Find elements which are present in first array and not in second
Last Updated :
20 Sep, 2023
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++
#include<bits/stdc++.h>
using namespace std;
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] << " ";
}
}
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
class GFG
{
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] + " ");
}
}
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);
}
}
|
Python 3
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 = " ")
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)
|
C#
using System;
class GFG {
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] + " ");
}
}
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);
}
}
|
Javascript
<script>
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] + " ");
}
}
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);
</script>
|
PHP
<?php
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] , " ";
}
}
$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);
?>
|
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++
#include<bits/stdc++.h>
using namespace std;
void findMissing(int a[], int b[],
int n, int m)
{
unordered_set <int> s;
for (int i = 0; i < m; i++)
s.insert(b[i]);
for (int i = 0; i < n; i++)
if (s.find(a[i]) == s.end())
cout << a[i] << " ";
}
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
import java.util.HashSet;
import java.util.Set;
public class GfG{
static void findMissing(int a[], int b[],
int n, int m)
{
HashSet<Integer> s = new HashSet<>();
for (int i = 0; i < m; i++)
s.add(b[i]);
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);
}
}
|
Python3
def findMissing(a, b, n, m):
s = dict()
for i in range(m):
s[b[i]] = 1
for i in range(n):
if a[i] not in s.keys():
print(a[i], end = " ")
a = [ 1, 2, 6, 3, 4, 5 ]
b = [ 2, 4, 3, 1, 0 ]
n = len(a)
m = len(b)
findMissing(a, b, n, m)
|
C#
using System;
using System.Collections.Generic;
class GfG
{
static void findMissing(int []a, int []b,
int n, int m)
{
HashSet<int> s = new HashSet<int>();
for (int i = 0; i < m; i++)
s.Add(b[i]);
for (int i = 0; i < n; i++)
if (!s.Contains(a[i]))
Console.Write(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);
}
}
|
Javascript
<script>
function findMissing(a,b,n,m)
{
let s = new Set();
for (let i = 0; i < m; i++)
s.add(b[i]);
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);
</script>
|
Time complexity : O(n+m)
Auxiliary Space : O(n)
Approach 3: Recursion
Algorithm:
- “findMissing” function takes four parameters, array “a” of size “n” and array “b” of size “m”.
- Base case : If n==0 , then there are no more elements left to check, so return from the function.
- 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”.
- If first element of array “a” is not in array “b”, print it.
- 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.
- 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) {
if (n == 0) {
return;
}
int i;
for (i = 0; i < m; i++) {
if (a[0] == b[i]) {
break;
}
}
if (i == m) {
cout << a[0] << " ";
}
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;
}
|
C
#include <stdio.h>
void findMissing(int a[], int b[], int n, int m) {
if (n == 0) {
return;
}
int i;
for (i = 0; i < m; i++) {
if (a[0] == b[i]) {
break;
}
}
if (i == m) {
printf("%d ", a[0]);
}
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;
}
|
Java
import java.util.*;
class Main {
public static void findMissing(int[] a, int[] b, int n, int m) {
if (n == 0) {
return;
}
int i;
for (i = 0; i < m; i++) {
if (a[0] == b[i]) {
break;
}
}
if (i == m) {
System.out.print(a[0] + " ");
}
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();
}
}
|
Python3
def find_missing(a, b, n, m):
if n == 0:
return
i = 0
while i < m:
if a[0] == b[i]:
break
i += 1
if i == m:
print(a[0], end=" ")
find_missing(a[1:], b, n - 1, m)
def main():
a = [1, 2, 6, 3, 4, 5]
b = [2, 4, 3, 1, 0]
n = len(a)
m = len(b)
find_missing(a, b, n, m)
print()
if __name__ == "__main__":
main()
|
C#
using System;
public class Program
{
public static void FindMissing(int[] a, int[] b, int n, int m)
{
if (n == 0)
{
return;
}
int i;
for (i = 0; i < m; i++)
{
if (a[0] == b[i])
{
break;
}
}
if (i == m)
{
Console.Write(a[0] + " ");
}
FindMissing(new ArraySegment<int>(a, 1, n - 1).ToArray(), 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);
Console.WriteLine();
}
}
|
Javascript
function findMissing(a, b, n, m) {
if (n === 0) {
return;
}
let i;
for (i = 0; i < m; i++) {
if (a[0] === b[i]) {
break;
}
}
if (i === m) {
console.log(a[0] + " ");
}
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();
|
This Recursive approach and code is contributed by Vaibhav Saroj .
The time and space complexity:
Time complexity : O(nm) .
Space complexity : O(1) .
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