4th answer: Tried in a different way and works newlist1=[] for i in range(0,30): if not newlist1: newlist1.append(i) print('first',i) elif len(newlist1)==1: print(i+newlist1[0]) newlist1.append(i) else: print(newlist1[len(newlist1)-1]+newlist1[len(newlist1)-2]) newlist1.append(newlist1[len(newlist1)-1]+newlist1[len(newlist1)-2])
Question on 3rd answer - no need to iterate over the list with a for-loop or did I miss something? Just get the smallest number, append to new, remove from existing and repeat using the while. data_list = [25, 55, 78, 64, 25, 12, 22, 11, 1, 2, 44, 3, 122, 23, 34] new_list = [] while data_list: minimum = min(data_list) new_list.append(minimum) data_list.remove(minimum) new_list.reverse(sort = True) print(new_list)
Questions are okay. But most of the solutions can be further optimised.. For example palindrome can be check while interating string .just check I and n-1-i char ..mis match then false I don't think reversed would take constant time..
Just to suggest a fancy solution to Question 8 in 8:10 sentence = "Adeptus Mechanicus. Glory to the Machine God." space_count = 0 for i in sentence: if i.isspace() == True: space_count += 1 word_count = space_count +1 print(word_count)
some nice questions to play around with shortest or interesnig answer. For example for first question. There is much shorter solution with if statment, and it possible to do it with oneliner: print(*(i for i in range(100, 200) if i % 2 != 0), sep=" ") just for fun.
Here is alternative for question 3 def sort_list(x): l = [] k = list(x) for j in range(0, len(k)): for i in k: if i == max(k): l.append(i) k.remove(i) print(l)
ye bhi chlega bro L = [23,45,10,9,11,66] final = [ ] while L : final.insert(0,max(L)) l.remove(max(L)) print(final) but l = [23,45,10,9,11,66] new = [ [ max(l),l.remove(max(l))]for i in range(len(l))] print(new) can we do this? i want in one line answer, but it returns [[66, None], [45, None], [23, None], [11, None], [10, None], [9, None]] is it right way to do this????
Okay but you time complexity is O(N^2) , not only solving the problem but also ALWAYS think of how to get the best optimized solution to a given problem.
Hi. Can you please confirm on the Question No: 14 solution. I think below code also produce the same result, then why you are using.... rstrip() etc. new_string = "How are you doing" output = new_string.split() print(output) Result = ['How', 'are', 'you', 'doing'] Thank you.
Last problem can be solved as below: s="missisippi" list1=[] for i in s: if (i not in list1) or (i==' '): list1.append(i) new_s=''.join(list1) print(new_s)
Hi Sir, In below code not getting how, lst1 is appended with element 6 when we are appending only lst2 only. Can you please Clarify, Thanks lst1=[1,2,3,4,5] lst2=lst1 lst2.append(6) print('lst1- ',lst1) print('lst2- ',lst2) Output- lst1- [1, 2, 3, 4, 5, 6] lst2- [1, 2, 3, 4, 5, 6]
In 3rd question, it should have been maximum instead of minimum as if x>minimum, minimum should not be equal to x. Please correct me if I am wrong guys !!
name is not matter : def sort_function(list): final = [ ] for i in range(len(list)): first_elem = list[0] for num in list: if num>first_elem: first_elem = num final.insert(0,first_elem) list.remove(first_elem) return final
What if we use built in fun i.e min() and max() to sort a list a=[5,1,8,10,3,7,2,4] b=[] while a: minimum = a[0] for i in a: if i==max(a): b.append(i) a.remove(i) print(b)
Use this template for print the questions (Can anybody solve this? pls help) Enter name: Enter department: Enter feedback: Enter year: The program should output the result with the entered values Name: Department: Feedback: Year:
9).Solution def Search (arr,x): for i in range(len(arr)): if arr[i]==x: return arr[i] print('Sorry This element is not Given in this array') list1=[1,2,3,4,5,6,7,8,9,10] Search(list1,10) Pardon Sir, if we can use this code for Solution (9). So that will give us the actual Values that are in the given array. instead of using iloc (Index Location).
Q15: #given a string , return recurring characters and display new string s = 'mississippi' lst = [] for i in s: if i not in lst: lst.append(i) newS = ''.join(lst) newS (or) #given a string , return recurring characters and display new string s = 'mississippi' setList = set() for i in s: setList.add(i) newS = ''.join(setList) newS
1.
for i in range(100,201):
y = []
for x in range(2,12):
if i%x == 0:
y.append(1)
else:
y.append(0)
if sum(y)==0:
print("prime number ",i)
4th answer: Tried in a different way and works
newlist1=[]
for i in range(0,30):
if not newlist1:
newlist1.append(i)
print('first',i)
elif len(newlist1)==1:
print(i+newlist1[0])
newlist1.append(i)
else:
print(newlist1[len(newlist1)-1]+newlist1[len(newlist1)-2])
newlist1.append(newlist1[len(newlist1)-1]+newlist1[len(newlist1)-2])
3rd problem.. Variable name should be maximum.. Right?
