Python 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 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.
Example
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.
Expected output:
1 3
-1
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.
Constraints
1 <= n <= 10000
1 <= m <= 100
1 <= length of each word in the input <= 100
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
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
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:
from collections import defaultdict
input_n, input_m = map(int, input().split())
d = defaultdict(list)
for i in range(input_n):
ans1 = input()
d[ans1].append(i+1)
for j in range(input_m):
ans2 = input()
if ans2 in d:
print(*d[ans2])
else:
print(-1)
Steps Used in solving the problem -
Step 1: First we imported defaultdict from collections.
Step 2: then we have taken the input of input_n and input_m.
Step 3: then we defined defaultdict as d.
Step 4: In the fourth line, we create a for loop in the range of input_n.
Step 5: Inside for loop, we have taken input and appended it into d.
Step 6: then we created another for loop in the range of input_m.
Step 7: Inside for loop, we have taken input. Then we used an if condition to check if ans2 is in d then print the index value of ans2 else print -1.
