**DefaultDict Tutorial python**

### Problem Statement :

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, m and n. 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, m and n 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.

### Solution :

` ````
Solution in C :
from collections import defaultdict
import sys
d = defaultdict(list)
lst = list(map(int, input().split()))
n = lst[0]
m = lst[1]
for i in range(n):
d[sys.stdin.readline().strip()].append(i+1)
for i in range(m):
l = d[sys.stdin.readline().strip()]
if l: print(*l)
else: print(-1)
```

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