Find Union and Intersection of two unsorted arrays

Given two unsorted arrays that represent two sets (elements in every array are distinct), find union and intersection of two arrays.

For example, if the input arrays are:
arr1[] = {7, 1, 5, 2, 3, 6}
arr2[] = {3, 8, 6, 20, 7}
Then your program should print Union as {1, 2, 3, 5, 6, 7, 8, 20} and Intersection as {3, 6, 7}. Note that the elements of union and intersection can be printed in any order.

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

Method 1 (Naive)
Union:
1) Initialize union U as empty.
2) Copy all elements of first array to U.
3) Do following for every element x of second array:
…..a) If x is not present in first array, then copy x to U.
4) Return U.

Intersection:
1) Initialize intersection I as empty.
2) Do following for every element x of first array
…..a) If x is present in second array, then copy x to I.
4) Return I.

Time complexity of this method is O(mn) for both operations. Here m and n are number of elements in arr1[] and arr2[] respectively.

Method 2 (Use Sorting)
1) Sort arr1[] and arr2[]. This step takes O(mLogm + nLogn) time.
2) Use O(m + n) algorithms to find union and intersection of two sorted arrays.

Overall time complexity of this method is O(mLogm + nLogn).

Method 3 (Use Sorting and Searching)
Union:
1) Initialize union U as empty.
2) Find smaller of m and n and sort the smaller array.
3) Copy the smaller array to U.
4) For every element x of larger array, do following
…….b) Binary Search x in smaller array. If x is not present, then copy it to U.
5) Return U.

Intersection:
1) Initialize intersection I as empty.
2) Find smaller of m and n and sort the smaller array.
3) For every element x of larger array, do following
…….b) Binary Search x in smaller array. If x is present, then copy it to I.
4) Return I.

Time complexity of this method is min(mLogm + nLogm, mLogn + nLogn) which can also be written as O((m+n)Logm, (m+n)Logn). This approach works much better than the previous approach when difference between sizes of two arrays is significant.

Thanks to use_the_force for suggesting this method in a comment here.

Below is the implementation of this method.

C++

// A C++ program to print union and intersection  
/// of two unsorted arrays 
#include <iostream> 
#include <algorithm> 
using namespace std; 
  
int binarySearch(int arr[], int l, int r, int x); 
  
// Prints union of arr1[0..m-1] and arr2[0..n-1] 
void printUnion(int arr1[], int arr2[], int m, int n) 
{ 
    // Before finding union, make sure arr1[0..m-1]  
    // is smaller 
    if (m > n) 
    { 
        int *tempp = arr1; 
        arr1 = arr2; 
        arr2 = tempp; 
  
        int temp = m; 
        m = n; 
        n = temp; 
    } 
  
    // Now arr1[] is smaller 
  
    // Sort the first array and print its elements (these two 
    // steps can be swapped as order in output is not important) 
    sort(arr1, arr1 + m); 
    for (int i=0; i<m; i++) 
        cout << arr1[i] << " "; 
  
    // Search every element of bigger array in smaller array 
    // and print the element if not found 
    for (int i=0; i<n; i++) 
        if (binarySearch(arr1, 0, m-1, arr2[i]) == -1) 
            cout << arr2[i] << " "; 
} 
  
// Prints intersection of arr1[0..m-1] and arr2[0..n-1] 
void printIntersection(int arr1[], int arr2[], int m, int n) 
{ 
    // Before finding intersection, make sure arr1[0..m-1]  
    // is smaller 
    if (m > n) 
    { 
        int *tempp = arr1; 
        arr1 = arr2; 
        arr2 = tempp; 
  
        int temp = m; 
        m = n; 
        n = temp; 
    } 
  
    // Now arr1[] is smaller 
  
    // Sort smaller array arr1[0..m-1] 
    sort(arr1, arr1 + m); 
  
    // Search every element of bigger array in smaller 
    // array and print the element if found 
    for (int i=0; i<n; i++) 
        if (binarySearch(arr1, 0, m-1, arr2[i]) != -1) 
            cout << arr2[i] << " "; 
} 
  
// A recursive binary search function. It returns  
// location of x in given array arr[l..r] is present,  
// otherwise -1 
int binarySearch(int arr[], int l, int r, int x) 
{ 
    if (r >= l) 
    { 
        int mid = l + (r - l)/2; 
  
        // If the element is present at the middle itself 
        if (arr[mid] == x)  return mid; 
  
        // If element is smaller than mid, then it can only 
        // be presen in left subarray 
        if (arr[mid] > x)  
          return binarySearch(arr, l, mid-1, x); 
  
        // Else the element can only be present in right subarray 
        return binarySearch(arr, mid+1, r, x); 
    } 
  
    // We reach here when element is not present in array 
    return -1; 
} 
  
/* Driver program to test above function */
int main() 
{ 
    int arr1[] = {7, 1, 5, 2, 3, 6}; 
    int arr2[] = {3, 8, 6, 20, 7}; 
    int m = sizeof(arr1)/sizeof(arr1[0]); 
    int n = sizeof(arr2)/sizeof(arr2[0]); 
    cout << "Union of two arrays is n"; 
    printUnion(arr1, arr2, m, n); 
    cout << "nIntersection of two arrays is n"; 
    printIntersection(arr1, arr2, m, n); 
    return 0; 
} 

