Union and Intersection of two sorted arrays

Given two sorted arrays, find their union and intersection.
Example:

Input: arr1[] = {1, 3, 4, 5, 7}
        arr2[] = {2, 3, 5, 6} 
Output: Union : {1, 2, 3, 4, 5, 6, 7} 
         Intersection : {3, 5}

Input: arr1[] = {2, 5, 6}
        arr2[] = {4, 6, 8, 10} 
Output: Union : {2, 4, 5, 6, 8, 10} 
         Intersection : {6}

We strongly recommend that you click here and practice it, before moving on to the solution.

Union of arrays arr1[] and arr2[]

To find union of two sorted arrays, follow the following merge procedure : 

1) Use two index variables i and j, initial values i = 0, j = 0 
2) If arr1[i] is smaller than arr2[j] then print arr1[i] and increment i. 
3) If arr1[i] is greater than arr2[j] then print arr2[j] and increment j. 
4) If both are same then print any of them and increment both i and j. 
5) Print remaining elements of the larger array.

Below is the implementation of the above approach : 

C++

// C++ program to find union of
// two sorted arrays
#include <bits/stdc++.h>
using namespace std;

/* Function prints union of arr1[] and arr2[]
   m is the number of elements in arr1[]
   n is the number of elements in arr2[] */
void printUnion(int arr1[], int arr2[], int m, int n)
{
    int i = 0, j = 0;
    while (i < m && j < n) {
        if (arr1[i] < arr2[j])
            cout << arr1[i++] << " ";

        else if (arr2[j] < arr1[i])
            cout << arr2[j++] << " ";

        else {
            cout << arr2[j++] << " ";
            i++;
        }
    }

    /* Print remaining elements of the larger array */
    while (i < m)
        cout << arr1[i++] << " ";

    while (j < n)
        cout << arr2[j++] << " ";
}

/* Driver program to test above function */
int main()
{
    int arr1[] = { 1, 2, 4, 5, 6 };
    int arr2[] = { 2, 3, 5, 7 };

    int m = sizeof(arr1) / sizeof(arr1[0]);
    int n = sizeof(arr2) / sizeof(arr2[0]);

    // Function calling
    printUnion(arr1, arr2, m, n);

    return 0;
}

C

// C program to find union of
// two sorted arrays
#include <stdio.h>

/* Function prints union of arr1[] and arr2[]
   m is the number of elements in arr1[]
   n is the number of elements in arr2[] */
void printUnion(int arr1[], int arr2[], int m, int n)
{
    int i = 0, j = 0;
    while (i < m && j < n) {
        if (arr1[i] < arr2[j])
            printf(" %d ", arr1[i++]);
        else if (arr2[j] < arr1[i])
            printf(" %d ", arr2[j++]);
        else {
            printf(" %d ", arr2[j++]);
            i++;
        }
    }

    /* Print remaining elements of the larger array */
    while (i < m)
        printf(" %d ", arr1[i++]);
    while (j < n)
        printf(" %d ", arr2[j++]);
}

/* Driver program to test above function */
int main()
{
    int arr1[] = { 1, 2, 4, 5, 6 };
    int arr2[] = { 2, 3, 5, 7 };
    int m = sizeof(arr1) / sizeof(arr1[0]);
    int n = sizeof(arr2) / sizeof(arr2[0]);
    printUnion(arr1, arr2, m, n);
    getchar();
    return 0;
}

Java

// Java program to find union of
// two sorted arrays

class FindUnion {
    /* Function prints union of arr1[] and arr2[]
    m is the number of elements in arr1[]
    n is the number of elements in arr2[] */
    static int printUnion(int arr1[], int arr2[], int m, int n)
    {
        int i = 0, j = 0;
        while (i < m && j < n) {
            if (arr1[i] < arr2[j])
                System.out.print(arr1[i++] + " ");
            else if (arr2[j] < arr1[i])
                System.out.print(arr2[j++] + " ");
            else {
                System.out.print(arr2[j++] + " ");
                i++;
            }
        }

        /* Print remaining elements of
         the larger array */
        while (i < m)
            System.out.print(arr1[i++] + " ");
        while (j < n)
            System.out.print(arr2[j++] + " ");

        return 0;
    }

    public static void main(String args[])
    {
        int arr1[] = { 1, 2, 4, 5, 6 };
        int arr2[] = { 2, 3, 5, 7 };
        int m = arr1.length;
        int n = arr2.length;
        printUnion(arr1, arr2, m, n);
    }
}

Python3

# Python program to find union of
# two sorted arrays
# Function prints union of arr1[] and arr2[]
# m is the number of elements in arr1[]
# n is the number of elements in arr2[]
def printUnion(arr1, arr2, m, n):
    i, j = 0, 0
    while i < m and j < n:
        if arr1[i] < arr2[j]:
            print(arr1[i],end=" ")
            i += 1
        elif arr2[j] < arr1[i]:
            print(arr2[j],end=" ")
            j+= 1
        else:
            print(arr2[j],end=" ")
            j += 1
            i += 1

