Coin Change - Leetcode 322 - Dynamic Programming (Python)

[@GregHogg]

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[@adilmolzhigitov4381]

Thanks, that was really good explanation. After watching your tutorials I also started actively using Paint when solving leetcode

[@yingxie9974]

Thanks for great explanation ! I like your smile !

[@ibraheem_Zain]

That was clear as sun and helpful even if you use python. Thanks sir you are awesome 👍😎❤

[@GregHogg]

Awesome super glad it was helpful 😊

[@tarilonte]

Your explanation is cristal clear! Thanks a lot for this great video!

[@dp8jl]

Used to think this is a backtracking question 😂 and failed multiple times, thanks for a great explanation

[@Alex-tm5hr]

Great job, imo better than neetcode's solution for the draw out part

[@GregHogg]

Yay!

[@internick_]

Why did we have to sort the coins in ascending order? Not sure why it becomes an issue if we always consider every coin, still

[@AI_STORY_JJANG]

I also don't think we need sorting

[@arnobchowdhury9641]

Sorting is important for breaking out of the loop early. Check line 15 and 16.

[@darshandani1]

Great explanation ! Thanks !

[@mansurischanov3204]

thanks dude, this makes so much sense now!

[@GregHogg]

Glad to hear it, this is a tricky one!

[@ayushdey5342]

what is the name of the blackboard app that you use to draw stuff? its really good looking

[@GregHogg]

Yes it's awesome. It's miro

[@new-anointingaremu3597]

This explanation is great thanks a lot

[@GregHogg]

Awesome happy to hear it!

[@carefree_ladka]

Sorting is O(NlogN). Why didn't we consider this in time complexity?

[@ArtyomKonoplya]

cause len(coins) way too smaller than amount value

[@Krish-hk1bt]

hei, you are saying "smallest number of ways to make a coin" , isn't it the "minimum number of coins"?

[@astronautgamer34]

plz explain coin change two it so hard for me to understand

[@GregHogg]

I haven't solved this, I'll have to take a look at some point :)

[@adesopekingsley9967]

I was thinking 🤔 We can sort reverse the list then pick the biggest number at index 0 then check if its greater than amount if false, then next but if true subtract from amount index. Check if result is present in the list [can use a set] if flase then call recursion using to repeat using result Example amount 12 List=[1,4,5] List.sort(reverse=True) 12>5.. therefore 12-5= 7 7 not in list and greater than 5 7>5 therefore 7-5=2 2

[@贪财]

$4x3 =$12. So fewest is 3 coins. 5+5+1+1 is 4coins. it cover most but edge

[@CGS9_]

@@贪财 always those edge cases that will get ya