Skip to main content

MaxCounters (Lesson 4)

Instructions​

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:

function solution(N, 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​

function solution(N, A) {
  let preMaxCount = false;
  let maxSequence = 0;
  const answer = new Array(N).fill(0);

  A.forEach((cur) => {
    if (cur === N + 1) {
      const result = preMaxCount ? answer : answer.fill(maxSequence);
      preMaxCount = true;
      return result;
    }
    preMaxCount = false;
    answer[cur - 1] += 1;
    maxSequence = Math.max(maxSequence, answer[cur - 1]);
  });
  return answer;
}

Review​

.

References​