Algorithm: Max Counters

Question

I have the following problem:

You are given N counters, initially set to 0, and you have two possible operations on them:

• increase(X) − counter X is increa

Answer 1

A 100% JavaScript solution

function solution(N, A) {
    // initialize all counters to 0
    let counters = Array(N).fill(0)

    // The maximum value of the counter is 0
    let max = 0

    // This variable will determine if an increment all operation has been encountered
    // and its value determines the maximum increment all operation encountered so far
    // for start it is 0, and we will set it to the value of max when A[i] == N + 1
    let max_all = 0

    for(let i = 0; i < A.length; i++) {

        if(A[i] <= counters.length) {

            // if the value of A[i] is 1, we have to increment c[0]
            // and hence the following index
            const c_index = A[i] - 1

            // if max all operation was found previously,
            // we have to set it here, because we are not setting anything in the else section
            // we are just setting a flag in the else section
            // if its value however is greater than max_all, it probably was already maxed
            // and later incremented, therefore we will skip it
            if(counters[c_index] < max_all) counters[c_index] = max_all

            // do the increment here
            const v = ++counters[c_index]

            // update the max if the current value is max
            max = v > max ? v : max
        }

        // this is straight forward
        else max_all = max
    }

    // if there are remaining items that have not been set to max_all inside the loop
    // we will update them here.
    // and we are updating them here instead of inside the for loop in the else section
    // to make the running time better. If updated inside the loop, we will have a running time of M * N
    // however here it's something like (M + N) ~ O(N)
    if(max_all) counters = counters.map(v => v < max_all ? max_all : v)

    // return the counters
    return counters
}