Java

// A Java program to print union and intersection  
/// of two unsorted arrays 
import java.util.Arrays; 
  
class UnionAndIntersection  
{ 
    // Prints union of arr1[0..m-1] and arr2[0..n-1] 
    void printUnion(int arr1[], int arr2[], int m, int n)  
    { 
        // Before finding union, make sure arr1[0..m-1]  
        // is smaller 
        if (m > n)  
        { 
            int tempp[] = arr1; 
            arr1 = arr2; 
            arr2 = tempp; 
  
            int temp = m; 
            m = n; 
            n = temp; 
        } 
  
        // Now arr1[] is smaller 
        // Sort the first array and print its elements (these two 
        // steps can be swapped as order in output is not important) 
        Arrays.sort(arr1); 
        for (int i = 0; i < m; i++) 
            System.out.print(arr1[i] + " "); 
  
        // Search every element of bigger array in smaller array 
        // and print the element if not found 
        for (int i = 0; i < n; i++)  
        { 
            if (binarySearch(arr1, 0, m - 1, arr2[i]) == -1) 
                System.out.print(arr2[i] + " "); 
        } 
    } 
  
    // Prints intersection of arr1[0..m-1] and arr2[0..n-1] 
    void printIntersection(int arr1[], int arr2[], int m, int n)  
    { 
        // Before finding intersection, make sure arr1[0..m-1]  
        // is smaller 
        if (m > n)  
        { 
            int tempp[] = arr1; 
            arr1 = arr2; 
            arr2 = tempp; 
  
            int temp = m; 
            m = n; 
            n = temp; 
        } 
  
        // Now arr1[] is smaller 
        // Sort smaller array arr1[0..m-1] 
        Arrays.sort(arr1); 
  
        // Search every element of bigger array in smaller array 
        // and print the element if found 
        for (int i = 0; i < n; i++)  
        { 
            if (binarySearch(arr1, 0, m - 1, arr2[i]) != -1)  
                System.out.print(arr2[i] + " "); 
        } 
    } 
  
    // A recursive binary search function. It returns location of x in 
    // given array arr[l..r] is present, otherwise -1 
    int binarySearch(int arr[], int l, int r, int x)  
    { 
        if (r >= l)  
        { 
            int mid = l + (r - l) / 2; 
  
            // If the element is present at the middle itself 
            if (arr[mid] == x) 
                return mid; 
  
            // If element is smaller than mid, then it can only  
            // be present in left subarray 
            if (arr[mid] > x) 
                return binarySearch(arr, l, mid - 1, x); 
  
            // Else the element can only be present in right subarray 
            return binarySearch(arr, mid + 1, r, x); 
        } 
  
        // We reach here when element is not present in array 
        return -1; 
    } 
  
    // Driver program to test above functions 
    public static void main(String[] args)  
    { 
        UnionAndIntersection u_i = new UnionAndIntersection(); 
        int arr1[] = {7, 1, 5, 2, 3, 6}; 
        int arr2[] = {3, 8, 6, 20, 7}; 
        int m = arr1.length; 
        int n = arr2.length; 
        System.out.println("Union of two arrays is "); 
        u_i.printUnion(arr1, arr2, m, n); 
        System.out.println(""); 
        System.out.println("Intersection of two arrays is "); 
        u_i.printIntersection(arr1, arr2, m, n); 
    } 
} 

Python3

# A Python3 program to print union and intersection  
# of two unsorted arrays 
  
# Prints union of arr1[0..m-1] and arr2[0..n-1] 
def printUnion(arr1, arr2, m, n): 
  
    # Before finding union, make sure arr1[0..m-1]  
    # is smaller 
    if (m > n): 
        tempp = arr1; 
        arr1 = arr2; 
        arr2 = tempp; 
  
        temp = m; 
        m = n; 
        n = temp; 
  
    # Now arr1[] is smaller 
  
    # Sort the first array and print its elements (these two 
    # steps can be swapped as order in output is not important) 
    arr1.sort(); 
    for i in range(0,m): 
        print(arr1[i],end=" "); 
  
    # Search every element of bigger array in smaller array 
    # and print the element if not found 
    for i in range(0,n): 
        if (binarySearch(arr1, 0, m - 1, arr2[i]) == -1): 
            print(arr2[i],end=" "); 
  
# Prints intersection of arr1[0..m-1] and arr2[0..n-1] 
def printIntersection(arr1, arr2, m, n): 
  
    # Before finding intersection, make sure arr1[0..m-1]  
    # is smaller 
    if (m > n): 
        tempp = arr1; 
        arr1 = arr2; 
        arr2 = tempp; 
  
        temp = m; 
        m = n; 
        n = temp; 
  
