Hacker Rank (Climbing the Leaderboard)

문제

An arcade game player wants to climb to the top of the leaderboard and track their ranking. The game uses Dense Ranking, so its leaderboard works like this:

• The player with the highest score is ranked number  on the leaderboard.
• Players who have equal scores receive the same ranking number, and the next player(s) receive the immediately following ranking number.

Example

The ranked players will have ranks , , , and , respectively. If the player's scores are ,  and , their rankings after each game are ,  and . Return .

Function Description

Complete the climbingLeaderboard function in the editor below.

climbingLeaderboard has the following parameter(s):

• int ranked[n]: the leaderboard scores
• int player[m]: the player's scores

Returns

• int[m]: the player's rank after each new score

Input Format

The first line contains an integer , the number of players on the leaderboard.
The next line contains  space-separated integers , the leaderboard scores in decreasing order.
The next line contains an integer, , the number games the player plays.
The last line contains  space-separated integers , the game scores.

 

def climbingLeaderboard(ranked, player):
    result = []
    cnt = 0
    duplicate_rank = set(ranked)
    sub = list(duplicate_rank)
    sub = sorted(sub, reverse=True)

    for i in range(len(player)):
        for j in range(len(sub)):
            if player[i] >= sub[j]:
                result.append(j + 1)
                cnt += 1
                break
            elif player[i] < sub[-1]:
                result.append(len(sub) + 1)
                cnt += 1
                break
                
    return result

 다음과 같이 중복된 점수를 제거하기위해 set로 표현한 뒤 이중포문을 이용하여 구현을 해보았지만 테스트케이스의 run time error가 나와서 player의 점수가 오름차순으로 나타낸 점을 포인트로 인덱스를 통한 해결방법을 생각해보았다.

def climbingLeaderboard(ranked, player):
    cnt = 0
    duplicate_rank = set(ranked)
    sub = list(duplicate_rank)
    sub = sorted(sub, reverse=True)
    idx = len(sub)

    for i in player:
        while sub[idx - 1] <= i and idx > 0:
            idx -= 1
        if(idx<0):
            result.append(1)
            continue
        result.append(idx+1)

 

www.hackerrank.com/challenges/climbing-the-leaderboard/problem

Climbing the Leaderboard | HackerRank