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[codility] Counting Elements - MaxCountersAlgorithm 2021. 9. 23. 23:41
문제
You are given N counters, initially set to 0, and you have two possible operations on them:
- increase(X) − counter X is increased by 1,
- max counter − all counters are set to the maximum value of any counter.
A non-empty array A of M integers is given. This array represents consecutive operations:
- if A[K] = X, such that 1 ≤ X ≤ N, then operation K is increase(X),
- if A[K] = N + 1 then operation K is max counter.
For example, given integer N = 5 and array A such that:
A[0] = 3 A[1] = 4 A[2] = 4 A[3] = 6 A[4] = 1 A[5] = 4 A[6] = 4
the values of the counters after each consecutive operation will be:
(0, 0, 1, 0, 0) (0, 0, 1, 1, 0) (0, 0, 1, 2, 0) (2, 2, 2, 2, 2) (3, 2, 2, 2, 2) (3, 2, 2, 3, 2) (3, 2, 2, 4, 2)
The goal is to calculate the value of every counter after all operations.
Write a function:
public func solution(_ N : Int, _ A : inout [Int]) -> [Int]
that, given an integer N and a non-empty array A consisting of M integers, returns a sequence of integers representing the values of the counters.
Result array should be returned as an array of integers.
For example, given:
A[0] = 3 A[1] = 4 A[2] = 4 A[3] = 6 A[4] = 1 A[5] = 4 A[6] = 4
the function should return [3, 2, 2, 4, 2], as explained above.
Write an efficient algorithm for the following assumptions:
- N and M are integers within the range [1..100,000];
- each element of array A is an integer within the range [1..N + 1].
Copyright 2009–2021 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.
문제 해설
A[K] = X, 1 <= X <= N 카운트 증가
A[K] = N + 1 이면 최대 카운트값 전체적용
구현 해설
X값이 1보다 크거나 같고 N보다 작거나 같으면 해당 인덱스 카운팅 값 증가
큰값이나 예측하지 못한 데이터가 들어오는 경우가 발생
arr 에 현재 카운팅중 가장 큰값을 초기화하는 [Int](repeating: arr.max()!, count: N) 구문인듯 하다.
다른 방법으로 계속 풀어봤으나 100%는 힘들어서 검색을 해봄
https://app.codility.com/demo/results/trainingC62947-485/
import Foundation import Glibc // you can write to stdout for debugging purposes, e.g. // print("this is a debug message") public func solution(_ N : Int, _ A : inout [Int]) -> [Int] { // write your code in Swift 4.2.1 (Linux) var arr = [Int](repeating: 0, count: N) for (index, _) in A.enumerated() { if 1 <= A[index] && A[index] <= N { arr[A[index]-1] += 1 } else if A[index] == N+1 { arr = [Int](repeating: arr.max()!, count: N) } } return arr }
해결 힌트는 return 시 maxCounter와 arr에 카운트값과 비교하여 작은 값에 초기화
https://app.codility.com/demo/results/trainingWYS26A-946/
import Foundation import Glibc // you can write to stdout for debugging purposes, e.g. // print("this is a debug message") public func solution(_ N : Int, _ A : inout [Int]) -> [Int] { // write your code in Swift 4.2.1 (Linux) var arr = [Int](repeating: 0, count: N) var max = 0 var maxCounter = 0 for (index, _) in A.enumerated() { if A[index] > N { maxCounter = max } else { if arr[A[index]-1] < maxCounter { arr[A[index]-1] = maxCounter } arr[A[index]-1] += 1 if max < arr[A[index]-1] { max = arr[A[index]-1] } } } return arr.map({ $0 < maxCounter ? maxCounter : $0 }) }
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