    # Now arr1[] is smaller 
  
    # Sort smaller array arr1[0..m-1] 
    arr1.sort(); 
  
    # Search every element of bigger array in smaller 
    # array and print the element if found 
    for i in range(0,n): 
        if (binarySearch(arr1, 0, m - 1, arr2[i]) != -1): 
            print(arr2[i],end=" "); 
  
# A recursive binary search function. It returns  
# location of x in given array arr[l..r] is present,  
# otherwise -1 
def binarySearch(arr, l, r,x): 
  
    if (r >= l): 
        mid = int(l + (r - l)/2); 
  
        # If the element is present at the middle itself 
        if (arr[mid] == x): 
            return mid; 
  
        # If element is smaller than mid, then it can only 
        # be presen in left subarray 
        if (arr[mid] > x): 
            return binarySearch(arr, l, mid - 1, x); 
  
        # Else the element can only be present in right subarray 
        return binarySearch(arr, mid + 1, r, x); 
  
    # We reach here when element is not present in array 
    return -1; 
  
# Driver program to test above function  
arr1 = [7, 1, 5, 2, 3, 6]; 
arr2 = [3, 8, 6, 20, 7]; 
m = len(arr1); 
n = len(arr2); 
print("Union of two arrays is "); 
printUnion(arr1, arr2, m, n); 
print("\nIntersection of two arrays is "); 
printIntersection(arr1, arr2, m, n); 
  
# This code is contributed by mits 

C#

// A C# program to print union and  
// intersection of two unsorted arrays 
using System; 
  
class GFG 
{ 
// Prints union of arr1[0..m-1] and arr2[0..n-1] 
static void printUnion(int []arr1, int []arr2, 
                                   int m, int n)  
{ 
    // Before finding union, make  
    // sure arr1[0..m-1] is smaller 
    if (m > n)  
    { 
        int []tempp = arr1; 
        arr1 = arr2; 
        arr2 = tempp; 
  
        int temp = m; 
        m = n; 
        n = temp; 
    } 
  
    // Now arr1[] is smaller 
    // Sort the first array and print 
    // its elements (these two steps can 
    // be swapped as order in output is  
    // not important) 
    Array.Sort(arr1); 
    for (int i = 0; i < m; i++) 
        Console.Write(arr1[i] + " "); 
  
    // Search every element of bigger  
    // array in smaller array and print  
    // the element if not found 
    for (int i = 0; i < n; i++)  
    { 
        if (binarySearch(arr1, 0, m - 1,  
                         arr2[i]) == -1) 
            Console.Write(arr2[i] + " "); 
    } 
} 
  
// Prints intersection of arr1[0..m-1]  
// and arr2[0..n-1] 
static void printIntersection(int []arr1, 
                              int []arr2,  
                              int m, int n)  
{ 
    // Before finding intersection,  
    // make sure arr1[0..m-1] is smaller 
    if (m > n)  
    { 
        int []tempp = arr1; 
        arr1 = arr2; 
        arr2 = tempp; 
  
        int temp = m; 
        m = n; 
        n = temp; 
    } 
  
    // Now arr1[] is smaller 
    // Sort smaller array arr1[0..m-1] 
    Array.Sort(arr1); 
  
    // Search every element of bigger array in  
    // smaller array and print the element if found 
    for (int i = 0; i < n; i++)  
    { 
        if (binarySearch(arr1, 0, m - 1, arr2[i]) != -1)  
            Console.Write(arr2[i] + " "); 
    } 
} 
  
// A recursive binary search function.  
// It returns location of x in given  
// array arr[l..r] is present, otherwise -1 
static int binarySearch(int []arr, int l,  
                        int r, int x)  
{ 
    if (r >= l)  
    { 
        int mid = l + (r - l) / 2; 
  
        // If the element is present at 
        // the middle itself 
        if (arr[mid] == x) 
            return mid; 
  
        // If element is smaller than mid, then it  
        // can only be present in left subarray 
        if (arr[mid] > x) 
            return binarySearch(arr, l, mid - 1, x); 
  
        // Else the element can only be  
        // present in right subarray 
        return binarySearch(arr, mid + 1, r, x); 
    } 
  
    // We reach here when element is 
    // not present in array 
    return -1; 
} 
  
// Driver Code 
static public void Main () 
{ 
    int []arr1 = {7, 1, 5, 2, 3, 6}; 
    int []arr2 = {3, 8, 6, 20, 7}; 
    int m = arr1.Length; 
    int n = arr2.Length; 
    Console.WriteLine("Union of two arrays is "); 
    printUnion(arr1, arr2, m, n); 
    Console.WriteLine(""); 
    Console.WriteLine("Intersection of two arrays is "); 
    printIntersection(arr1, arr2, m, n); 
} 
} 
  
// This code is contributed  
// by Sach_Code 

PHP

<?php 
// A PHP program to print union and intersection  
/// of two unsorted arrays 
  
