프로그래머스 - 완주하지 못한 선수

첫 코드 - 효율성 문제(시간복잡도 [O(n*n)])

class Solution {
    public String solution(String[] participant, String[] completion) {
        String answer = "";

        String str;
        int cntPart;
        int cntComp;
        
        for (int i =  0; i < participant.length; i++) {
            str = participant[i];
            cntPart = 0;
            cntComp = 0;
            
            for (int j =  0; j < participant.length; j++) {
                if (str.equals(participant[j])) {
                    cntPart++;
                }
            }

            for (int j =  0; j < completion.length; j++) {
                if (str.equals(completion[j])) {
                    cntComp++;
                }
            }
            
            if (cntPart > cntComp) {
                answer = participant[i];
                break;
            }
        }
        
        return answer;
    }
}

해결 코드 - HashMap 사용 [O(n)]

import java.util.HashMap;

class Solution {
    public String solution(String[] participant, String[] completion) {
        String answer = "";

        HashMap<String, Integer> map = new HashMap<>();
        int val = 0;

        for (String part : participant) {
            if (map.get(part) == null) {
                map.put(part, 1);
            } else {
                val = map.get(part) + 1;
                map.put(part, val);
            }
        }
        
        for (String comp : completion) {
            val = map.get(comp) - 1;
            map.put(comp, val);
        }
        
        for (String key : map.keySet()) {
            if (map.get(key) == 1) {
                answer = key;
            }
        }

        return answer;
    }
}

다른사람 코드 - Arrays 사용

import java.util.Arrays;

class Solution {
    public String solution(String[] participant, String[] completion) {
        String answer = "";

        Arrays.sort(participant);
        Arrays.sort(completion);

        for (int i = 0; i < completion.length; i++) {
            if (!participant[i].equals(completion[i])) {
                answer = participant[i];
                break;
            }
        }

        return answer;
    }
}