Exactly
Question on 3rd answer - no need to iterate over the list with a for-loop or did I miss something? Just get the smallest number, append to new, remove from existing and repeat using the while.
data_list = [25, 55, 78, 64, 25, 12, 22, 11, 1, 2, 44, 3, 122, 23, 34]
new_list = []
while data_list:
minimum = min(data_list)
new_list.append(minimum)
data_list.remove(minimum)
new_list.reverse(sort = True)
print(new_list)
I think the idea is not to use min/max method, but yeah, for isnt that needed.
Your're not wasting anyone's time -- we all come here to learn so don't skip anything, no matter how minute it is.
TO LEARN AND CRACK COMPETITIVE PROGRAMS
th-cam.com/channels/yoFQsVztx2oHWq14nY8A6A.html
Questions are okay. But most of the solutions can be further optimised..
For example palindrome can be check while interating string .just check I and n-1-i char ..mis match then false
I don't think reversed would take constant time..
def palindrom(s):
return "Yes it is palindrom" if s[::-1]==s else "not a palindrom"
if we write Madam , it returns False, Very case sensitive bro, so here we convert to lower( ) first then run. will it be okay?
How about this
Def ss(a)
A=[]
B=[]
A.append (a)
B.append(a)
B.reverse()
If A==B:
print("yes")
Elif A!=B:
print("no")
Else:
print("error")
@@noxnpc still nlog time complexity if you can do it by converting that will take long which significantly very optimised
Just to suggest a fancy solution to Question 8 in 8:10
sentence = "Adeptus Mechanicus. Glory to the Machine God."
space_count = 0
for i in sentence:
if i.isspace() == True:
space_count += 1
word_count = space_count +1
print(word_count)
Q1:
For num in range(100, 200) :
Cpt = 0
For i in range(2, num) :
If num % i != 0:
Cpt += 1
If cpt == num - 2:
Print(num)
For i in range(2, num) what is the range of i ? please explain this for loop
PALINDROME
s = input()
if s[0:] == s[::-1]:
print("palindrome")
else:
print("not palidrome")
Exactly it is working ✨
not optimised
Very good work. I just to saying that Fibonacci first number is 0
Good one💯
But it said to write a function to sort not lib function..l.sort() is method of a list
List in reverse :
for i in range(len(input_list)-1,-1,-1) :
reverse_list.append(input_list[i])
print(reverse_list)
yes this one is also correct, but print(L[::-1]) it will save time bro .
l=[12,43,2,35,354,5,521]
for i in range(0,len(l)):
for j in range(i+1,len(l)):
if l[i]
Qus 1:
for x in range(100,200,2):
print(x)
wrong..returning even numbers not prime
explain in 5th question sir... I understand logic of slice operations.. But output?
he just forgot to put it and somehow Google prints the result anyways.
In real Python code you SHOULD write the print() func
Thx for this video
Great, thanks for the video
th-cam.com/channels/yoFQsVztx2oHWq14nY8A6A.html
last question:
a='arrrbbb'
b=''
for i in a:
if i not in b:
b=b+i
b
Thanks a lot for video. I refreshed so many concepts
q.5>
li = [1, 2, 3, 4, 5, 6, 7, 8, 100]
l = li[::-1]
print(l)
this is also show same result ..is that right ?
some nice questions to play around with shortest or interesnig answer. For example for first question. There is much shorter solution with if statment, and it possible to do it with oneliner: print(*(i for i in range(100, 200) if i % 2 != 0), sep="
") just for fun.
this only gives you odd numbers, not prime numbers.
@@srdjan780 damn, you're right :D failed on first question :)
Thanks! Great practice
q1 better solution
for i in range(101,200):
if i %i==0:
print(i)
Here is alternative for question 3
def sort_list(x):
l = []
k = list(x)
for j in range(0, len(k)):
for i in k:
if i == max(k):
l.append(i)
k.remove(i)
print(l)
ye bhi chlega bro
L = [23,45,10,9,11,66]
final = [ ]
while L :
final.insert(0,max(L))
l.remove(max(L))
print(final)
but
l = [23,45,10,9,11,66]
new = [ [ max(l),l.remove(max(l))]for i in range(len(l))]
print(new)
can we do this? i want in one line answer, but it returns
[[66, None], [45, None], [23, None], [11, None], [10, None], [9, None]]
is it right way to do this????
