Minimum Number of Pushes to Type Word II - Leetcode 3016 - Python

[@adiyansd4356]

First comment pin me bro📌

[@NeetCodeIO]

Fine but you better not edit it

[@Nishanth_S]

😂😂 ​@@NeetCodeIO

[@ameyakhot4458]

That "back in my day" hits different since we're not even that old to use this phrase.

[@chrischika7026]

im 21 and this also applies to me

[@BikerInsights]

I just saw the problem was II then solved first one and then came straight to Neetcode for explanation

[@yang5843]

The problem description is too long

[@omega_sine]

These interns need to have their writing skills vetted 😂

[@dipakmaity2205]

reading the problem makes you feel like a nightmare but we actually its not ... well hurrah solved it w/o seeing ur video or intuition

[@ajitpalsingh606]

Done on my own

[@Trippynix]

Damn Ascii values to solve this problem was genius!!

[@pastori2672]

i think you should start a codeforces series leetcode is getting easy

[@danlan4591]

Thank you broski I had the right idea and was working on it but the explanation really helped me out. However, one small suggestion maybe include the previous problem in the description since I did that one first and it helped out when moving onto the second. Great video!!!

first daily medium I solved by myself, all thanks to your channel

[@freecourseplatformenglish2829]

Solved it on my own. A bit easy problem.

[@hagaii5858]

for counting you can use a hash map with characters as keys and then get the sorted array of values by using list(dict.values()) and then sort

[@plsgivemecat]

really loved the explanation. thank you neetcode!

[@JainNamit]

I wanted to know is there any improvement or advantages of using heap way rather than just sorting?

[@Film-Screen]

in this problem no, but in othere problems if u need to make alot of del, insert and keep maintaining a sorted array heap would be much better in time complixty

[@arunpranavat]

Could you please explain this problem - Time Taken to Mark All Nodes (LeetCode - 3241). Please...

[@meetverma4812]

Thanks a Lot G!

[@mohammedkaifullakazim]

class Solution:
def minimumPushes(self, word: str) -> int:
d=dict(Counter(word))
m=[]
for i,j in d.items():
m.append((j,i))
m.sort(reverse=True)
print(m)
c=2
s=0
a=0
for p,q in m:
if c9 and c=18 and c=26:
s+=(p*4)
c+=1
return s
we could solve the problem with this approach as well @NeetCodeIO

[@apodikoglou]

what the hell

[@mohammedkaifullakazim]

​@@apodikoglou Means ?

[@MP-ny3ep]

Thank you for the daily

[@JamesBond-mq7pd]

super clever. !

[@krupakarreddy1758]

broo isn't the time complexity just O(n)?

[@krupakarreddy1758]

@@7akon yeah he said O(NlogN)..... But its just O(N)

[@guy_cod]

sorting is O(NlogN)

[@krupakarreddy1758]

@@guy_cod yeah like for an ideal case it is O(NlogN)... But we are sorting a constant space list ie, 26 so it takes constant time.. So overall just O(N)

[@guy_cod]

@@krupakarreddy1758 thank you for the explanation

[@maheshjamdade1]

You just understand the examples for this question, you will find the approach

[@janaSdj]

Nice

[@Provehitum]

class Solution:
def minimumPushes(self, word: str) -> int:
MAX_BUTTONS = 8
count_keys = 0
pushes = 0
for key in Counter(word).most_common():
pushes += key[1] * (1 + count_keys // MAX_BUTTONS)
count_keys += 1
return pushes

[@vikaskumarnitrr]

Sometimes, I feel like I am God before watching the video,I was able to come with a solution, but I came here to see his approach. the approach is same❤

[@Ishtathuku]

How about this?
class Solution:
def minimumPushes(self, word: str) -> int:
c=list(Counter(word).items())
c.sort(key=lambda x:x[1],reverse=True)
count=0
k=1
for i,j in enumerate(c,start=1):
count+= j[1]*k
if i%8==0:
k+=1
return count

[@qulinxao]

return sum(c*(1+i//8)for i,c in enumerate(sorted(Counter(w).values(),reverse=1)))

[@ameyakhot4458]

This sol is from larry.

[@IamAWESOME3980]

The rent is too damn high!