04. Counting Elements. MaxCounters.swift

import Foundation
import Glibc

// Solution @ Sergey Leschev, Belarusian State University

// 04. Counting Elements. MaxCounters.

// 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:
// class Solution { public int[] solution(int N, int[] A); }
// 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].

public func solution(_ N: Int, _ A: inout [Int]) -> [Int] {
    var result = Array<Int>.init(repeating: 0, count: N)
    var repeatedMaxValue = 0
    var maxValue = 0
    
    for e in A {
        if e <= N {
            result[e - 1] += 1
            maxValue = max(result[e - 1], maxValue)
        } else if maxValue > repeatedMaxValue {
            result = Array<Int>.init(repeating: maxValue, count: N)
            repeatedMaxValue = maxValue
        }
    }
    return result
}