Okay but you time complexity is O(N^2) , not only solving the problem but also ALWAYS think of how to get the best optimized solution to a given problem.
Should we be using recursion for fibonacci?
bro he has used count > 1 to see duplicate but why he used set ??
For SDE 1 role, can we use Python to solve
Hi.
Can you please confirm on the Question No: 14 solution. I think below code also produce the same result, then why you are using.... rstrip() etc.
new_string = "How are you doing"
output = new_string.split()
print(output)
Result = ['How', 'are', 'you', 'doing']
Thank you.
for Palindrome :
def palindrome(s):
return "Yes it is palindrome" if s[::-1].lower() ==s.lower() else "not a palindrome"
Very very small letters.... Not visible at all.... Plz do lil bigger so that it will good for us to view
Osm Video bro
Last problem can be solved as below:
s="missisippi"
list1=[]
for i in s:
if (i not in list1) or (i==' '):
list1.append(i)
new_s=''.join(list1)
print(new_s)
Hi Sir,
In below code not getting how, lst1 is appended with element 6 when we are appending only lst2 only.
Can you please Clarify, Thanks
lst1=[1,2,3,4,5]
lst2=lst1
lst2.append(6)
print('lst1- ',lst1)
print('lst2- ',lst2)
Output-
lst1- [1, 2, 3, 4, 5, 6]
lst2- [1, 2, 3, 4, 5, 6]
Use this - lst2 =lst1.copy(); and then append to lst2. it will not append to lst1
can you pls provide link for colab
Developer questions or production support questions?
bro he has used count > 1 to see duplicate but why he used set ??
In 3rd question, it should have been maximum instead of minimum as if x>minimum, minimum should not be equal to x. Please correct me if I am wrong guys !!
name is not matter :
def sort_function(list):
final = [ ]
for i in range(len(list)):
first_elem = list[0]
for num in list:
if num>first_elem:
first_elem = num
final.insert(0,first_elem)
list.remove(first_elem)
return final
for x in range(1,101):
if x % 10 in (1,3,5,7,9):
print(x)
if what you are trying to achieve is printing primes, then this logic is not correct. For eg. 99 % 10 will give you 9.
Right approach but should not include 9 as it isn't a prime and 2 shld be added
The written code prints odd numbers between 0 and 101 but not all the primes
can u please share that google drive doc you have shown with answers
What if we use built in fun i.e min() and max() to sort a list
a=[5,1,8,10,3,7,2,4]
b=[]
while a:
minimum = a[0]
for i in a:
if i==max(a):
b.append(i)
a.remove(i)
print(b)
Extra Space you are using.
Sir ...could u please tell me which compiler do u use here ??
It is a notebook of Google colab
@@madhusudanverma6564 its Jupiter
You could just use this instead of the palindrome one:
def palindrome(word):
return word[::-1] == word
Use this template for print the questions (Can anybody solve this? pls help)
Enter name:
Enter department:
Enter feedback:
Enter year:
The program should output the result with the entered values
Name:
Department:
Feedback:
Year:
Name = input("enter name")
Depart = input("Enter department:")
feed = input("Enter feedback:")
year = input("Enter year:")
print(f"Name: {Name} Department: {Depart} Feedback: {feed} Year: {year}.")
this what you wanted?
@@natemillner7199 bro he has used count > 1 to see duplicate but why he used set ??
1st problom
For i in range (100,200):
If i%2==1:
Print(i)
It's also CRT or not??
will return odd numbers not prime
In 3 it seems he copied the code... the minumum should be maximum for the problem he is trying to solve but good stuff overall for quick revision
9).Solution
def Search (arr,x):
for i in range(len(arr)):
if arr[i]==x:
return arr[i]
print('Sorry This element is not Given in this array')
list1=[1,2,3,4,5,6,7,8,9,10]
Search(list1,10)
Pardon Sir,
if we can use this code for Solution (9). So that will give us the actual Values that are in the given array. instead of using iloc (Index Location).
Q15:
#given a string , return recurring characters and display new string
s = 'mississippi'
lst = []
for i in s:
if i not in lst:
lst.append(i)
newS = ''.join(lst)
newS
(or)
#given a string , return recurring characters and display new string
s = 'mississippi'
setList = set()
for i in s:
setList.add(i)
newS = ''.join(setList)
newS
why in range is given (2, num) the range as been give from 100 -200;😏😏
prime number : a number is said to be prime number if it is divisible by 1 and itself....so (2, num) so here we should be start with range(2 ,num)
Writ a program to extact the digit?
a="1w3er5t678"
b= list(filter(lambda i:i.isdigit(),a))
str="".join(b)
print(str)