Problem: DefaultDict Tutorial – HackerRank Python 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.
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
('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 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.
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.
1 3 -1
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.
1<=length of each word in the input<=100
Output m lines.
The ith line should contain the 1-indexed positions of the occurrences of the ith word separated by spaces.
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
1 2 4 3 5
‘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.
# DefaultDict hackerrank solution 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 hackerrank solution