[HackerRank] Sherlock and the Valid String

이 문제는 단순한 구현 문제로 문제에 대한 세부 조건을 잘 읽지 않으면 틀릴 수 있는 문제이다.

    public static String isValid(String s) {
        // Write your code here
        HashMap<Character, Integer> map = new HashMap<>();

        pushCharToMap(s, map);

        return checkValidation(map);
    }

    public static void pushCharToMap(String s, HashMap<Character, Integer> map) {
        for(int i = 0; i < s.length(); i++) {
            map.putIfAbsent(s.charAt(i), 0);
            map.computeIfPresent(s.charAt(i),(k,v) -> v+1);
        }
    }

    public static String checkValidation(HashMap<Character, Integer> map) {
        TreeMap<Integer, Integer> sizeMap = new TreeMap<>();

        for (Map.Entry<Character, Integer> entry : map.entrySet()) {
            sizeMap.putIfAbsent(entry.getValue(), 0);
            sizeMap.computeIfPresent(entry.getValue(), (k, v) -> v + 1);
        }

		//All alphabet count is 1 case
        if (sizeMap.size() == 1) return "YES";
        //alphabet's counts are over 2 case
        if (sizeMap.keySet().size() > 2) return "NO";
        //alphabet's counts are 2 and has only one alphabet
        else if(sizeMap.firstEntry().getKey() == 1 && sizeMap.firstEntry().getValue() == 1) return "YES";
        //alphabet's counts are 2 and can delete one alphabet
        else if(checkHasOnlyOneCharacterSize(sizeMap) && (sizeMap.firstEntry().getKey() + 1 == sizeMap.lastEntry().getKey())) return "YES";
        else return "NO";
    }
    
    public static boolean checkHasOnlyOneCharacterSize(TreeMap<Integer, Integer> sizeMap) {
        return sizeMap.entrySet().stream().anyMatch(entry->entry.getValue() == 1);
    }
}

https://www.hackerrank.com/challenges/sherlock-and-valid-string/

Sherlock and the Valid String | HackerRank