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Find Union and Intersection of two unsorted arrays
  • Difficulty Level : Easy
  • Last Updated : 25 May, 2021

Given two unsorted arrays that represent two sets (elements in every array are distinct), find the 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.

Method 0 (Using Set in C++ STL):

Union of two arrays we can get with the Set data structure very easily. Set is a data structure that allows only the distinct elements in it. So, when we put the elements of both the array into the set we will get only the distinct elements that are equal to the union operation over the arrays. Let’s code it now –> 

C++




// C++ program for the union of two arrays using Set
#include <bits/stdc++.h>
using namespace std;
void getUnion(int a[], int n, int b[], int m)
{
     
    // Defining set container s
    set<int> s;
   
    // Inserting array elements in s
    for (int i = 0; i < n; i++)
      s.insert(a[i]);
   
    for (int i = 0; i < m; i++)
        s.insert(b[i]);
    cout << "Number of elements after union operation: " << s.size() << endl;
      cout << "The union set of both arrays is :" << endl;
    for (auto itr = s.begin(); itr != s.end(); itr++)
        cout << *itr
             << " "; // s will contain only distinct
                     // elements from array a and b
}
 
// Driver Code
int main()
{
    int a[9] = { 1, 2, 5, 6, 2, 3, 5, 7, 3 };
    int b[10] = { 2, 4, 5, 6, 8, 9, 4, 6, 5, 4 };
 
    getUnion(a, 9, b, 10);
}
 
// contributed by Anirban Chand

Output:

Number of elements after union operation: 9
The union set of both arrays is :
1 2 3 4 5 6 7 8 9 

The above method has a time complexity of –> O(m+n)



Method 1 (Using map data structure)

From the knowledge of data structures, we know that map stores distinct keys only. So if we insert any key appearing more than one time it gets stored only once. The idea is to insert both the arrays in one common map which would then store the distinct elements of both arrays (union of both the array).

Below is the implementation of the above method:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
void printUnion(int* a, int n, int* b, int m)
{
     
    // Defining map container mp
    map<int, int> mp;
   
    // Inserting array elements in mp
    for (int i = 0; i < n; i++)
        mp.insert({ a[i], i });
   
    for (int i = 0; i < m; i++)
        mp.insert({ b[i], i });
    cout << "The union set of both arrays is :" << endl;
    for (auto itr = mp.begin(); itr != mp.end(); itr++)
        cout << itr->first
             << " "; // mp will contain only distinct
                     // elements from array a and b
}
 
// Driver Code
int main()
{
    int a[7] = { 1, 2, 5, 6, 2, 3, 5 };
    int b[9] = { 2, 4, 5, 6, 8, 9, 4, 6, 5 };
 
    printUnion(a, 7, b, 9);
}

Java




// Java program for the above approach
import java.io.*;
import java.util.*;
 
class GFG{
 
static void printUnion(int[] a, int n,
                       int[] b, int m)
{
    Map<Integer,
        Integer> mp = new HashMap<Integer,
                                  Integer>();
     
    // Inserting array elements in mp
    for(int i = 0; i < n; i++)
    {
        mp.put(a[i], i);
    }
    for(int i = 0; i < m; i++)
    {
        mp.put(b[i], i);
    }
     
    System.out.println("The union set of both arrays is :");
    for(Map.Entry mapElement : mp.entrySet())
    {
        System.out.print(mapElement.getKey() + " ");
         
        // mp will contain only distinct
        // elements from array a and b
    }
}
 
// Driver Code
public static void main (String[] args)
{
    int a[] = { 1, 2, 5, 6, 2, 3, 5 };
    int b[] = { 2, 4, 5, 6, 8, 9, 4, 6, 5 };
     
    printUnion(a, 7, b, 9);
}
}
 
// This code is contributed by avanitrachhadiya2155

Output

The union set of both arrays is :
1 2 3 4 5 6 8 9

The above method has time complexity O(m+n).

*This method is suggested by Vinay Verma

Method 2 (Naive) 
Union: 

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

Intersection: 

  1. Initialize intersection I as empty.
  2. Do the following for every element x of the first array
    1. If x is present in the second array, then copy x to I.
  3. Return I.

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



Method 3 (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.