    # Print remaining elements of the larger array
    while i < m:
        print(arr1[i],end=" ")
        i += 1

    while j < n:
        print(arr2[j],end=" ")
        j += 1

# Driver program to test above function
arr1 = [1, 2, 4, 5, 6]
arr2 = [2, 3, 5, 7]
m = len(arr1)
n = len(arr2)
printUnion(arr1, arr2, m, n)

# This code is contributed by Pratik Chhajer

C#

// C# program to find union of
// two sorted arrays

using System;

class GFG {
    /* Function prints union of arr1[] and arr2[]
    m is the number of elements in arr1[]
    n is the number of elements in arr2[] */
    static int printUnion(int[] arr1,
                          int[] arr2, int m, int n)
    {
        int i = 0, j = 0;

        while (i < m && j < n) {
            if (arr1[i] < arr2[j])
                Console.Write(arr1[i++] + " ");
            else if (arr2[j] < arr1[i])
                Console.Write(arr2[j++] + " ");
            else {
                Console.Write(arr2[j++] + " ");
                i++;
            }
        }

        /* Print remaining elements of
        the larger array */
        while (i < m)
            Console.Write(arr1[i++] + " ");
        while (j < n)
            Console.Write(arr2[j++] + " ");

        return 0;
    }

    public static void Main()
    {
        int[] arr1 = { 1, 2, 4, 5, 6 };
        int[] arr2 = { 2, 3, 5, 7 };
        int m = arr1.Length;
        int n = arr2.Length;

        printUnion(arr1, arr2, m, n);
    }
}

// This code is contributed by Sam007

PHP

<?php
// PHP program to find union of
// two sorted arrays

/* Function prints union of
  arr1[] and arr2[] m is the
  number of elements in arr1[]
  n is the number of elements
  in arr2[] */
function printUnion($arr1, $arr2,
                         $m, $n)
{
    $i = 0; $j = 0;
    while ($i < $m && $j < $n)
    {
        if ($arr1[$i] < $arr2[$j])
            echo($arr1[$i++] . " ");

        else if ($arr2[$j] < $arr1[$i])
            echo($arr2[$j++] . " ");

        else
        {
            echo($arr2[$j++] . " ");
            $i++;
        }
    }

    // Print remaining elements
    // of the larger array
    while($i < $m)
        echo($arr1[$i++] . " ");

    while($j < $n)
        echo($arr2[$j++] . " ");
}

// Driver Code
$arr1 = array(1, 2, 4, 5, 6);
$arr2 = array(2, 3, 5, 7);

$m = sizeof($arr1);
$n = sizeof($arr2);

// Function calling
printUnion($arr1, $arr2, $m, $n);

// This code is contributed by Ajit.
?>

Javascript

<script>
// JavaScript program to find union of
// two sorted arrays

    /* Function prints union of arr1[] and arr2[]
    m is the number of elements in arr1[]
    n is the number of elements in arr2[] */
    function printUnion( arr1,  arr2,  m,  n)
    {
        var i = 0, j = 0;
        while (i < m && j < n) {
            if (arr1[i] < arr2[j])
                document.write(arr1[i++] + " ");
            else if (arr2[j] < arr1[i])
                document.write(arr2[j++] + " ");
            else {
                document.write(arr2[j++] + " ");
                i++;
            }
        }

        /* Print remaining elements of
        the larger array */
        while (i < m)
            document.write(arr1[i++] + " ");
        while (j < n)
            document.write(arr2[j++] + " ");

        return 0;
    }

        var arr1 = [ 1, 2, 4, 5, 6 ];
        var arr2 = [ 2, 3, 5, 7 ];
        var m = arr1.length;
        var n = arr2.length;
        printUnion(arr1, arr2, m, n);

// this code is contributed by shivanisinghss2110
</script>

Output

1 2 3 4 5 6 7 

Time Complexity : O(m + n)
Auxiliary Space: O(1)

Handling duplicates in any of the arrays: Above code does not handle duplicates in any of the arrays. To handle the duplicates, just check for every element whether adjacent elements are equal. 

Below is the implementation of this approach. 

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

static void UnionArray(int arr1[], int arr2[], int l1,
                       int l2)
{
    // Taking max element present in either array
    int m = arr1[l1 - 1];
    int n = arr2[l2 - 1];
    int ans = 0;
    if (m > n)
        ans = m;
    else
        ans = n;

    // Finding elements from 1st array (non duplicates
    // only). Using another array for storing union elements
    // of both arrays Assuming max element present in array
    // is not more than 10^7
    int newtable[ans + 1];
    memset(newtable, 0, sizeof(newtable));
    // First element is always present in final answer
    cout << arr1[0] << " ";

    // Incrementing the First element's count in it's
    // corresponding index in newtable
    ++newtable[arr1[0]];

    // Starting traversing the first array from 1st index
    // till last
    for (int i = 1; i < l1; i++) {
        // Checking whether current element is not equal to
        // it's previous element
        if (arr1[i] != arr1[i - 1]) {
            cout << arr1[i] << " ";
            ++newtable[arr1[i]];
        }
    }

    // Finding only non common elements from 2nd array
    for (int j = 0; j < l2; j++) {
        // By checking whether it's already resent in
        // newtable or not
        if (newtable[arr2[j]] == 0) {
            cout << arr2[j] << " ";
            ++newtable[arr2[j]];
        }
    }
}

