In your solution, if you remove the 'Walrus operator',
nothing will change.
Everything worked,
and will continue to work without the 'Walrus operator'

'Walrus operator' capable make this solution more efficient oneliner. Open for arguments...

Is this mathematically correct ? I don't think it will work for all integers..

Neither best practice or clever. Same goes for the other solutions which are highly rated.

better use an enumerate's argument for the start index

630ms still seems like a long time. My solution is repeatedly under 75ms. I notice that your solution, like many others, creates an extra array. You are also summing the list, then substracting from that sum. Best to work through the list backwards, summing as you go with each iteration. And then just removing the list items that aren't leaders - this makes for less manipulation of the string.

Not only it's slow but this creates a new list instad of removing stuff from the original as stated in description

Thank Guido for Python :D

Now imagine we cannot index lists beyond their actual sizes :)

nice

it's still O(n²), even in this case:

• "sum on one element less each time"...
• boils down to: "sum n, then n-1, n-2, ...,1)"
• boils down to: n*(n+1)//2 = n²//2 + ... terms added overall (as you actually stated)
• boils down to: O(n²)

The other solution is O(n)

yes it is.

is your solution more efficient than that?