// Prints union of arr1[0..m-1] and arr2[0..n-1] 
function printUnion($arr1, $arr2, $m, $n) 
{ 
    // Before finding union, make sure arr1[0..m-1]  
    // is smaller 
    if ($m > $n) 
    { 
        $tempp = $arr1; 
        $arr1 = $arr2; 
        $arr2 = $tempp; 
  
        $temp = $m; 
        $m = $n; 
        $n = $temp; 
    } 
  
    // Now arr1[] is smaller 
  
    // Sort the first array and print its elements (these two 
    // steps can be swapped as order in output is not important) 
    sort($arr1); 
    for ($i = 0; $i < $m; $i++) 
        echo $arr1[$i]." "; 
  
    // Search every element of bigger array in smaller array 
    // and print the element if not found 
    for ($i = 0; $i < $n; $i++) 
        if (binarySearch($arr1, 0, $m - 1, $arr2[$i]) == -1) 
            echo $arr2[$i]." "; 
} 
  
// Prints intersection of arr1[0..m-1] and arr2[0..n-1] 
function printIntersection($arr1, $arr2, $m, $n) 
{ 
    // Before finding intersection, make sure arr1[0..m-1]  
    // is smaller 
    if ($m > $n) 
    { 
        $tempp = $arr1; 
        $arr1 = $arr2; 
        $arr2 = $tempp; 
  
        $temp = $m; 
        $m = $n; 
        $n = $temp; 
    } 
  
    // Now arr1[] is smaller 
  
    // Sort smaller array arr1[0..m-1] 
    sort($arr1); 
  
    // Search every element of bigger array in smaller 
    // array and print the element if found 
    for ($i = 0; $i < $n; $i++) 
        if (binarySearch($arr1, 0, $m - 1, $arr2[$i]) != -1) 
            echo $arr2[$i]." "; 
} 
  
// A recursive binary search function. It returns  
// location of x in given array arr[l..r] is present,  
// otherwise -1 
function binarySearch($arr, $l, $r,$x) 
{ 
    if ($r >= $l) 
    { 
        $mid = (int)($l + ($r - $l)/2); 
  
        // If the element is present at the middle itself 
        if ($arr[$mid] == $x) return $mid; 
  
        // If element is smaller than mid, then it can only 
        // be presen in left subarray 
        if ($arr[$mid] > $x)  
        return binarySearch($arr, $l, $mid - 1, $x); 
  
        // Else the element can only be present in right subarray 
        return binarySearch($arr, $mid + 1, $r, $x); 
    } 
  
    // We reach here when element is not present in array 
    return -1; 
} 
  
/* Driver program to test above function */
    $arr1 = array(7, 1, 5, 2, 3, 6); 
    $arr2 = array(3, 8, 6, 20, 7); 
    $m = count($arr1); 
    $n = count($arr2); 
    echo "Union of two arrays is \n"; 
    printUnion($arr1, $arr2, $m, $n); 
    echo "\nIntersection of two arrays is \n"; 
    printIntersection($arr1, $arr2, $m, $n); 
  
// This code is contributed by mits 
?> 

Output:

Union of two arrays is
3 6 7 8 20 1 5 2
Intersection of two arrays is
7 3 6

Another Approach (When elements in the array may not be distinct) :

C

// C code to find intersection when 
// elements may not be distinct 
#include<bits/stdc++.h> 
  
using namespace std; 
  
// Function to find intersection 
void intersection(int a[], int b[], int n, int m) 
{ 
    int i = 0, j = 0; 
      
    while (i < n && j < m)  
    { 
                  
        if (a[i] > b[j])  
        { 
            j++; 
        }  
                  
        else 
        if (b[j] > a[i])  
        { 
            i++; 
        }  
        else 
        { 
            // when both are equal 
            cout << a[i] << " "; 
            i++; 
            j++; 
        } 
    } 
} 
  
// Driver Code 
int main() 
{ 
    int a[] = {1, 2, 3, 3, 4, 5, 5, 6}; 
    int b[] = {3, 3, 5}; 
      
    int n = sizeof(a)/sizeof(a[0]); 
    int m = sizeof(b)/sizeof(b[0]); 
    intersection(a, b, n, m); 
} 

Java

// Java code to find intersection when 
// elements may not be distinct 
  
import java.io.*; 
  
class GFG { 
      
// Function to find intersection 
static void intersection(int a[], int b[], int n, int m) 
{ 
    int i = 0, j = 0; 
      
    while (i < n && j < m)  
    { 
                  
        if (a[i] > b[j])  
        { 
            j++; 
        }  
                  
        else
        if (b[j] > a[i])  
        { 
            i++; 
        }  
        else
        { 
            // when both are equal 
            System.out.print(a[i] + " "); 
            i++; 
            j++; 
        } 
    } 
} 
  
// Driver Code 
    public static void main (String[] args) { 
    int a[] = {1, 2, 3, 3, 4, 5, 5, 6}; 
    int b[] = {3, 3, 5}; 
      
    int n = a.length; 
    int m = b.length; 
    intersection(a, b, n, m); 
    } 
} 