// Driver Code
int main()
{
    int arr1[] = { 1, 2, 2, 2, 3 };
    int arr2[] = { 2, 3, 4, 5 };
    int n = sizeof(arr1) / sizeof(arr1[0]);
    int m = sizeof(arr2) / sizeof(arr2[0]);

    UnionArray(arr1, arr2, n, m);

    return 0;
}

// This code is contributed by Sania Kumari Gupta (kriSania804)

C

// C program for the above approach
#include <stdio.h>
#include <string.h>

static void UnionArray(int arr1[], int arr2[], int l1,
                       int l2)
{
    // Taking max element present in either array
    int m = arr1[l1 - 1];
    int n = arr2[l2 - 1];
    int ans = 0;
    if (m > n)
        ans = m;
    else
        ans = n;

    // Finding elements from 1st array (non duplicates
    // only). Using another array for storing union elements
    // of both arrays Assuming max element present in array
    // is not more than 10^7
    int newtable[ans + 1];
    for (int i = 0; i < ans + 1; i++)
        newtable[i] = 0;
    // First element is always present in final answer
    printf("%d ", arr1[0]);

    // Incrementing the First element's count in it's
    // corresponding index in newtable
    ++newtable[arr1[0]];

    // Starting traversing the first array from 1st index
    // till last
    for (int i = 1; i < l1; i++) {
        // Checking whether current element is not equal to
        // it's previous element
        if (arr1[i] != arr1[i - 1]) {
            printf("%d ", arr1[i]);
            ++newtable[arr1[i]];
        }
    }

    // Finding only non common elements from 2nd array
    for (int j = 0; j < l2; j++) {
        // By checking whether it's already resent in
        // newtable or not
        if (newtable[arr2[j]] == 0) {
            printf("%d ", arr2[j]);
            ++newtable[arr2[j]];
        }
    }
}

// Driver Code
int main()
{
    int arr1[] = { 1, 2, 2, 2, 3 };
    int arr2[] = { 2, 3, 4, 5 };
    int n = sizeof(arr1) / sizeof(arr1[0]);
    int m = sizeof(arr2) / sizeof(arr2[0]);

    UnionArray(arr1, arr2, n, m);

    return 0;
}

// This code is contributed by Sania Kumari Gupta (kriSania804)

Java

// Java program to find union of two
// sorted arrays (Handling Duplicates)
class FindUnion {

    static void UnionArray(int arr1[],
                           int arr2[])
    {
        // Taking max element present in either array
        int m = arr1[arr1.length - 1];
        int n = arr2[arr2.length - 1];

        int ans = 0;

        if (m > n) {
            ans = m;
        }
        else
            ans = n;

        // Finding elements from 1st array
        // (non duplicates only). Using
        // another array for storing union
        // elements of both arrays
        // Assuming max element present
        // in array is not more than 10^7
        int newtable[] = new int[ans + 1];

        // First element is always
        // present in final answer
        System.out.print(arr1[0] + " ");

        // Incrementing the First element's count
        // in it's corresponding index in newtable
        ++newtable[arr1[0]];

        // Starting traversing the first
        // array from 1st index till last
        for (int i = 1; i < arr1.length; i++) {
            // Checking whether current element
            // is not equal to it's previous element
            if (arr1[i] != arr1[i - 1]) {
                System.out.print(arr1[i] + " ");
                ++newtable[arr1[i]];
            }
        }

        // Finding only non common
        // elements from 2nd array
        for (int j = 0; j < arr2.length; j++) {
            // By checking whether it's already
            // present in newtable or not
            if (newtable[arr2[j]] == 0) {
                System.out.print(arr2[j] + " ");
                ++newtable[arr2[j]];
            }
        }
    }

    // Driver Code
    public static void main(String args[])
    {
        int arr1[] = { 1, 2, 2, 2, 3 };
        int arr2[] = { 2, 3, 4, 5 };

        UnionArray(arr1, arr2);
    }
}

Python3

# Python3 program to find union of two
# sorted arrays (Handling Duplicates)
def union_array(arr1, arr2):
    m = len(arr1)
    n = len(arr2)
    i = 0
    j = 0

    # keep track of last element to avoid duplicates
    prev = None

    while i < m and j < n:
        if arr1[i] < arr2[j]:
            if arr1[i] != prev:
                print(arr1[i], end=' ')
                prev = arr1[i]
            i += 1
        elif arr1[i] > arr2[j]:
            if arr2[j] != prev:
                print(arr2[j], end=' ')
                prev = arr2[j]
            j += 1
        else:
            if arr1[i] != prev:
                print(arr1[i], end=' ')
                prev = arr1[i]
            i += 1
            j += 1

    while i < m:
        if arr1[i] != prev:
            print(arr1[i], end=' ')
            prev = arr1[i]
        i += 1

    while j < n:
        if arr2[j] != prev:
            print(arr2[j], end=' ')
            prev = arr2[j]
        j += 1

# Driver Code
if __name__ == "__main__":
    arr1 = [1, 2, 2, 2, 3]
    arr2 = [2, 3, 4, 5]

    union_array(arr1, arr2)

