HackerRank DefaultDict Tutorial solution in Python | python problem solution

August 25, 2022 blogTech

Table of Contents

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 .

Example

Group A contains ‘a’, ‘b’, ‘a’ Group B contains ‘a’, ‘c’

For the first word in group B, ‘a’, it appears at positions and in group A. The second word, ‘c’, does not appear in group A, so print .

Expected output:

1 3
-1

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 .

Constraints

Output Format

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

Problem solution in Python 2 programming.

from collections import defaultdict
d = defaultdict(list)
n,m = map(int, raw_input().split())
for i in range(1,n+1):
    d[raw_input()].append(str(i))

for _ in range(m):
    w = raw_input()
    if w in d:
        print " ".join(d[w])
    else:
        print -1

Problem solution in Python 3 programming.

# Enter your code here. Read input from STDIN. Print output to STDOUT
from collections import defaultdict
d = defaultdict(list)
list1=[]
n, m = map(int,input().split())
for i in range(1, n+1):
    d[input()].append(str(i))


for i in range(m):
    b = input()
    if b in d: print(' '.join(d[b]))
    else: print(-1)

Problem solution in pypy programming.

# Enter your code here. Read input from STDIN. Print output to STDOUT
from collections import defaultdict
d, n = defaultdict(list), list(map(int, raw_input().split()))

for i in xrange(n[0]):
    d[raw_input()].append(i + 1)

for i in xrange(n[1]):
    print ' '.join(map(str, d[raw_input()])) or -1

Problem solution in pypy3 programming.

from collections import defaultdict
d=defaultdict(list)
n,m=map(int,input().split())
x=[]
for i in range(1,n+m+1):
    c=input()
    if(i>n):
        if(d[c]==[]):
            d[c].append(-1)
            x.append(d[c])
        else:
            x.append(d[c])
    else:
        d[c].append(i)

for i in range(m):
    print(" ".join(list(map(str,(x[i])))))