# Union and Intersection of two sorted arrays

• Difficulty Level : Easy
• Last Updated : 24 Jun, 2022

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 ``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);``    ``int` `n = ``sizeof``(arr2) / ``sizeof``(arr2);` `    ``// Function calling``    ``printUnion(arr1, arr2, m, n);` `    ``return` `0;``}`

## C

 `// C program to find union of``// two sorted arrays``#include ` `/* 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);``    ``int` `n = ``sizeof``(arr2) / ``sizeof``(arr2);``    ``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

 ``

## Javascript

 ``

Output

`1 2 3 4 5 6 7 `

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

Handling duplicates in any of the array : Above code does not handle duplicates in any of the array. 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 ``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 << ``" "``;` `    ``// Incrementing the First element's count in it's``    ``// corresponding index in newtable``    ``++newtable[arr1];` `    ``// 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);``    ``int` `m = ``sizeof``(arr2) / ``sizeof``(arr2);` `    ``UnionArray(arr1, arr2, n, m);` `    ``return` `0;``}` `// This code is contributed by Sania Kumari Gupta (kriSania804)`

## C

 `// C program for the above approach``#include ``#include ` `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);` `    ``// Incrementing the First element's count in it's``    ``// corresponding index in newtable``    ``++newtable[arr1];` `    ``// 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);``    ``int` `m = ``sizeof``(arr2) / ``sizeof``(arr2);` `    ``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 + ``" "``);` `        ``// Incrementing the First element's``        ``// count in it's corresponding``        ``// index in newtable``        ``++newtable[arr1];` `        ``// 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

 ``

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``    ``Unionarray(``int` `arr1[], ``int` `arr2[],``               ``int` `n, ``int` `m)``    ``{``        ``TreeSet 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 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 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 + n)

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 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``    ``Unionarray(``int` `arr1[], ``int` `arr2[],``               ``int` `n, ``int` `m)``    ``{``        ``HashMap map = ``new` `HashMap();``         ` `        ``// Remove the duplicates from arr1[]``        ``for` `(``int` `i =``0``;i 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 uni``            ``= Unionarray(arr1, arr2, n, m);``        ``for` `(``int` `i : uni) {``            ``System.out.print(i + ``" "``);``        ``}``    ``}``}`` ` `// This code is contributed by Aarti_Rathi`

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 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 ``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 ``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);``    ``int` `n = ``sizeof``(arr2) / ``sizeof``(arr2);` `    ``// Function calling``    ``printIntersection(arr1, arr2, m, n);` `    ``return` `0;``}`

## C

 `// C program to find intersection of``// two sorted arrays``#include ` `/* 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);``    ``int` `n = ``sizeof``(arr2) / ``sizeof``(arr2);``    ``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

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## Javascript

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Output

`2 5 `

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

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

## C++

 `// C++ program to find intersection of two sorted arrays``#include ``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<

## 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 s = ``new` `TreeSet();``        ``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

 ``

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``    ``Intersection(``int` `arr1[], ``int` `arr2[],``                 ``int` `n, ``int` `m)``    ``{``        ``TreeSet set = ``new` `TreeSet<>();``        ``// Removing duplicates from first array``        ``for` `(``int` `i : arr1)``            ``set.add(i);` `        ``ArrayList 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 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.

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