Python 3

# Python 3 code to find intersection  
# when elements may not be distinct 
  
# Function to find intersection 
def intersection(a, b, n, m): 
  
    i = 0
    j = 0
      
    while (i < n and j < m) : 
                  
        if (a[i] > b[j]) : 
            j += 1
                  
        else: 
            if (b[j] > a[i]) : 
                i += 1
  
            else: 
                # when both are equal 
                print(a[i], end = " ") 
                i += 1
                j += 1
  
# Driver Code 
if __name__ =="__main__": 
      
    a = [1, 2, 3, 3, 4, 5, 5, 6] 
    b = [3, 3, 5] 
      
    n = len(a) 
    m = len(b) 
    intersection(a, b, n, m) 
  
# This code is contributed by Ita_c 

C#

// C# code to find intersection when 
// elements may not be distinct 
  
using System; 
  
class GFG { 
      
// Function to find intersection 
static void intersection(int[] a, int[] b, int n, int m) 
{ 
    int i = 0, j = 0; 
      
    while (i < n && j < m)  
    { 
                  
        if (a[i] > b[j])  
        { 
            j++; 
        }  
                  
        else
        if (b[j] > a[i])  
        { 
            i++; 
        }  
        else
        { 
            // when both are equal 
            Console.Write(a[i] + " "); 
            i++; 
            j++; 
        } 
    } 
} 
  
// Driver Code 
    public static void Main () { 
    int[] a = {1, 2, 3, 3, 4, 5, 5, 6}; 
    int[] b = {3, 3, 5}; 
      
    int n = a.Length; 
    int m = b.Length; 
    intersection(a, b, n, m); 
    } 
} 
// this code is contributed by mukul singh 

PHP

<?php 
// PHP code to find intersection when 
// elements may not be distinct 
  
// Function to find intersection 
function intersection($a, $b, $n, $m) 
{ 
    $i = 0; $j = 0; 
      
    while ($i < $n && $j < $m)  
    { 
                  
        if ($a[$i] > $b[$j])  
        { 
            $j++; 
        }  
                  
        else
        if ($b[$j] > $a[$i])  
        { 
            $i++; 
        }  
        else
        { 
            // when both are equal 
            echo($a[$i] . " "); 
            $i++; 
            $j++; 
        } 
    } 
} 
  
// Driver Code 
$a = array(1, 2, 3, 3, 4, 5, 5, 6); 
$b = array(3, 3, 5); 
  
$n = sizeof($a); 
$m = sizeof($b); 
intersection($a, $b, $n, $m); 
  
// This code is contributed  
// by Mukul Singh 
?> 

Output :

3 3 5

Thanks Sanny Kumar for suggesting the above method.

Method 4 (Use Hashing)
Union:
Union
1. Initialize an empty hash set hs.
2. Iterate through the first array and put every element of the first array in the set S.
3. Repeat the process for the second array.
4. Print the set hs.

Intersection
1. Initialize an empty set hs.
2. Iterate through the first array and put every element of the first array in the set S.
3. For every element x of the second array, do the following :
Search x in the set hs. If x is present, then print it.

Time complexity of this method is Θ(m+n) under the assumption that hash table search and insert operations take Θ(1) time.

C++

// CPP program to find union and intersection 
// using sets 
#include <bits/stdc++.h> 
using namespace std; 
  
// Prints union of arr1[0..n1-1] and arr2[0..n2-1] 
void printUnion(int arr1[], int arr2[], int n1, int n2) 
{ 
    set<int> hs; 
  
    // Inhsert the elements of arr1[] to set hs 
    for (int i = 0; i < n1; i++) 
        hs.insert(arr1[i]); 
  
    // Insert the elements of arr2[] to set hs 
    for (int i = 0; i < n2; i++) 
        hs.insert(arr2[i]); 
  
    // Print the content of set hs 
    for (auto it = hs.begin(); it != hs.end(); it++) 
        cout << *it << " "; 
    cout << endl; 
} 
  
// Prints intersection of arr1[0..n1-1] and 
// arr2[0..n2-1] 
void printIntersection(int arr1[], int arr2[], 
                               int n1, int n2) 
{ 
    set<int> hs; 
  
    // Insert the elements of arr1[] to set S 
    for (int i = 0; i < n1; i++) 
        hs.insert(arr1[i]); 
  
    for (int i = 0; i < n2; i++) 
  
        // If element is present in set then 
        // push it to vector V 
        if (hs.find(arr2[i]) != hs.end()) 
            cout << arr2[i] << " "; 
} 
  
// Driver Program 
int main() 
{ 
    int arr1[] = { 7, 1, 5, 2, 3, 6 }; 
    int arr2[] = { 3, 8, 6, 20, 7 }; 
    int n1 = sizeof(arr1) / sizeof(arr1[0]); 
    int n2 = sizeof(arr2) / sizeof(arr2[0]); 
    printUnion(arr1, arr2, n1, n2); 
    printIntersection(arr1, arr2, n1, n2); 
  
    return 0; 
} 

Java

// Java program to find union and intersection 
// using Hashing 
import java.util.HashSet; 
  
class Test 
{    
    // Prints union of arr1[0..m-1] and arr2[0..n-1] 
    static void printUnion(int arr1[], int arr2[]) 
    { 
        HashSet<Integer> hs = new HashSet<>(); 
          
        for (int i = 0; i < arr1.length; i++)  
            hs.add(arr1[i]);         
        for (int i = 0; i < arr2.length; i++)  
            hs.add(arr2[i]); 
        System.out.println(hs);         
    } 
      