# This code is contributed by Sanjay Kumar

C#

// C# program to find union of two
// sorted arrays (Handling Duplicates)
using System;

class GFG {

    static void UnionArray(int[] arr1,
                           int[] arr2)
    {

        // Taking max element present
        // in either array
        int m = arr1[arr1.Length - 1];
        int n = arr2[arr2.Length - 1];

        int ans = 0;

        if (m > n)
            ans = m;
        else
            ans = n;

        // Finding elements from 1st array
        // (non duplicates only). Using
        // another array for storing union
        // elements of both arrays
        // Assuming max element present
        // in array is not more than 10^7
        int[] newtable = new int[ans + 1];

        // First element is always
        // present in final answer
        Console.Write(arr1[0] + " ");

        // Incrementing the First element's
        // count in it's corresponding
        // index in newtable
        ++newtable[arr1[0]];

        // Starting traversing the first
        // array from 1st index till last
        for (int i = 1; i < arr1.Length; i++) {
            // Checking whether current
            // element is not equal to
            // it's previous element
            if (arr1[i] != arr1[i - 1]) {
                Console.Write(arr1[i] + " ");
                ++newtable[arr1[i]];
            }
        }

        // Finding only non common
        // elements from 2nd array
        for (int j = 0; j < arr2.Length; j++) {
            // By checking whether it's already
            // present in newtable or not
            if (newtable[arr2[j]] == 0) {
                Console.Write(arr2[j] + " ");
                ++newtable[arr2[j]];
            }
        }
    }

    // Driver Code
    public static void Main()
    {
        int[] arr1 = { 1, 2, 2, 2, 3 };
        int[] arr2 = { 2, 3, 4, 5 };

        UnionArray(arr1, arr2);
    }
}

// This code is contributed by anuj_67.

Javascript

<script>
// javascript program to find union of two
// sorted arrays (Handling Duplicates)

    function UnionArray(arr1 , arr2) {
        // Taking max element present in either array
        var m = arr1[arr1.length - 1];
        var n = arr2[arr2.length - 1];

        var ans = 0;

        if (m > n) {
            ans = m;
        } else
            ans = n;

        // Finding elements from 1st array
        // (non duplicates only). Using
        // another array for storing union
        // elements of both arrays
        // Assuming max element present
        // in array is not more than 10^7
        var newtable = Array(ans+1).fill(0);

        // First element is always
        // present in final answer
        document.write(arr1[0] + " ");

        // Incrementing the First element's count
        // in it's corresponding index in newtable
        newtable[arr1[0]]+=1;

        // Starting traversing the first
        // array from 1st index till last
        for (var i = 1; i < arr1.length; i++) {
            // Checking whether current element
            // is not equal to it's previous element
            if (arr1[i] != arr1[i - 1]) {
                document.write(arr1[i] + " ");
                newtable[arr1[i]]+= 1;
            }
        }

        // Finding only non common
        // elements from 2nd array
        for (var j = 0; j < arr2.length; j++) {
            // By checking whether it's already
            // present in newtable or not
            if (newtable[arr2[j]] == 0) {
                document.write(arr2[j] + " ");
                ++newtable[arr2[j]];
            }
        }
    }

    // Driver Code
        var arr1 = [ 1, 2, 2, 2, 3 ];
        var arr2 = [ 2, 3, 4, 5 ];

        UnionArray(arr1, arr2);

// This code is contributed by gauravrajput1
</script>

Output

1 2 3 4 5 

Time Complexity: O(l1 + l2) 
Auxiliary Space: O(n)

Thanks to Sanjay Kumar for suggesting this solution.

Another Approach using TreeSet in Java: The idea of the approach is to build a TreeSet and insert all the elements from both arrays into it. As a tree set stores only unique values, it will only keep all the unique values of both arrays.

Below is the implementation of the approach.

Java

// Java code to implement the approach

import java.io.*;
import java.util.*;

class GFG {

    // Function to return the union of two arrays
    public static ArrayList<Integer>
    Unionarray(int arr1[], int arr2[],
               int n, int m)
    {
        TreeSet<Integer> set = new TreeSet<>();

        // Remove the duplicates from arr1[]
        for (int i : arr1)
            set.add(i);

        // Remove duplicates from arr2[]
        for (int i : arr2)
            set.add(i);

        // Loading set to array list
        ArrayList<Integer> list
            = new ArrayList<>();
        for (int i : set)
            list.add(i);

        return list;
    }

    // Driver code
    public static void main(String[] args)
    {
        int arr1[] = { 1, 2, 2, 2, 3 };
        int arr2[] = { 2, 3, 3, 4, 5, 5 };
        int n = arr1.length;
        int m = arr2.length;

        // Function call
        ArrayList<Integer> uni
            = Unionarray(arr1, arr2, n, m);
        for (int i : uni) {
            System.out.print(i + " ");
        }
    }
}

//  Contributed by ARAVA SAI TEJA

Output

1 2 3 4 5 

Time Complexity: O(m + n) where ‘m’ and ‘n’ are the size of the arrays
Auxiliary Space: O(m*log(m)+n*log(n)) because adding element into TreeSet takes O(logn) time adding n elements will take (nlogn)

Thanks to Arava Sai Teja for suggesting this solution.

Another Approach using HashMap in Java: The idea of the approach is to build a HashMap and insert all the elements.  As a HashMap has the complexity of O(1) for insertion and lookup.

