릿코드 1838. Frequency of the Most Frequent Element 코틀린

일단 정렬치고,리스트 집합에 넣고(같은숫자처리),집합 루프돌려서 리스트의 라스트인덱스뽑고
리스트의 라스트인덱스부터 거꾸로 올라가면서 i-j만큼 k에서 뺴고,k가 없으면 컨티뉴

정렬치고 맨뒤부터 루프돌리면서,자기자신을 포함해서 1개씩 늘려가면서 arr[0]+arr[1]+k가 arr[0]*2보다 크면 max값 2로 변경하는걸 반복
->시간초과

class Solution {
    fun maxFrequency(nums: IntArray, k: Int): Int {
        val numList = nums.sorted().asReversed()
        var maxFrequency = 0
        for (i in numList.indices){
            var forMaxFrequency=0
            for (j in 0 until numList.size-i){
                val numSubList = numList.subList(i, i+1+ j)
                val sum = numSubList.sum()+k
                //println("""i:${numList[i]} j:${j} list:$numSubList sum:${sum} if:${numList[i]*(j+1)}""")
                if (sum<numList[i]*(j+1)){
                    break
                }


                forMaxFrequency=j+1
                //println(forMaxFrequency)
            }
            maxFrequency= maxOf(maxFrequency,forMaxFrequency)
        }
        //println(maxFrequency)
        return maxFrequency
    }
}

 

정답

class Solution {
   fun maxFrequency(nums: IntArray, k: Int): Int {
      nums.sort()
      var from = 0
      var inc = 0
      var max = 1
      for (to in 1..<nums.size) {
        inc += (to - from) * (nums[to] - nums[to - 1])
        while (from <= to && inc > k)
          inc -= nums[to] - nums[from++]
        max = max(max, to - from + 1)
      }
      return max
    }
}

그냥 포인터를 2개두고 그거가지고 밀고당기고 하는게 맞았던듯

일단 저 서브리스트같은거도 그냥 포인터가지고 해결했어야했고