# Union of arrays arr1[] and arr2[] using JAVA

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}

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

`// 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);`

` ` `}`

`}`

Output:

1 2 3 4 5 6 7

Time Complexity : O(m + n)