#10 : Nested List | hackerrank Python Solutions

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[spanningTree]

The question has to done using Nested Lists and not using the dictionary. Appreciate your video
Here is the correct solution
if __name__ == '__main__':
students = []
scores = []
for N in range(int(input())):
name = input()
score = float(input())
scores.append(score)
students.append([name, score])


count = scores.count(min(scores))
for i in range(count):
scores.remove(min(scores))

secondHigh = min(scores)

students.sort()
output = [x for x in students if x[1] == secondHigh]
for i in output:
print(i[0])

[tayus_animations]

you have explained the logic in a very good way.

[Bhongaon]

if __name__ == '__main__':
students = []
for _ in range(int(input())):
name = input()
score = float(input())
students.append([name, score])

second_lowest = sorted(list(set([x[1] for x in students])))[1]
for name in sorted([x[0] for x in students if x[1] == second_lowest]):
print(name)

[NightlyNewsChat]

Thank you for the video, I am from Taiwan, it will be great if you can speak English throughout for the future tutorials ~

[HappyHour]

Keep doing this, you are really doing amazing

[priyashakya]

List = []
s = []
for _ in range(int( input())):
name = input()
score= float( input())
List.append([name, score])
s.append( score)
a = min(s)
while(a in s):
s.remove(a)

b = min(s)
List.sort()
for i in range(len(List)):
if(List[i][1]==b):
print(List[i][0])

[preetmehta]

dic={}
if __name__ == '__main__':
names=list()
scores=list()
l=list()
for i in range(0, int(input())):
name = input()
score = float(input())
if score in dic:
dic[score].append(name)
else:
dic[score]=[name]
names.append(name)
scores.append(score)
l.append([name, score])

s=sorted(set(scores))[1]
for i in sorted(dic[s]):
print(i)

[shivamparashar]

bhai volume badha lo yrr ! kuch samaj ni aa rha...

[niharsampath]

if __name__ == '__main__':

ls = []
n = int(input())
for _ in range(n):
name = input()
score = float(input())
ls.append([name, score)
ls.sort(key=lambda x: x[1])
lowest_score = ls[0][1]
second_lowest_score = None
for i in range(1, n):
if ls[i][1] > lowest_score:
second_lowest_score = ls[i][1]
break
result = [name for name, score in ls if score == second_lowest_score]
result.sort()
for name in result:
print(name)

[Yoohooooo]

Did you just copy the exact same solution from the editorial?

[JashSingh-bvge]

the Problem has to be solved using the nested list. that’s why my and you’re code is so different :)

[luciferm-ue]

l1=[]
l2=[]
for i in range(int(input("Enter how many students: "))):
name=input("Enter your names: ")
marks=float(input("Enter numbers: "))
l1.append([name, marks])
l2.append(marks)
l2=list(sorted(set(l2)))
lo=l2[1]
l3=[]
for i in l1:
if lo==i[1]:
l3.append(i[0])
l3.sort()
for i in l3:
print(i)


BRO TRY THIS :)

[machour]

#Create arrays according to the number mentioned in the first input
n = int(input())

students_list = [ ]

#Distribute the first (name) and second (grade) value to the first array and so on until all the arrays are finished
for _ in range(0, n):
name = input()
grade = float(input())
students_list.append([name, grade])

#Sort grades by lowest to highest
students_list = sorted(students_list, key=lambda student: student[1])

#Assign value to the lowest grade
lowest_grade = students_list[0][1]

#Assign value to the second lowest grade
for name, grade in students_list:
if grade > lowest_grade:
second_lowest_grade = grade
break

#Filter and sort names alphabetically
second_lowest_names = sorted([name for name, grade in students_list if grade == second_lowest_grade])

#Print the sorted names
if second_lowest_names:
for name in second_lowest_names:
print(name)

[OMKARDESHMUKH-kj]

bhai hindi may bol if you dont know english, their is not a problem in speaking in hindi bhai tu dhono language may bolra a na tho kuch nahi sumuj aara english to bol tura or hindi to bol

[IdleDeathGambling]

I understand Hindi/Urdu a little bit to know what youre saying :)

[ksquarev-kkv]

Can we use append method in dictionary?