    // Prints intersection of arr1[0..m-1] and arr2[0..n-1] 
    static void printIntersection(int arr1[], int arr2[]) 
    { 
        HashSet<Integer> hs = new HashSet<>(); 
        HashSet<Integer> hs1 = new HashSet<>(); 
          
        for (int i = 0; i < arr1.length; i++)  
            hs.add(arr1[i]); 
          
        for (int i = 0; i < arr2.length; i++)  
            if (hs.contains(arr2[i])) 
               System.out.print(arr2[i] + " "); 
    } 
      
    // Driver method to test the above function 
    public static void main(String[] args)  
    { 
        int arr1[] = {7, 1, 5, 2, 3, 6}; 
        int arr2[] = {3, 8, 6, 20, 7}; 
  
        System.out.println("Union of two arrays is : "); 
        printUnion(arr1, arr2); 
          
        System.out.println("Intersection of two arrays is : "); 
        printIntersection(arr1, arr2);         
    } 
} 

Python

# Python program to find union and intersection  
# using sets 
def printUnion(arr1, arr2, n1, n2): 
    hs = set() 
      
    # Inhsert the elements of arr1[] to set hs  
    for i in range(0, n1): 
        hs.add(arr1[i]) 
          
    # Inhsert the elements of arr1[] to set hs  
    for i in range(0, n2): 
        hs.add(arr2[i]) 
    print("Union:") 
    for i in hs: 
        print(i, end=" ") 
    print("\n") 
      
    # Prints intersection of arr1[0..n1-1] and  
    # arr2[0..n2-1]  
def printIntersection(arr1, arr2, n1, n2): 
    hs = set() 
      
    # Insert the elements of arr1[] to set S  
    for i in range(0, n1): 
        hs.add(arr1[i]) 
    print("Intersection:") 
    for i in range(0,n2): 
          
        # If element is present in set then  
        # push it to vector V 
        if arr2[i] in hs: 
            print(arr2[i],end=" ") 
              
# Driver Program  
arr1 = [ 7, 1, 5, 2, 3, 6 ]  
arr2 = [ 3, 8, 6, 20, 7 ]  
n1 = len(arr1) 
n2 = len(arr2) 
printUnion(arr1, arr2, n1, n2)  
printIntersection(arr1, arr2, n1, n2)  
          
# This artice is contributed by Kumar Suman . 

C#

// C# program to find union and intersection 
// using Hashing 
using System; 
using System.Linq; 
using System.Collections.Generic; 
  
class GFG 
{  
    // Prints union of arr1[0..m-1] and arr2[0..n-1] 
    static void printUnion(int []arr1, int []arr2) 
    { 
        HashSet<int> hs = new HashSet<int>(); 
          
        for (int i = 0; i < arr1.Length; i++)  
            hs.Add(arr1[i]);      
        for (int i = 0; i < arr2.Length; i++)  
            hs.Add(arr2[i]); 
      
            Console.WriteLine(string.Join(", ", hs));      
    } 
      
    // Prints intersection of arr1[0..m-1] and arr2[0..n-1] 
    static void printIntersection(int []arr1, int []arr2) 
    { 
        HashSet<int> hs = new HashSet<int>(); 
          
        for (int i = 0; i < arr1.Length; i++)  
            hs.Add(arr1[i]); 
          
        for (int i = 0; i < arr2.Length; i++)  
            if (hs.Contains(arr2[i])) 
            Console.Write(arr2[i] + " "); 
    } 
      
    // Driver Code 
    static void Main()  
    { 
        int []arr1 = {7, 1, 5, 2, 3, 6}; 
        int []arr2 = {3, 8, 6, 20, 7}; 
  
        Console.WriteLine("Union of two arrays is : "); 
        printUnion(arr1, arr2); 
      
        Console.WriteLine("\nIntersection of two arrays is : "); 
        printIntersection(arr1, arr2);  
    } 
} 
  
// This code is contributed by mits 

This method is contributed by Ankur Singh.

