Birthday Paradox

[@adityajha2342]

Sir your energy removes my lazyness😁thanks for this awesome lecture

[@ayushshaw1755]

The clicks of keyboard is relaxing 😌

[@UdaykumarKUday-yd9ut]

Ur way of telling is nice, more interesting 🎉🎉BINGOOO

[@akshaykumart.r.3523]

10:50 Listen to Sudarshan Sir's evil laugh😂😂. Play it at different speeds at 0.5, normal or 1.75, hard to determine where it sounds more evil. 😂😂😂

[@nandinisingh_3504]

awesome it was sir , till date I enjoyed this lecture most and not regretting to be the part of this learning .

[@avenumadhav3568]

random.random(): 2:22
random.randint(): 3:44
birthday paradox: 4:49 5:08 5:30 5:43 8:25 9:12 9:47 10:25 11:00 12:30 13:00 13:43 16:45 18:56
l.sort: 7:22
repetition: 7:27 7:46 (proof)
simulation: 18:56 19:39 20:31 21:00

[@nikhilatiitm]

why are you wasting your time bro.

[@avenumadhav3568]

@@nikhilatiitm I will try not to :)

[@tatvakeswani8483]

Since we know the length of the list, a for loop is a better option isn't it?

[@sudarshaniyengar1814]

Absolutely. 👍

[@avenumadhav3568]

Does this work?
l = []
for i in range(60):
l.append(random.randint(1,365))
l.sort()
for i in range(len(l)-1):
if(l[i] == l[i+1]):
print('repeats', l[i], l[i+1])

[@nikulpd8137]

sir ,you just have to remove break or else use continue instead , inorder to get get an output of one or more repeatations, right?

[@manideep2411]

Sir should use 1,366 in randrange so as to include 365 i.e. to add Dec 31,right?

[@TusharMahalya]

To get all the repeating no. in a list, we just need to remove "break" from the if-block..

[@UECSoumyaRay]

also the break keyword

[@rashmihiremath191]

how to know when to use different brackets like (), {}, and []

[@SimpleSage19]

just learn or memorise it

[@chinnareddy143]

Respected Sir,
I have done the home work. Kindly can you please validate.
l=[1,1,5,5,6]
#create empty list
#for i in range(30):
#l.append(random.randint(1,365))
#append random numbers from 1 to 365
#append 30 of them
l.sort()
print(l)
i=0
count=0
while(i

[@yashj1072]

I like your style

[@abhinayrawat6838]

Remove 'break' and get the all repeating elements printing in kernal

[@Apex7official]

use of flag??

[@manishpatankar7020]

Surprised to see why code did not error out for len(l-1), It has to be len(l-2)....any reason why code worked on your machine professor?

[@ZampGaming]

because there is no equality at len(i-1), rewatch at 15:50

[@manishpatankar7020]

@@ZampGaming Thanks.... missed it... actually '

[@ZampGaming]

@@manishpatankar7020 then subscribe me 😂

[@sudarshaniyengar1814]

Minor slippages. Please adjust 😄

[@manishpatankar7020]

@@sudarshaniyengar1814 Thank you sir...

[@ayushshaw1755]

18:31 just simply remove break to get all repetition

[@kumarprateek238]

Which video were we taught about import math?

[@AgentSnapMi]

exactly my doubt too

[@TheUnknownU]

23rd video of the list - "import library"

[@tanvijot]

do we really need to use flag here?

[@umeshmakode5309]

yep

[@HIMANSHURASTOGI-f8o]

105 105 to hai hi, hindi bhi hai hi.

[@ashishsaini1443]

from random import randint
date_collection=[]
for i in range(30):
date_collection.append(randint(1,365))
print(f"The various 30 random date between are
{date_collection}")
data_collection=date_collection.sort(reverse=True)
print(f"The decending order random date between are
{date_collection}")
no_repeat_condition=True
for i in range(0,len(date_collection)-1):
if(date_collection[i]==date_collection[i+1]):
no_repeat_condition=False
print(f"we found repeating values {date_collection[i]} and {date_collection[i+1]}")
if(no_repeat_condition):
print(f"we didnt find any repeating values")

[@chinmay7417]

For people using PyCharm as their IDE, add a " ; " at the end of the break statement.

[@054_nirajkumar3]

थैंक्यू

[@kajolsingh7259]

Great

[@alexioscarlos7727]

At 23:20 sir spoke in hindi mistakenly😂😂😂

[@LAVANYATADAVARTI]

Sir used a little bit of Hindi here😅

[@harshitchhipa]

i heard that "105 to h hi" 😂

[@LAVANYATADAVARTI]

@@harshitchhipa yeah