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

Software Development
#1

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

// 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);

}

}

Output:

2 5

Time Complexity : O(m + n)

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.

  • 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)

Output:

2 4

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