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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