Max Counters

07 Apr, 2021

Problem

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].

My Solution

failed O(n*m)

function solution(N, A) {
	let result = new Array(N).fill(0);

	for (let i = 0; i < A.length; i++) {
		if (A[i] === N + 1) {
			result = new Array(N).fill(Math.max(...result));
		} else {
			result[A[i] - 1] += 1;
		}
	}

	return result;
}

success

function solution(N, A) {
	let counters = new Array(N).fill(0);
	let maxVal = 0;
	let lastMax = 0;

	for (var j = 0; j < A.length; j++) {
		if (A[j] > N) {
			lastMax = maxVal;
		} else {
			let currentMax = Math.max(lastMax, counters[A[j] - 1]);
			counters[A[j] - 1] = currentMax += 1;
			maxVal = Math.max(counters[A[j] - 1], maxVal);
		}

		console.log(lastMax, maxVal, counters);
	}

	for (var l = 0; l < N; l++) {
		counters[l] = Math.max(counters[l], lastMax);
	}

	return counters;
}