Output:

Union of two arrays is : 
[1, 2, 3, 20, 5, 6, 7, 8]
Intersection of two arrays is : 
3 6 7

Method 5 (Kind of hashing technique without using any predefined Java Collections)
1. Intialize array with size of m+n
2. Fill first array value in resultant array by doing hashing(to find appropriate position)
3. Repeat for second array
4. While doing hashing if collision happens increment the position in recursive way

Java

// Java program to find union and intersection  
// using similar Hashing Technique  
// without using any predefined Java Collections 
// Time Complexity best case & avg case = O(m+n) 
// Worst case = O(nlogn) 
  
//package com.arrays.math; 
  
public class UnsortedIntersectionUnion { 
  
        // Prints intersection of arr1[0..n1-1] and 
        // arr2[0..n2-1] 
    public void findPosition(int a[], int b[]) { 
        int v = (a.length + b.length); 
        int ans[] = new int[v]; 
  
        int zero1 = 0; 
        int zero2 = 0; 
  
        System.out.print("Intersection : "); 
        // Iterate first array 
        for (int i = 0; i < a.length; i++) 
            zero1 = iterateArray(a, v, ans, i); 
  
        // Iterate second array 
        for (int j = 0; j < b.length; j++) 
            zero2 = iterateArray(b, v, ans, j); 
  
        int zero = zero1 + zero2; 
        placeZeros(v, ans, zero); 
        printUnion(v, ans, zero); 
  
    } 
      
    // Prints union of arr1[0..n1-1] and arr2[0..n2-1] 
    private void printUnion(int v, int[] ans, int zero) { 
        int zero1 = 0; 
        System.out.print("\nUnion : "); 
        for (int i = 0; i < v; i++) { 
            if ((zero == 0 && ans[i] == 0) || (ans[i] == 0 && zero1 > 0)) 
                continue; 
            if (ans[i] == 0) 
                zero1++; 
            System.out.print(ans[i] + ","); 
        } 
    } 
  
    private void placeZeros(int v, int[] ans, int zero) { 
        if (zero == 2) { 
            System.out.println("0"); 
            int d[] = { 0 }; 
            placeValue(d, ans, 0, 0, v); 
        } 
        if (zero == 1) { 
            int d[] = { 0 }; 
            placeValue(d, ans, 0, 0, v); 
        } 
    } 
  
    // Function to itreate array 
    private int iterateArray(int[] a, int v, int[] ans, int i) { 
        if (a[i] != 0) { 
            int p = a[i] % v; 
            placeValue(a, ans, i, p, v); 
        } else
            return 1; 
        return 0; 
    } 
  
    private void placeValue(int[] a, int[] ans, int i, int p, int v) { 
        p = p % v; 
        if (ans[p] == 0) 
            ans[p] = a[i]; 
        else { 
            if (ans[p] == a[i]) 
                System.out.print(a[i] + ","); 
            else { 
                  
                //Hashing collision happened increment position and do recursive call 
                p = p + 1; 
                placeValue(a, ans, i, p, v); 
            } 
        } 
    } 
  
    public static void main(String args[]) { 
        int a[] = { 7, 1, 5, 2, 3, 6 }; 
        int b[] = { 3, 8, 6, 20, 7 }; 
  
        UnsortedIntersectionUnion uiu = new UnsortedIntersectionUnion(); 
        uiu.findPosition(a, b); 
    } 
} 
// This code is contributed by Mohanakrishnan S. 

Python3

# Python3 program to find union and intersection  
# using similar Hashing Technique  
# without using any predefined Java Collections 
# Time Complexity best case & avg case = O(m+n) 
# Worst case = O(nlogn) 
  
  
# Prints intersection of arr1[0..n1-1] and 
# arr2[0..n2-1] 
def findPosition(a, b): 
    v = len(a) + len(b); 
    ans = [0]*v; 
    zero1 = zero2 = 0; 
    print("Intersection :",end=" "); 
      
    # Iterate first array 
    for i in range(len(a)): 
        zero1 = iterateArray(a, v, ans, i); 
      
    # Iterate second array 
    for j in range(len(b)): 
        zero2 = iterateArray(b, v, ans, j); 
      
    zero = zero1 + zero2; 
    placeZeros(v, ans, zero); 
    printUnion(v, ans, zero); 
      
# Prints union of arr1[0..n1-1] and arr2[0..n2-1] 
def printUnion(v, ans,zero): 
    zero1 = 0; 
    print("\nUnion :",end=" "); 
    for i in range(v): 
        if ((zero == 0 and ans[i] == 0) or
            (ans[i] == 0 and zero1 > 0)): 
            continue; 
        if (ans[i] == 0): 
            zero1+=1; 
        print(ans[i],end=","); 
  
def placeZeros(v, ans, zero): 
    if (zero == 2): 
        print("0"); 
        d = [0]; 
        placeValue(d, ans, 0, 0, v); 
    if (zero == 1): 
        d=[0]; 
        placeValue(d, ans, 0, 0, v); 
  
# Function to itreate array 
def iterateArray(a,v,ans,i): 
    if (a[i] != 0): 
        p = a[i] % v; 
        placeValue(a, ans, i, p, v); 
    else: 
        return 1; 
      
    return 0; 
  
def placeValue(a,ans,i,p,v): 
    p = p % v; 
    if (ans[p] == 0): 
        ans[p] = a[i]; 
    else: 
        if (ans[p] == a[i]): 
            print(a[i],end=","); 
        else: 
            # Hashing collision happened increment  
            # position and do recursive call 
            p = p + 1; 
            placeValue(a, ans, i, p, v); 
  
