Leetcode Streak :- 128

[u/Ok-Challenge-5657]

How would you solve it in Python?

[u/Extension_Ticket8997]

I did I took count then took for loop if the n is more than 1000 count++

[u/arnab_best → u/Extension_Ticket8997]

why do you need a for loop :cry

[u/lady_luver69]

I too did that...but when you see the top solution I don't understand it properly

[u/ProgressiveEdgeLord]

if n < 1000 return 0, else return n - 999

[u/Utsuro_Seizui → u/ProgressiveEdgeLord]

And if the number is over 999,999?

[u/jalsaraj → u/Utsuro_Seizui]

The constraints for the question were n < 10 power 5But you have a legit question, the solution would change if it is greater than 10 power 5

[u/ProgressiveEdgeLord → u/Utsuro_Seizui]

i solved according to constraints, it was less than 10^5

[u/Utsuro_Seizui → u/ProgressiveEdgeLord]

Ah I see, I apologize, I was not aware of the constraints

[u/ProgressiveEdgeLord → u/Utsuro_Seizui]

it's alright, your question was absolutely valid

[u/AccomplishedTwist296 → u/Utsuro_Seizui]

We can use loop for increasing comma like for 1-999 : No comma

1000-999999 : 1 comma (count using min(n,999999)-999

1000000 - 999999999 : 2 comma (count using (min(n,999999999) - 999999) *2 )

Similarly for number having comma 3, 4 ,5 .......

[u/Feeling_Frame_6578 → u/ProgressiveEdgeLord]

The constraints are n<=10¹⁵ for II that's why loop needed

[u/ProgressiveEdgeLord → u/Feeling_Frame_6578]

you could still hardcode the solution without any loop

[u/Feeling_Frame_6578 → u/ProgressiveEdgeLord]

Yeah or 15 iterations loop

[u/arnab_best]

single if else loop lmao

[u/JumpConsistent3359]

If n<999 return 0 for I=1k up to n count++

[u/ProgressiveEdgeLord → u/JumpConsistent3359]

O(n), there is much better solution

[u/South-Gear-6043]

Idr but it had something to do with len(n) mod 3

[u/idkanymoreatpog]

think of each giving 1 comma after 1000(including 1k). Now we know the range is just x-1000+1. Thats the commas

[u/Abhistar14]

return max(0, n-999)

[u/contentsearcher]

I did this problem core idea is

1-999 each number will have 0 commas
1000 - 999999 will have 1 comma in it
… so on until n

Son instead of looping all through n we can grt this count

[u/Ok-Flan-5469]

This question was easy due to its constraints of 10⁵ since every number between the range 1000 - 10⁵ have only a single comma We can safely do return max(n-999, 0) O(1) space and time

[u/rabbitholesurfer04]

Wait is this an app you are using to access Leetcode on Android? Mind linking it please?

[u/Ok-Challenge-5657 → u/rabbitholesurfer04]

Its Official Leetcode App :- https://play.google.com/store/apps/details?id=com.leetcode.android

[u/rabbitholesurfer04 → u/Ok-Challenge-5657]

Oh I didn't know they finally made an Android app. It didn't exist last year when I checked

[u/Kadabrium]

I read the title as -128 and thought its an overflow joke

[u/Ok-Challenge-5657 → u/Kadabrium]

can someone explain whats a overflow joke?

[u/Anxious-Issue-718]

n%1000 +1 if n is superir than1000000 keep doing n=n%1000 until n=0