DefaultDict Tutorial in Python - HackerRank Solution

DefaultDict Tutorial in Python - HackerRank Solution
DefaultDict Tutorial in Python - HackerRank Solution


Problem :


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 words, 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.



Constraints :

  • 1 <= n <= 10000
  • 1 <= m <= 100
  • 1 <= length of each word in the input <= 100

Input Format :

The first line contains integers, n and m separated by a space.
The next n lines contains the words belonging to group A.
The next m lines contains the words belonging to group B.

Output Format :

Output m lines.
The ith line should contain the 1-indexed positions of the occurrences of the ith word separated by spaces.



Sample Input :

5 2
a
a
b
a
b
a
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 :


# DefaultDict Tutorial in Python - Hacker Rank Solution
# Python 3
# Enter your code here. Read input from STDIN. Print output to STDOUT
# DefaultDict Tutorial in Python - Hacker Rank Solution START
from collections import defaultdict

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

a = defaultdict(list)
for i in range(1, n + 1):
    a[input()].append(i)

for i in range(1, m + 1):
    key = input()
    if len(a[key]) > 0:
        print(" ".join(str(c) for c in a[key]))
    else:
        print(-1)
# DefaultDict Tutorial in Python - Hacker Rank Solution END





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1 Comments
  • Success Guru
    Success Guru Tuesday, May 10, 2022

    I think we can also use for i in range(m):
    There is no need to use for i in range(1,m+1):

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