Below is the implementation of the approach.

Java

// Java code to implement the approach
import java.io.*;
import java.util.*;
import java.util.HashMap;

class GFG {

    // Function to return the union of two arrays
    public static ArrayList<Integer>
    Unionarray(int arr1[], int arr2[],
               int n, int m)
    {
        HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();

        // Remove the duplicates from arr1[]
        for (int i =0;i<arr1.length;i++)
        {
            if(map.containsKey(arr1[i]))
            {
                map.put(arr1[i], map.get(arr1[i]) + 1);
            }
            else
            {
                map.put(arr1[i], 1);
            }
        }

        // Remove duplicates from arr2[]
        for (int i =0;i<arr2.length;i++)
        {
            if(map.containsKey(arr2[i]))
            {
                map.put(arr2[i], map.get(arr2[i]) + 1);
            }
            else
            {
                map.put(arr2[i], 1);
            }
        }

        // Loading set to array list
        ArrayList<Integer> list = new ArrayList<>();
        for (int i : map.keySet())
        {
            list.add(i);;
        }

        return list;
    }

    // Driver code
    public static void main(String[] args)
    {
        int arr1[] = { 1, 2, 2, 2, 3 };
        int arr2[] = { 2, 3, 3, 4, 5, 5 };
        int n = arr1.length;
        int m = arr2.length;
           System.out.println("Union is :");
        // Function call
        ArrayList<Integer> uni
            = Unionarray(arr1, arr2, n, m);
        for (int i : uni) {
            System.out.print(i + " ");
        }
    }
}

// This code is contributed by Aarti_Rathi

C#

// C# code to implement the approach

using System;
using System.Collections;
using System.Collections.Generic;

public class GFG {

  public static ArrayList
    Unionarray(int[] arr1, int[] arr2, int n, int m)
  {
    Dictionary<int, int> map
      = new Dictionary<int, int>();

    // Remove the duplicates from arr1[]
    for (int i = 0; i < n; i++) {
      if (map.ContainsKey(arr1[i])) {
        map[arr1[i]] += 1;
      }
      else {
        map.Add(arr1[i], 1);
      }
    }

    // Remove duplicates from arr2[]
    for (int i = 0; i < m; i++) {
      if (map.ContainsKey(arr2[i])) {
        map[arr2[i]] += 1;
      }
      else {
        map.Add(arr2[i], 1);
      }
    }

    // Loading set to array list
    ArrayList list = new ArrayList();
    foreach(int i in map.Keys) { list.Add(i); }
    return list;
  }

  static public void Main()
  {

    // Code
    int[] arr1 = { 1, 2, 2, 2, 3 };
    int[] arr2 = { 2, 3, 3, 4, 5, 5 };
    int n = arr1.Length;
    int m = arr2.Length;
    Console.WriteLine("Union is :");
    // Function call
    ArrayList uni = Unionarray(arr1, arr2, n, m);
    foreach(int i in uni) { Console.Write(i + " "); }
  }
}

// This code is contributed by lokeshmvs21.

Output

Union is :
1 2 3 4 5 

Time Complexity: O(m + n) where ‘m’ and ‘n’ are the size of the arrays
Auxiliary Space: O(m + n)

Thanks to Aarti Rathi for suggesting this solution.

Another optimized approach: In the above code we use some extra auxiliary space by creating newtable[ ]. We can reduce the space complexity program to test the above function to constant by checking adjacent elements when incrementing i or j such that i or j directly move to the next distinct element. We can perform this operation in place (i.e. without using any extra space).

C++

// This implementation uses vectors but can be easily modified to adapt arrays
#include <bits/stdc++.h>
using namespace std;

/* Helper function for printUnion().
   This same function can also be implemented as a lambda function inside printUnion().
*/
void next_distinct(const vector<int> &arr, int &x) // Moving to next distinct element
{
  // vector CAN be passed by reference to avoid unnecessary copies.
  // x(index) MUST be passed by reference so to reflect the change in the original index parameter

  /* Checks whether the previous element is equal to the current element,
       if true move to the element at the next index else return with the current index
  */
    do
    {
        ++x;
    } while (x < arr.size() && arr[x - 1] == arr[x]);
}

void printUnion(vector<int> arr1, vector<int> arr2)
{
    int i = 0, j = 0;
    while (i < arr1.size() && j < arr2.size())
    {
        if (arr1[i] < arr2[j])
        {
            cout << arr1[i] << " ";
            next_distinct(arr1, i); // Incrementing i to next distinct element
        }
        else if (arr1[i] > arr2[j])
        {
            cout << arr2[j] << " ";
            next_distinct(arr2, j); // Incrementing j to next distinct element
        }
        else
        {
            cout << arr1[i] << " ";
            // OR cout << arr2[j] << " ";
            next_distinct(arr1, i); // Incrementing i to next distinct element
            next_distinct(arr2, j); // Incrementing j to next distinct element
        }
    }
    // Remaining elements of the larger array
    while (i < arr1.size())
    {
        cout << arr1[i] << " ";
        next_distinct(arr1, i); // Incrementing i to next distinct element
    }
    while (j < arr2.size())
    {
        cout << arr2[j] << " ";
        next_distinct(arr2, j); // Incrementing j to next distinct element
    }
}

int main()
{
    vector<int> arr1 = {1, 2, 2, 2, 3};    // Duplicates Present
    vector<int> arr2 = {2, 3, 3, 4, 5, 5}; // Duplicates Present

    printUnion(arr1, arr2);

    return 0;
}
// This code is contributed by ciphersaini.

