*TLDR: Input: bids = list of lists of ints representing [userid, # shares, $bid, timestamp] totalShares = total # of shares to be distributed. TODO: distribute shares amongst bidders and return userids of bidders that got 0 shares. Share distribution logic: 1) Bidder with highest offer gets all the shares they bid for, and then 2) if there are ties in $ bid price, assign shares to earlier bidder For the above question, I feel like the solution I came up with is relatively simple but it seems to pass all edge cases I can think of. The only questionable situation I can think of is like Bid price and times are the same and there aren't enough shares for all bidders ie: bids = [[0,2,10,0], [1,2,10,0]] and totalShares = 2. I'm unclear if like 1 share should be given to each or userid 0 just gets both. Does anyone see if my solution could be optimized in anyway? def getUnallocatesUsers(bids, totalShares): s = 0 for b in bids: s += b[1] if totalShares >= s: return [] # no losers because enough shares to go around bids.sort(key = lambda x: (-x[2],x[3])) # sort by highest bid, then timestamp for ties losers = [] for b in bids: if totalShares <= 0: losers.append(b[0]) else: totalShares -= b[1] return losers Continue reading...
