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

Software Development
#1

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)