Output

1 2 3 4 5 

Time Complexity:  where m & n are the sizes of the arrays.
Auxiliary Space: 

Intersection of arrays arr1[] and arr2[]

To find intersection of 2 sorted arrays, follow the below approach : 

1) Use two index variables i and j, initial values i = 0, j = 0 
2) If arr1[i] is smaller than arr2[j] then increment i. 
3) If arr1[i] is greater than arr2[j] then increment j. 
4) If both are same then print any of them and increment both i and j.

Below is the implementation of the above approach :

C++

// C++ program to find intersection of
// two sorted arrays
#include <bits/stdc++.h>
using namespace std;

/* Function prints Intersection of arr1[] and arr2[]
m is the number of elements in arr1[]
n is the number of elements in arr2[] */
void printIntersection(int arr1[], int arr2[], int m, int n)
{
    int i = 0, j = 0;
    while (i < m && j < n) {
        if (arr1[i] < arr2[j])
            i++;
        else if (arr2[j] < arr1[i])
            j++;
        else /* if arr1[i] == arr2[j] */
        {
            cout << arr2[j] << " ";
            i++;
            j++;
        }
    }
}

/* Driver program to test above function */
int main()
{
    int arr1[] = { 1, 2, 4, 5, 6 };
    int arr2[] = { 2, 3, 5, 7 };

    int m = sizeof(arr1) / sizeof(arr1[0]);
    int n = sizeof(arr2) / sizeof(arr2[0]);

    // Function calling
    printIntersection(arr1, arr2, m, n);

    return 0;
}

C

// C program to find intersection of
// two sorted arrays
#include <stdio.h>

/* Function prints Intersection of arr1[] and arr2[]
   m is the number of elements in arr1[]
   n is the number of elements in arr2[] */
void printIntersection(int arr1[], int arr2[], int m, int n)
{
    int i = 0, j = 0;
    while (i < m && j < n) {
        if (arr1[i] < arr2[j])
            i++;
        else if (arr2[j] < arr1[i])
            j++;
        else /* if arr1[i] == arr2[j] */
        {
            printf(" %d ", arr2[j++]);
            i++;
        }
    }
}

/* Driver program to test above function */
int main()
{
    int arr1[] = { 1, 2, 4, 5, 6 };
    int arr2[] = { 2, 3, 5, 7 };
    int m = sizeof(arr1) / sizeof(arr1[0]);
    int n = sizeof(arr2) / sizeof(arr2[0]);
    printIntersection(arr1, arr2, m, n);
    getchar();
    return 0;
}

Java

// Java program to find intersection of
// two sorted arrays

class FindIntersection {
    /* Function prints Intersection of arr1[] and arr2[]
       m is the number of elements in arr1[]
       n is the number of elements in arr2[] */
    static void printIntersection(int arr1[], int arr2[], int m, int n)
    {
        int i = 0, j = 0;
        while (i < m && j < n) {
            if (arr1[i] < arr2[j])
                i++;
            else if (arr2[j] < arr1[i])
                j++;
            else {
                System.out.print(arr2[j++] + " ");
                i++;
            }
        }
    }

    public static void main(String args[])
    {
        int arr1[] = { 1, 2, 4, 5, 6 };
        int arr2[] = { 2, 3, 5, 7 };
        int m = arr1.length;
        int n = arr2.length;
        printIntersection(arr1, arr2, m, n);
    }
}

Python3

# Python program to find intersection of
# two sorted arrays
# Function prints Intersection of arr1[] and arr2[]
# m is the number of elements in arr1[]
# n is the number of elements in arr2[]
def printIntersection(arr1, arr2, m, n):
    i, j = 0, 0
    while i < m and j < n:
        if arr1[i] < arr2[j]:
            i += 1
        elif arr2[j] < arr1[i]:
            j+= 1
        else:
            print(arr2[j],end=" ")
            j += 1
            i += 1

# Driver program to test above function
arr1 = [1, 2, 4, 5, 6]
arr2 = [2, 3, 5, 7]
m = len(arr1)
n = len(arr2)
printIntersection(arr1, arr2, m, n)

# This code is contributed by Pratik Chhajer

C#

// C# program to find Intersection of
// two sorted arrays

using System;

class GFG {

    /* Function prints Intersection of arr1[]
    and arr2[] m is the number of elements in arr1[]
    n is the number of elements in arr2[] */
    static void printIntersection(int[] arr1,
                                  int[] arr2, int m, int n)
    {
        int i = 0, j = 0;

        while (i < m && j < n) {
            if (arr1[i] < arr2[j])
                i++;
            else if (arr2[j] < arr1[i])
                j++;
            else {
                Console.Write(arr2[j++] + " ");
                i++;
            }
        }
    }

