DefaultDict Tutorial-HackerRank Solution

 DefaultDict Tutorial-HackerRank Solution

  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  integers,  and . There are  words, which might repeat, in word group . There are  words belonging to word group . For each  words, check whether the word has appeared in group  or not. Print the indices of each occurrence of  in group . If it does not appear, print .

Constraints

Input Format

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

Output Format

Output  lines.
The  line should contain the -indexed positions of the occurrences of the  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  times in positions ,  and .
'b' appeared  times in positions  and .
In the sample problem, if 'c' also appeared in word group , you would print .

DefaultDict Tutorial Solution

from collections import defaultdict
n,m=map(int,input().split())
d=defaultdict(list)
check=[]
for i in range(n):
    n=str(input())
    d[n].append(i+1) 
for i in range(m):
    n=input()
    check.append(n)
for i in check:
    if i in d:
         print (" ".join( map(str,d[i]) ))
    else:
         print("-1")

Explantion-

In defaultdict 'd' we are storing the element like-{'a': [1, 2, 4], 'b': [3, 5]}). Here we are doing (i+1) because we are starting our loop from 0 and in the output, the positions of elements are starting from 1. Whereas, the check list is usually like-['a', 'b']. In the final loop if the element is in check we will join the method in python to display the output. If the element not in check we will print -1