π“Ÿ host1.bitchan.one

𓀐 @cqNbz~Mu

⏱ 5d seenβ›“ 0πŸ‘ 0β§– 1

filter [-]

languages:onother

What's the smart way to solve it? The logic I think is necessary is: Every alloy ID should have a list of alloy IDs that are weaker than it. For every pair of alloys N and M, where N is stronger than M, update list N to include alloy M, update any list K that includes N to also include M, update any list J that includes K to also include M, repeat as necessary. Is there a simpler way to do it?⏱ 4d↩ 1
mine solution was basically: >make a dict where the key is the allow and the value is a list of alloys known to be weaker than it >make a list of all alloys >find the alloy that isn't in any other alloy's "weaker than" list >that's the strongest >now iterate through the list again, ignoring the strongest alloy, to find the other alloy not in anyone's list >that's the second strongest alloy >keep iterating til you've assigned them all probably a faster way of doing it, but it's pretty simple conceptually⏱ 4d↩