HackerRank Python Solution - Collections Topic - DefaultDict

The defaultdict tool is a container in the collections class of Python. It's similar to the usual dictionary (dict) container, but the only difference is that a defaultdict will have a default value if that key has not been set yet. If you didn't use a defaultdict you'd have to check to see if that key exists, and if it doesn't, set it to what you want.

For example:

 

from collections import defaultdict
d = defaultdict(list)
d['python'].append("awesome")
d['something-else'].append("not relevant")
d['python'].append("language")
for i in d.items():
    print i

This prints:

('python', ['awesome', 'language'])
('something-else', ['not relevant'])

In this challenge, you will be given 2 integers, n, and m. There are n words, which might repeat, in word group A. There are m words belonging to word group B. For each m word, check whether the word has appeared in group A or not. Print the indices of each occurrence of m in group A. If it does not appear, print -1.

Example:

• Group A contains 'a', 'b', 'a'

• Group B contains 'a', 'c'

• For the first word in group B, 'a', it appears at positions 1 and 3 in group A. The second word, 'c', does not appear in group A, so print -1. 

Expected output:

1 3
-1

Input Format:

• The first line contains integers, n, and m separated by a space.

• The next n lines contain the words belonging to group A.

• The next m lines contain the words belonging to group B.

Constraints:

• 1 <= n <=10000

• 1 <= m <= 100

• 1 <= length of each word in input <= 100

Output Format:

• Output m lines.

• The i^th line should contain the 1-indexed positions of the occurrences of the i^th word separated by spaces.

Sample Input:

STDIN   Function
-----   --------
5 2     group A size n = 5, group B size m = 2
a       group A contains 'a', 'a', 'b', 'a', 'b'
a
b
a
b
a       group B contains 'a', 'b'
b

Sample Output:

1 2 4
3 5

Explanation: 

• 'a' appeared 3 times in positions 1, 2, and 4.

• 'b' appeared 2 times in positions 3 and 5.

• In the sample problem, if 'c' also appeared in word group B, you would print -1.

Solution:

from collections import defaultdict

n, m = map(int,input().split())

A = [input() for _ in range(n)]
B = [input() for _ in range(m)]

d = defaultdict(list)

for index,item in enumerate(A,1):
    d[item].append(index)

for item in B:
    if d[item]:
        print(*d[item])
    else:
        print(-1)

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