Handling duplicate in Arrays

Software Development
#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.

  • 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))