return sorted array in lexicographical order python

Question

a1 = ['arp', 'bull', 'mice']
a2 = ["lively", "alive", "harp", "sharp", "armstrong"]

I need it to return:

['arp']

My code is:

def in_array(array1, array2):
    x = array1
    return sorted(x)

which works for:

a1 = ["live", "arp", "strong"] 
a2 = ["lively", "alive", "harp", "sharp", "armstrong"]
r = ['arp', 'live', 'strong']

How do I sort through an array and only match the sorted elements that are substrings of a2?

Answer 1

You can simply use a filter and check whether there is any(..) element in a2 that contains such a substring:

def in_array(a1,a2):
    return filter(lambda e1: any(e1 in e2 for e2 in a2),a1)

In case you want the result to be sorted as well, you can use sorted(..) :

def in_array(a1,a2):
    return sorted(filter(lambda e1: any(e1 in e2 for e2 in a2),a1))

If you want to eliminate duplicates, you can use set(..) :

def in_array(a1,a2):
    return sorted(set(filter(lambda e1: any(e1 in e2 for e2 in a2),a1)))

The algorithm will run in O(n×m+n×log(n)) with n the number of elements in a1 , and m the number of elements in a2 .

You can boost the algorithm complexity in terms of O(n) in case you can do preprocessing on ar2 (and generate for instance a trie). But this is only beneficial if the number of elements in a1 is huge compared to the number of elements in a2 .