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 increased by 1,
max_counter − all counters are set to the maximum value of any counter.

A non-empty zero-indexed 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.

I did the following solution but it runs at O(NK) where K = length of array A.

public int[] solution(int N, int[] A) {
    int[] result = new int[N];
    int maximum = 0;

    for (int K = 0; K < A.Length; K++)
    {
        if (A[K] < 1 || A[K] > N + 1)
            throw new InvalidOperationException();

        if (A[K] >= 1 && A[K] <= N)
        {
            result[A[K] - 1]++;

            if (result[A[K] - 1] > maximum)
            {
                maximum = result[A[K] - 1];
            }
        }
        else
        {
            // inefficiency here
            for (int i = 0; i < result.Length; i++)
                result[i] = maximum;
        }
    }

    return result;
}

Could anyone show me how this can be better done with O(N + K) where K is the length of array A? Sorry for may terrible coding, I am doing these exercises to improve my programming. Thanks!

I don't know too much about complexity analysis, but your code/algorithm looks like O(N) to me -- you iterate over the `A` array only once and if all counters are to be maxed sometimes, the for-loop you chose seems efficient enough. — גלעד ברקן, Sep 20 '13 at 02:59
I tried this in codility and scored only 71. I know there is a much better solution than mine. — Randal Cunanan, Sep 20 '13 at 03:00
Its not O(n) since when setting everything to maximum is O(n) and assuming every call is to set Maximum, O(n) * O(n) = O(N^2). Btw code is wrong since maximun is updated on the run, and its not the global maximun. — Sinn, Sep 20 '13 at 03:08
What do you mean by global maximum? I need to update maximum on the run so as to not introduce another loop outside that will check for the maximum. Also the code passed several input data, the only problem is performance with large arrays. — Randal Cunanan, Sep 20 '13 at 03:12
well I converted your code to JavaScript and received a score of 66 on codility...ha ha I guess it's not as straightforward as it may seem. — גלעד ברקן, Sep 20 '13 at 03:24
I think the complexity class of your solution is actually O(kn) since K (the length of your input) is independent of N (the number of counters). — syazdani, Sep 20 '13 at 04:12
@syazdani You are right about the complexity. The best solution was targeting O(N + K) where K is the length of the array A. — Randal Cunanan, Sep 20 '13 at 04:15

score 17 · Accepted Answer · edited Sep 30 '17 at 20:16

This is what I came up with, but I am not sure if it works 100%:

public int[] solution(int N, int[] A) {
    int[] result = new int[N];
    int maximum = 0;
    int resetLimit = 0;

    for (int K = 0; K < A.Length; K++)
    {
        if (A[K] < 1 || A[K] > N + 1)
            throw new InvalidOperationException();

        if (A[K] >= 1 && A[K] <= N)
        {
            if (result[A[K] - 1] < resetLimit) {
                result[A[K] - 1] = resetLimit + 1;
            } else {
                result[A[K] - 1]++;
            }

            if (result[A[K] - 1] > maximum)
            {
                maximum = result[A[K] - 1];
            }
        }
        else
        {
            // inefficiency here
            //for (int i = 0; i < result.Length; i++)
            //    result[i] = maximum;
            resetLimit = maximum;
        }
    }

    for (int i = 0; i < result.Length; i++)
        result[i] = Math.Max(resetLimit, result[i]);

    return result;
}

congratulations! I tested your adjustment on codility and it scored 100% — גלעד ברקן, Sep 20 '13 at 04:58
Thanks! I am learning new programming tricks because of you guys. — Randal Cunanan, Sep 20 '13 at 05:09

fabriciorissetto · Answer 2 · 2016-07-02T23:17:13.423

Remember:

"Making your code readable is as important as making it executable."

-- Robert C Martin

Even when trying to solve a hard problem...

So trying to achieve a better readability I've created a class to encapsulate the counters array and its operations (Law of Demeter). Sadly my first solution got only 60% in the performance test, so at the cost of a bit of readability I've improved it with a smarter solution and finally got 100%.