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

Method 4 (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 the following
    1. Binary Search x in the 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 the following
    1. Binary Search x in the 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 the difference between the 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 <algorithm>
#include <iostream>
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]);
   
    // Function call
    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 code
    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;
       
        // Function call
        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 code
arr1 = [7, 1, 5, 2, 3, 6]
arr2 = [3, 8, 6, 20, 7]
m = len(arr1)
n = len(arr2)
 
# Function call
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;
       
        // Function call
        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
?>

Javascript




<script>
 
// A JavaScript program to
// print union and intersection
// of two unsorted arrays
 
// 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) {
        let mid = l + Math.floor((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;
}
 
// 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) {
        let tempp = arr1;
        arr1 = arr2;
        arr2 = tempp;
 
        let 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((a, b) => a - b);
    for (let i = 0; i < m; i++)
        document.write(arr1[i] + " ");
 
    // Search every element of bigger array in smaller array
    // and print the element if not found
    for (let i = 0; i < n; i++)
        if (binarySearch(arr1, 0, m - 1, arr2[i]) == -1)
            document.write(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) {
        let tempp = arr1;
        arr1 = arr2;
        arr2 = tempp;
 
        let temp = m;
        m = n;
        n = temp;
    }
 
    // Now arr1[] is smaller
 
    // Sort smaller array arr1[0..m-1]
    arr1.sort((a, b) => a - b);
 
    // Search every element of bigger array in smaller
    // array and print the element if found
    for (let i = 0; i < n; i++)
        if (binarySearch(arr1, 0, m - 1, arr2[i]) != -1)
            document.write(arr2[i] + " ");
}
 
 
 
/* Driver program to test above function */
    let arr1 = [ 7, 1, 5, 2, 3, 6 ];
    let arr2 = [ 3, 8, 6, 20, 7 ];
    let m = arr1.length;
    let n = arr2.length;
 
    // Function call
    document.write("Union of two arrays is <br>");
    printUnion(arr1, arr2, m, n);
    document.write("<br>Intersection of two arrays is<br>");
    printIntersection(arr1, arr2, m, n);
 
// This code is contributed by Surbhi Tyagi.
</script>
Output
Union of two arrays is n3 6 7 8 20 1 5 2 nIntersection of two arrays is n7 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, 3, 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]);
   
    // sort
    sort(a, a + n);
    sort(b, b + m);
   
    // Function call
    intersection(a, b, n, m);
}

Java




// Java code to find intersection when
// elements may not be distinct
 
import java.io.*;
import java.util.Arrays;
 
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, 3, 2, 3, 4, 5, 5, 6 };
        int b[] = { 3, 3, 5 };
 
        int n = a.length;
        int m = b.length;
       
        // sort
        Arrays.sort(a);
        Arrays.sort(b);
       
        // Function call
        intersection(a, b, n, m);
    }
}

Python3




# 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, 3, 2, 3, 4, 5, 5, 6]
    b = [3, 3, 5]
 
    n = len(a)
    m = len(b)
     
    # sort
    a.sort()
    b.sort()
     
    # function call
    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, 3, 2, 3, 4, 5, 5, 6 };
        int[] b = { 3, 3, 5 };
 
        int n = a.Length;
        int m = b.Length;
       
        // sort
        Array.Sort(a);
        Array.Sort(b);
       
        // Function call
        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, 3, 2, 3, 4, 5, 5, 6);
$b = array(3, 3, 5);
 
$n = sizeof($a);
$m = sizeof($b);
 
// sort
sort($a);
sort($b);
 
// Function call
intersection($a, $b, $n, $m);
 
// This code is contributed
// by Mukul Singh
?>

Javascript




<script>
// Javascript code to find intersection when
// elements may not be distinct
 
    // Function to find intersection
    function intersection(a,b,n,m)
    {
        let 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
                document.write(a[i] + " ");
                i++;
                j++;
            }
        }
    }
     
    // Driver Code
    let a = [1, 3, 2, 3, 4, 5, 5, 6 ];
    let b = [3, 3, 5 ]
     
    let n = a.length;
    let m = b.length;
     
    // sort
    a.sort(); 
    b.sort();
     
    // Function call
    intersection(a, b, n, m);
     
    // This code is contributed by rag2127
</script>
Output
3 3 5

 Thanks Sanny Kumar for suggesting the above method.



Method 5 (Use Hashing) 
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 ?(m+n) under the assumption that hash table search and insert operations take ?(1) time.
 Below is the implementation of the above idea:

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]);
 
    // Function call
    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 code
    public static void main(String[] args)
    {
        int arr1[] = { 7, 1, 5, 2, 3, 6 };
        int arr2[] = { 3, 8, 6, 20, 7 };
 
        // Function call
        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)
 
# Function call
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
Output
1 2 3 5 6 7 8 20 
3 6 7

This method is contributed by Ankur Singh.

Method 6 (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

Below is the implementation of the above code: 

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);
            }
        }
    }
 
    // Driver code
    public static void main(String args[])
    {
        int a[] = { 7, 1, 5, 2, 3, 6 };
        int b[] = { 3, 8, 6, 20, 7 };
 
        // Function call
        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,

Python3




#  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)
 
# Function call
printUnion(arr1, arr2, n1, n2)
printIntersection(arr1, arr2, n1, n2)
 
# This artice is contributed by Kumar Suman .

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

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