    // driver code
    public static void Main()
    {
        int[] arr1 = { 1, 2, 4, 5, 6 };
        int[] arr2 = { 2, 3, 5, 7 };
        int m = arr1.Length;
        int n = arr2.Length;

        printIntersection(arr1, arr2, m, n);
    }
}

// This code is contributed by Sam007

PHP

<?php
// PHP program to find intersection of
// two sorted arrays

/* Function prints Intersection
   of arr1[] and arr2[] m is the
   number of elements in arr1[]
   n is the number of elements
   in arr2[] */
function printIntersection($arr1, $arr2,
                                $m, $n)
{
    $i = 0 ;
    $j = 0;
    while ($i < $m && $j < $n)
    {
        if ($arr1[$i] < $arr2[$j])
            $i++;
        else if ($arr2[$j] < $arr1[$i])
            $j++;

        /* if arr1[i] == arr2[j] */
        else
        {
            echo $arr2[$j], " ";
            $i++;
            $j++;
        }
    }
}

// Driver Code
$arr1 = array(1, 2, 4, 5, 6);
$arr2 = array(2, 3, 5, 7);

$m = count($arr1);
$n = count($arr2);

// Function calling
printIntersection($arr1, $arr2, $m, $n);

// This code is contributed by anuj_67.
?>

Javascript

<script>

// JavaScript program to find intersection of
// two sorted arrays

// Function prints Intersection of arr1[] and arr2[]
// m is the number of elements in arr1[]
// n is the number of elements in arr2[]
function printIntersection(arr1, arr2, m, n)
{
    var i = 0, j = 0;
    while (i < m && j < n)
    {
        if (arr1[i] < arr2[j])
            i++;
        else if (arr2[j] < arr1[i])
            j++;
        else
        {
            document.write(arr2[j++] + " ");
            i++;
        }
    }
}

// Driver code
var arr1 = [ 1, 2, 4, 5, 6 ];
var arr2 = [ 2, 3, 5, 7 ];

var m = arr1.length;
var n = arr2.length;

printIntersection(arr1, arr2, m, n);

// This code is contributed by shivanisinghss2110

</script>

Output

2 5 

Time Complexity : O(m + n)
Auxiliary Space: O(1)

Handling duplicate in Arrays : 
The above code does not handle duplicate elements in arrays. The intersection should not count duplicate elements. To handle duplicates just check whether the current element is already present in the intersection list. Below is the implementation of this approach.

C++

// C++ program to find intersection of two sorted arrays
#include <bits/stdc++.h>
using namespace std;

/* Function prints Intersection of arr1[] and arr2[]
m is the number of elements in arr1[]
n is the number of elements in arr2[] */
void print_intersection(int arr1[], int arr2[], int m, int n)
{
    int i = 0, j = 0;
    set<int> s;  //set for handling duplicate elements in intersection list
    while (i < m && j < n) {
        if (arr1[i] < arr2[j])
            i++;
        else if (arr2[j] < arr1[i])
            j++;
        else /* if arr1[i] == arr2[j] */
        {
            s.insert(arr2[j]);   //insertion in set s
            i++;
            j++;
        }
    }
    for(auto itr: s)  //printing intersection set list
    {
        cout<<itr<<" ";
        }

}

/* Driver code */
int main()
{
    int arr1[] = { 1, 2, 2, 3, 4 };
    int arr2[] = { 2, 2, 4, 6, 7, 8 };

    int m = sizeof(arr1) / sizeof(arr1[0]);
    int n = sizeof(arr2) / sizeof(arr2[0]);

    // Function calling
    print_intersection(arr1, arr2, m, n);

    return 0;
}
// This code is contributed by Goldentiger.

Java

// Java program to find intersection of two sorted arrays
import java.io.*;
import java.util.*;
class Print_Intersection {
    /* Function prints Intersection of arr1[] and arr2[]
    m is the number of elements in arr1[]
    n is the number of elements in arr2[] */
    static void print_intersection(int arr1[], int arr2[],
                                   int m, int n)
    {
        // set for handling duplicate elements in
        // intersection list
        Set<Integer> s = new TreeSet<Integer>();
        int i = 0, j = 0;
        while (i < m && j < n) {
            if (arr1[i] < arr2[j])
                i++;
            else if (arr2[j] < arr1[i])
                j++;
            else {

                s.add(arr2[j++]); // insertion in set s
                i++;
            }
        }
        for (int element :
             s) // printing intersection set list
        {

            System.out.print(element + " ");
        }
        System.out.println("");
    }

    public static void main(String args[])
    {
        int arr1[] = { 1, 2, 2, 3, 4 };
        int arr2[] = { 2, 2, 4, 6, 7, 8 };
        int m = arr1.length;
        int n = arr2.length;
        print_intersection(arr1, arr2, m, n);
    }
}
// This code is contributed by CipherBhandari

Python3

# Python3 program to find Intersection of two
# Sorted Arrays (Handling Duplicates)
def IntersectionArray(a, b, n, m):
    '''
    :param a: given sorted array a
    :param n: size of sorted array a
    :param b: given sorted array b
    :param m: size of sorted array b
    :return: array of intersection of two array or -1
    '''

    Intersection = []
    i = j = 0

    while i < n and j < m:
        if a[i] == b[j]:

