### Max Counters

January 23, 2020

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: (5, [3, 4, 4, 6, 1, 4, 4] )

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: [3, 4, 4, 6, 1, 4, 4]

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].

```
function solution(N, A) {
// write your code in JavaScript (Node.js 8.9.4)
const counters = []
let i = 0
let j = 0
let max = 0
let update = 0
for(j=0; j<N; j++) counters[j] = 0 // initialized with zeros
for(i=0; i<A.length; i++){
if(A[i]===N+1){
max = update
//max = Math.max(...counters) bad performance for large array
//for(j=0; j<N; j++) counters[j] = max
}
//if(A[i]>=1 && A[i]<=N) counters[ A[i]-1 ] ++
if(A[i]>=1 && A[i]<=N){
if(counters[ A[i]-1 ]>max) counters[ A[i]-1 ] ++
else counters[ A[i]-1 ] = max + 1
if(counters[ A[i]-1 ]>update) update = counters[ A[i]-1 ]
}
//console.log(counters)
}
for(j=0; j<N; j++){
if(counters[j]<max) counters[j]=max
}
return counters
}
```