# Driver code 
a = [ 7, 1, 5, 2, 3, 6 ]; 
b = [ 3, 8, 6, 20, 7 ]; 
findPosition(a, b); 
  
# This code is contributed by mits 

C#

// C# program to find union and intersection  
// using similar Hashing Technique  
// without using any predefined Java Collections 
// Time Complexity best case & avg case = O(m+n) 
// Worst case = O(nlogn) 
  
//package com.arrays.math; 
using System; 
class UnsortedIntersectionUnion  
{ 
  
        // Prints intersection of arr1[0..n1-1] and 
        // arr2[0..n2-1] 
    public void findPosition(int []a, int []b) 
    { 
        int v = (a.Length + b.Length); 
        int []ans = new int[v]; 
  
        int zero1 = 0; 
        int zero2 = 0; 
  
        Console.Write("Intersection : "); 
          
        // Iterate first array 
        for (int i = 0; i < a.Length; i++) 
            zero1 = iterateArray(a, v, ans, i); 
  
        // Iterate second array 
        for (int j = 0; j < b.Length; j++) 
            zero2 = iterateArray(b, v, ans, j); 
  
        int zero = zero1 + zero2; 
        placeZeros(v, ans, zero); 
        printUnion(v, ans, zero); 
  
    } 
      
    // Prints union of arr1[0..n1-1]  
    // and arr2[0..n2-1] 
    private void printUnion(int v, int[] ans, int zero) 
    { 
        int zero1 = 0; 
        Console.Write("\nUnion : "); 
        for (int i = 0; i < v; i++)  
        { 
            if ((zero == 0 && ans[i] == 0) ||  
                    (ans[i] == 0 && zero1 > 0)) 
                continue; 
            if (ans[i] == 0) 
                zero1++; 
            Console.Write(ans[i] + ","); 
        } 
    } 
  
    private void placeZeros(int v, int[] ans, int zero) 
    { 
        if (zero == 2)  
        { 
            Console.WriteLine("0"); 
            int []d = { 0 }; 
            placeValue(d, ans, 0, 0, v); 
        } 
        if (zero == 1)  
        { 
            int []d = { 0 }; 
            placeValue(d, ans, 0, 0, v); 
        } 
    } 
  
    // Function to itreate array 
    private int iterateArray(int[] a, int v,  
                            int[] ans, int i) 
    { 
        if (a[i] != 0)  
        { 
            int p = a[i] % v; 
            placeValue(a, ans, i, p, v); 
        } else
            return 1; 
        return 0; 
    } 
  
    private void placeValue(int[] a, int[] ans,  
                                int i, int p, int v)  
    { 
        p = p % v; 
        if (ans[p] == 0) 
            ans[p] = a[i]; 
        else { 
            if (ans[p] == a[i]) 
                Console.Write(a[i] + ","); 
            else
            { 
                  
                //Hashing collision happened increment  
                // position and do recursive call 
                p = p + 1; 
                placeValue(a, ans, i, p, v); 
            } 
        } 
    } 
  
    // Driver code 
    public static void Main()  
    { 
        int []a = { 7, 1, 5, 2, 3, 6 }; 
        int []b = { 3, 8, 6, 20, 7 }; 
  
        UnsortedIntersectionUnion uiu = new UnsortedIntersectionUnion(); 
        uiu.findPosition(a, b); 
    } 
} 
  
// This code is contributed by PrinciRaj1992 

Output:

Intersection : 3,6,7,
Union : 1,2,3,5,6,7,8,20,

See following post for sorted arrays.
Find Union and Intersection of two sorted arrays

Please write comments if you find any bug in above codes/algorithms, or find other ways to solve the same problem.

#  Python program to find union and intersection  
#  using sets 
def printUnion(arr1,arr2,n1,n2): 
    hs=set() 
    #  Inhsert the elements of arr1[] to set hs  
    for i in range(0,n1): 
        hs.add(arr1[i]) 
    #  Inhsert the elements of arr1[] to set hs  
    for i in range(0,n2): 
        hs.add(arr2[i]) 
    print("Union:") 
    for i in hs: 
        print(i,end=" ") 
    print("\n") 
    #  Prints intersection of arr1[0..n1-1] and  
    #  arr2[0..n2-1]  
def printIntersection(arr1,arr2,n1,n2): 
    hs=set() 
    #  Insert the elements of arr1[] to set S  
    for i in range(0,n1): 
        hs.add(arr1[i]) 
    print("Intersection:") 
    for i in range(0,n2): 
        #  If element is present in set then  
        #  push it to vector V 
        if arr2[i] in hs: 
            print(arr2[i],end=" ") 
#  Driver Program  
arr1 = [ 7, 1, 5, 2, 3, 6 ]  
arr2 = [ 3, 8, 6, 20, 7 ]  
n1=len(arr1) 
n2=len(arr2) 
printUnion(arr1, arr2, n1, n2)  
printIntersection(arr1, arr2, n1, n2)  
          
# This artice is contributed by Kumar Suman . 

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My Personal Notes

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

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Java

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