            # If duplicate already present in Intersection list
            if len(Intersection) > 0 and Intersection[-1] == a[i]:
                i+= 1
                j+= 1

            # If no duplicate is present in Intersection list
            else:
                Intersection.append(a[i])
                i+= 1
                j+= 1
        elif a[i] < b[j]:
            i+= 1
        else:
            j+= 1

    if not len(Intersection):
        return [-1]
    return Intersection

# Driver Code
if __name__ == "__main__":

    arr1 = [1, 2, 2, 3, 4]
    arr2 = [2, 2, 4, 6, 7, 8]

    l = IntersectionArray(arr1, arr2, len(arr1), len(arr2))
    print(*l)

# This code is contributed by Abhishek Kumar

C#

// C# program to find Intersection of
// two sorted arrays

using System;
using System.Collections.Generic;

class GFG {

    /* Function prints Intersection of arr1[] and arr2[]
    m is the number of elements in arr1[]
    n is the number of elements in arr2[] */
    static void print_intersection(int []arr1, int []arr2, int m, int n)
    {
        int i = 0, j = 0;
        HashSet<int> s = new HashSet<int>();  //set for handling duplicate elements in intersection list
        while (i < m && j < n) {
            if (arr1[i] < arr2[j])
                i++;
            else if (arr2[j] < arr1[i])
                j++;
            else /* if arr1[i] == arr2[j] */
            {
                s.Add(arr2[j]);   //insertion in set s
                i++;
                j++;
            }
        }
        foreach(int k in s)  //printing intersection set list
        {
            Console.Write(k + " ");
        }

    }

    // driver code
    public static void Main()
    {
        int[] arr1 = { 1, 2, 2, 3, 4 };
        int[] arr2 = { 2, 2, 4, 6, 7, 8};
        int m = arr1.Length;
        int n = arr2.Length;

        print_intersection(arr1, arr2, m, n);

    }
}

// This code is contributed by Aarti_Rathi

Javascript

<script>

// Python3 program to find Intersection of two
// Sorted Arrays (Handling Duplicates)
function IntersectionArray(a, b, n, m){
    // :param a: given sorted array a
    // :param n: size of sorted array a
    // :param b: given sorted array b
    // :param m: size of sorted array b
    // :return: array of intersection of two array or -1

    Intersection = []
    let i = j = 0

    while(i < n && j < m){
        if(a[i] == b[j]){

            // If duplicate already present in Intersection list
            if(Intersection.length > 0 && Intersection[Intersection.length-1] == a[i]){
                i+= 1
                j+= 1
            }

            // If no duplicate is present in Intersection list
            else{
                Intersection.push(a[i])
                i+= 1
                j+= 1
            }
        }
        else if(a[i] < b[j])
            i+= 1
        else
            j+= 1
    }

    if(!Intersection.length)
        return [-1]
    return Intersection

}

// Driver Code

let arr1 = [1, 2, 2, 3, 4]
let arr2 = [2, 2, 4, 6, 7, 8]

let l = IntersectionArray(arr1, arr2, arr1.length, arr2.length)
document.write(l)

// This code is contributed by shinjanpatra


</script>

Output

2 4 

Time Complexity : O(m + n) 
Auxiliary Space : O(min(m, n))

Another Approach using Tree Set: The idea of this approach is to build a tree set to store the unique elements of arri[]. Then compare the elements arr2[] with the tree set and also check if that is considered in the intersection to avoid duplicates.

Below is the implementation of the approach.

Java

// Java code to implement the approach

import java.io.*;
import java.util.*;

class GFG {

    // Function to find the intersection
    // of two arrays
    public static ArrayList<Integer>
    Intersection(int arr1[], int arr2[],
                 int n, int m)
    {
        TreeSet<Integer> set = new TreeSet<>();
        // Removing duplicates from first array
        for (int i : arr1)
            set.add(i);

        ArrayList<Integer> list
            = new ArrayList<>();

        // Avoiding duplicates and
        // adding intersections
        for (int i : arr2)
            if (set.contains(i)
                && !list.contains(i))
                list.add(i);

        // Sorting
        Collections.sort(list);
        return list;
    }

    // Driver code
    public static void main(String[] args)
    {
        int arr1[] = { 1, 2, 4, 5, 6 };
        int arr2[] = { 2, 3, 5, 7 };
        int n = arr1.length;
        int m = arr2.length;

        // Function call
        ArrayList<Integer> inter
            = Intersection(arr1, arr2, n, m);
        for (int i : inter) {
            System.out.print(i + " ");
        }
    }
}

//  Contributed by ARAVA SAI TEJA

Output

2 5 

Time Complexity: O(m+n)
Auxiliary Space: O(m+n)

Thanks to Arava Sai Teja for suggesting this solution. 

Another approach that is useful when difference between sizes of two given arrays is significant.
The idea is to iterate through the shorter array and do a binary search for every element of short array in big array (note that arrays are sorted). Time complexity of this solution is O(min(mLogn, nLogm)). This solution works better than the above approach when ratio of larger length to smaller is more than logarithmic order.

See following post for unsorted arrays. 
Find Union and Intersection of two unsorted arrays
Please write comments if you find any bug in above codes/algorithms, or find other ways to solve the same problem.