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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Climbing the Leaderboard
  5. Discussions

Climbing the Leaderboard

  • emiryilmaz_dev
     + 0 comments

    I believe this is the most optimized way to do it. Maybe we can remove the sort parts but it is still efficient. The key point is: the parameters are already sorted. Basically, it is possible to iterate through player list and keep the last compared index to not iterate all ranked array repeatedly.

    def climbingLeaderboard(ranked, player):
    ranked = list(set(ranked))
    scores = []
    length = len(ranked)
    
    player.reverse() # We use reversed list, we basically want the ranked and player arrays in the same order.
    matched = {}
    current_rank = 0
    for i in player:

        if matched.get(i, None): # remove duplicated computations.
            scores.append(matched[i])
            continue

        if current_rank >= length: # in last elements, it is likely to rank as last. 
            scores.append(current_rank + 1)
            continue

        if i > ranked[current_rank] or i == ranked[current_rank]: # if score is higher than current rank, assign the rank.
            matched[i] = current_rank + 1
            scores.append(current_rank + 1)
        else: # if it is lesser, we need an iteration since it is getting higher than current_rank. In short, we skip unnecessary parts. This is the optimization part.
            while current_rank < length and i < ranked[current_rank]:
                current_rank += 1
            matched[i] = current_rank + 1
            scores.append(current_rank + 1)
    scores.reverse() # we have reversed the player list, so we need to reverse scores.
    print(scores)
    return scores