릿코드 1657. Determine if Two Strings Are Close 코틀린

두 문자열을 같은문자열로 만들수있는지
1번 연산으로 모든 문자열의 자리는 다 바꿀수있고(n<2),
2번 연산으로 1자리 문자열을 다 통일시킬수있음

한문자열을 다른문자열로 부을수있으니까
일단 문자열크기가 안맞으면 false
두 문자열을 해시맵에 넣고,
같은숫자들을 뺴고
남은것들을 비교해서 갯수맞출수있으면 true

bbcc 2,2
bbbc 3,1
1 1 3
2 3
aabb
abcd

해시맵은 영어단문갯수내에서 놀기떄문에(32)2중루프써도됨

서로 랭스가 다르면 false
둘중 한쪽이라도 서로 글자를 포함하지않으면 false(특정문자를 자기배열내에있는 특정문자로 바꾸는거기떄문에 없으면 못바꿈)


구현
전제조건 어긋나는거 다 false로 미리 처리하고
두개 문자열 해시맵에 넣은다음
두 해시맵으로 이중반복돌려서 서로 겹치는애들 제거(이제 겹치지않은애들만 남음)
이후 두 해시맵 모두 값으로 정렬쳐두고
이중반복돌려서 1번해시맵에서의 숫자와 2번해시맵의 루프합이 같아지면 패스하고,적으면 계속 더하고,커지면 false리턴
다성공했으면 true리턴


정답

class Solution {
    fun closeStrings(word1: String, word2: String): Boolean {
        if (word1.length!=word2.length) return false
        val word1HashMap = hashMapOf<Char, Int>()
        val word2HashMap = hashMapOf<Char, Int>()
        for (i in word1){
            word1HashMap[i]=word1HashMap[i]?.plus(1)?:1
        }
        for (i in word2){
            word2HashMap[i]=word2HashMap[i]?.plus(1)?:1
        }
        
        val word1Set = word1HashMap.keys.toSet()
        val word2Set = word2HashMap.keys.toSet()
        if (!(word1Set.containsAll(word2Set) || word2Set.containsAll(word1Set))){
            return false
        }


        val word1List = word1HashMap.values.sorted().toMutableList()
        val word2List = word2HashMap.values.sorted().toMutableList()

        val tempList= mutableListOf<Int>()
        val passList = mutableListOf<Int>()
        for ((iIndex,iValue) in word1List.withIndex()){

            for ((jIndex,jValue) in word2List.withIndex()){
                if (passList.contains(jIndex)){
                    continue
                }
                if (iValue==jValue){
                    tempList.add(iValue)
                    passList.add(jIndex)
                    break;
                }
           }
        }


        for (i in tempList) {
            word1List.remove(i)
            word2List.remove(i)
        }

        for (i in word1List){
            var addNum=0
            for (j in word2List){
                addNum+=j
                if (i==addNum){
                    break
                }
                if(addNum>i){
                    println("false")
                    return false
                }
            }
        }
        return true
    }
}