Coin Change - Leetcode 322 - Dynamic Programming (Python)

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[@GregHogg]

Master Data Structures & Algorithms For FREE at AlgoMap.io!

[@tarilonte]

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

[@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 😊

[@darshandani1]

Great explanation ! Thanks !

[@Alex-tm5hr]

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

[@GregHogg]

Yay!

[@mansurischanov3204]

thanks dude, this makes so much sense now!

[@GregHogg]

Glad to hear it, this is a tricky one!

[@new-anointingaremu3597]

This explanation is great thanks a lot

[@GregHogg]

Awesome happy to hear it!

[@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 :)

[@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

[@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

[@user-pn6kq5mx4z]

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

[@CGS9_]

@@user-pn6kq5mx4z always those edge cases that will get ya