] == b[i]) {
                break;
            }
        }

        // If the first element of a[] is not in b[], print it
        if (i == m) {
            System.out.print(a[0] + " ");
        }

        // Recursively call findMissing with the remaining elements of a[] and b[]
        findMissing(Arrays.copyOfRange(a, 1, n), b, n-1, m);
    }

    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);
        System.out.println();
    }
}



// This code is contributed by Vaibhav Saroj

Javascript

function findMissing(a, b, n, m) {
  // Base case: if n is zero, then there are no more elements to check
  if (n === 0) {
    return;
  }

  // Recursive case: check if the first element of a[] is in b[]
  let i;
  for (i = 0; i < m; i++) {
    if (a[0] === b[i]) {
      break;
    }
  }

  // If the first element of a[] is not in b[], print it
  if (i === m) {
    console.log(a[0] + " ");
  }

  // Recursively call findMissing with the remaining elements of a[] and b[]
  findMissing(a.slice(1), b, n - 1, m);
}

const a = [1, 2, 6, 3, 4, 5];
const b = [2, 4, 3, 1, 0];
const n = a.length;
const m = b.length;

findMissing(a, b, n, m);
console.log(); // add a newline at the end

// This code is contributed by Vaibhav Saroj

Output

6 5 

This Recursive approach and code is contributed by Vaibhav Saroj .

The time and space complexity:

 Time complexity :  O(nm) .

 Space complexity :  O(1) .