The output of the code is in different order than expected - python-3.x

Question

n = int(input())
a_name = []
a_score = []
for _ in range(n):
name = input()
score = float(input())
a_name.append(name)
a_score.append(score)
a_score1 = set(a_score)
a_score1.remove(min(a_score1))
x = min(a_score1)
for i in range(n):
if a_score[i]==x:
print(a_name[i])
For the following code input is:
5
Harry
37.21
Berry
37.21
Tina
37.2
Akriti
41
Harsh
39
My Output:
Harry
Berry
Expected Output:
Berry
Harry
I am a newbie in the programming world and would appreciate help to solve this.
This is a hacker rank problem and the problem is to find the names with a second minimum score. I was able to pass 8/10 test cases but 2 test cases are failing with this solution. I am able to find the correct names(i.e. names with second-lowest score) however it is printed in wrong order)

Answer 1

you can find first the second minim and then find the names that have the second minim using a list comprehension and the built-in function zip:
second_minim = sorted(a_score)[1]
[n for n, s in zip(a_name, a_score) if s == second_minim]
output:
['Harry', 'Berry']
related to why should be ['Berry', 'Harry'] instead of ['Harry', 'Berry'] according to the requirements that you pointed:
If there are multiple students with the same grade, order their names alphabetically and print each name on a new line
so you should use:
result = sorted(n for n, s in zip(a_name, a_score) if s == second_minim)
print(*result, sep='\n')
output:
Berry
Harry

Answer 2

So instead of using two lists, you can map the scores to a name with a dict. Like so:
scores_dict = {'Harry': 37.21,
'Berry': 37.21,
'Tina': 37.2,
'Akriti': 41,
'Harsh': 39
}
Next remove all matching keys to the starting min value so that we can get the second min
x = min(scores_dict, key=scores_dict.get)
filtered_scores = {k:v for k,v in scores_dict.items() if v != scores_dict[x]}
Then we grab all the keys of any of the new min value.
x = min(filtered_scores, key=scores_dict.get)
res = [key for key in filtered_scores if filtered_scores[key] == filtered_scores[x]]
Full Code
n = int(input)
scores_dict = {}
for _ in range(n):
name = input()
score = int(input())
scores_dict[name] = score
# The loop will produce this dict
'''
scores_dict = {'Harry': 37.21,
'Berry': 37.21,
'Tina': 37.2,
'Akriti': 41,
'Harsh': 39
}
'''
x = min(scores_dict, key=scores_dict.get)
filtered_scores = {k:v for k,v in scores_dict.items() if v != scores_dict[x]}
x = min(filtered_scores, key=scores_dict.get)
res = [key for key in filtered_scores if filtered_scores[key] == filtered_scores[x]]
print(res)
If you really want to use two lists, it can be done, but it adds potential for bugs by adding unneeded complexity.
Dictionaries are just associative (named) arrays (lists).

Answer 3

n = int(input())
a_name = []
a_score = []
for _ in range(n):
name = input()
score = float(input())
a_name.append(name)
a_score.append(score)
a_score1 = set(a_score)
a_score1.remove(min(a_score1))
x = min(a_score1)
y =[]
for i in range(n):
if a_score[i]==x:
y.append(a_name[i])
y.sort()
for j in range(len(y)):
print(y[j])
I got the output with this code. Thank you everyone for the help!