[leetCode/JS] 2225. Find Players With Zero or One Losses

문제 설명

You are given an integer array matches where matches[i] = [winneri, loseri] indicates that the player winneri defeated player loseri in a match.

Return a list answer of size 2 where:

• answer[0] is a list of all players that have not lost any matches.
• answer[1] is a list of all players that have lost exactly one match.

The values in the two lists should be returned in increasing order.

Note:

• You should only consider the players that have played at least one match.
• The testcases will be generated such that no two matches will have the same outcome.

matchs = [winner, loser] 형식으로 param이 온다.
answer[0]은 절대 진적 없는 winner들만.
answer[1]은 1번만 진 플레이어들만.
sort((a,b)=>a-b) 오름차순 하여 return answer 하면 된다.

 

입출력 예

Example 1:

Input: matches = [[1,3],[2,3],[3,6],[5,6],[5,7],[4,5],[4,8],[4,9],[10,4],[10,9]]
Output: [[1,2,10],[4,5,7,8]]
Explanation:
Players 1, 2, and 10 have not lost any matches.
Players 4, 5, 7, and 8 each have lost one match.
Players 3, 6, and 9 each have lost two matches.
Thus, answer[0] = [1,2,10] and answer[1] = [4,5,7,8].

Example 2:

Input: matches = [[2,3],[1,3],[5,4],[6,4]]
Output: [[1,2,5,6],[]]
Explanation:
Players 1, 2, 5, and 6 have not lost any matches.
Players 3 and 4 each have lost two matches.
Thus, answer[0] = [1,2,5,6] and answer[1] = [].

Constraints

• 1 <= matches.length <= 105
• matches[i].length == 2
• 1 <= winneri, loseri <= 105
• winneri != loseri
• All matches[i] are unique.

---

내 솔루션

• Object를 사용해도 비슷하지만, 중복을 고려한다면 Set이 편하다.
• loser[] 를 만들어서 nerverLost[]나 oneLost[]에 add하기 전에 비교하였다.
• 마지막에 Set을 Array로 변환하여 리턴하면 끝.

/**
 * @param {number[][]} matches
 * @return {number[][]}
 */
var findWinners = function(matches) {
    const neverLost = new Set();
    const oneLost = new Set();
    const losers = new Set();

    for(let i = 0; i < matches.length; i++) {
        const [winner, loser] = matches[i];

        if(oneLost.has(loser)){
            oneLost.delete(loser);
        } else if(!losers.has(loser)) {
            oneLost.add(loser);
        }

        losers.add(loser);

        if(!losers.has(winner)) {
            neverLost.add(winner);
        }
        if(neverLost.has(loser)){
            neverLost.delete(loser);
        }
    }
    return [[...neverLost].sort((a,b)=>a-b), [...oneLost].sort((a,b)=>a-b)]
};

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감상평

• 아침에 첫 문제로 풀기 딱 좋은 가볍고 재미있는 문제였다.