MaxCounters

tab1

Task Score     :100
Correctness   :100
Performance : 100

Task description

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].

Solution

public class MaxCounters {

 public static void main(String[] args) {
  int[] A = { 3, 4, 4, 6, 1, 4, 4 };

  System.out.print("[");
  int[] solution = new MaxCounters().solution(5, A);
  for (int i = 0; i < solution.length; i++) {

   System.out.print(solution[i] + (i == solution.length - 1 ? "" : ","));
  }
  System.out.print("]");
 }

 public int[] solution(int N, int[] A) {
  
  int[] b = new int[N];
  int max = 0;
  int appliedMax = 0;
  boolean applyMaxCount = false;
  for (int i = 0; i < A.length; i++) {
   if (A[i] > N) {
    applyMaxCount = true;
    appliedMax = max;
   } else {
    if (applyMaxCount && (b[A[i] - 1] < appliedMax)) {
     b[A[i] - 1] = appliedMax + 1;
    } else {
     b[A[i] - 1] += 1;
    }
    if (b[A[i] - 1] > max) {
     max = b[A[i] - 1];
    }
   }
  }
  for (int i = 0; i < b.length; i++) {
   if (b[i] < appliedMax) {
    b[i] = appliedMax;
   }
  }
  return b;
 }
}