Here are my two implementations with comments:

O(N*M) Correctness 100% / Performance 60% (high redability)

//I didn't refactored the names of the variables N and A
//to maintain it aligned with the question description
public int[] solution(int N, int[] A)
{
    var counters = new Counters(N);

    for (int k = 0; k < A.Length; k++)
    {
        if (A[k] <= N)
            counters.IncreaseCounter(A[k]);
        else
            counters.MaxAllCounters();
    }

    return counters.ToArray();
}

public class Counters
{
    private int[] counters;
    private int greaterValueInCounter = 0;

    public Counters(int length)
    {
        counters = new int[length];
    }

    public void MaxAllCounters()
    {
        for (int i = 0; i < counters.Length; i++)
        {
            counters[i] = greaterValueInCounter;
        }
    }

    public void IncreaseCounter(int counterPosition)
    {
        //The counter is one-based, but our array is zero-based
        counterPosition--;

        //Increments the counter
        counters[counterPosition]++;

        if (counters[counterPosition] > greaterValueInCounter)
            greaterValueInCounter = counters[counterPosition];
    }

    //The counters array is encapsuled in this class so if we provide external 
    //acess to it anyone could modify it and break the purpose of the encapsulation
    //So we just exposes a copy of it :)
    public int[] ToArray()
    {
        return (int[])counters.Clone();
    }
}

Codility result

O(N+M) Correctness 100% / Performance 100% (not so high redability)

Note the beauty of the encapsulation: to improve the algorithm I just have to edit some methods of the Counters class without changing a single character on the solution method.

Methods edited in the Counters class:

IncreaseCounter()
MaxAllCounters()
ToArray()

Final code:

//Exactly the same code
public int[] solution(int N, int[] A)
{
    var counters = new Counters(N);

    for (int k = 0; k < A.Length; k++)
    {
        if (A[k] <= N)
            counters.IncreaseCounter(A[k]);
        else
            counters.MaxAllCounters();
    }

    return counters.ToArray();
}

public class Counters
{
    private int[] counters;
    private int greaterValueInCounter = 0;
    private int currentEquilibratedScore = 0;

    public Counters(int length)
    {
        counters = new int[length];
    }

    public void MaxAllCounters()
    {
        //We don't update the entire array anymore - that was what caused the O(N*M)
        //We just save the current equilibrated score value
        currentEquilibratedScore = greaterValueInCounter;
    }

    public void IncreaseCounter(int counterPosition)
    {
        //The counter is one-based, but our array is zero-based
        counterPosition--;

        //We need to add this "if" here because with this new solution the array
        //is not always updated, so if we detect that this position is lower than
        //the currentEquilibratedScore, we update it before any operation
        if (counters[counterPosition] < currentEquilibratedScore)
            counters[counterPosition] = currentEquilibratedScore + 1;
        else
            counters[counterPosition]++;

        if (counters[counterPosition] > greaterValueInCounter)
            greaterValueInCounter = counters[counterPosition];
    }

    //The counters array is encapsuled in this class so if we provide external 
    //acess to it anyone could modify it and break the purpose of the encapsulation
    //So we just exposes a copy of it :)
    public int[] ToArray()
    {
        //Now we need to fix the unupdated values in the array
        //(the values that are less than the equilibrated score)
        for (int i = 0; i < counters.Length; i++)
        {
            if (counters[i] < currentEquilibratedScore)
                counters[i] = currentEquilibratedScore;
        }

        return (int[])counters.Clone();
    }
}

Codility result

score 4 · Answer 3 · answered Nov 11 '13 at 12:07

def solution(N, A):
    # write your code in Python 2.6
    res = [0] * N
    m = 0
    minValue = 0
    for x in A:
        if 1 <= x <= N:
            res[x - 1] = max(res[x - 1], minValue) + 1
            if res[x - 1] > m:
                m = res[x - 1]
        else:
            minValue = m
    for i in xrange(N):
        res[i] = max(res[i], minValue)
    return res

score 2 · Answer 4 · answered Feb 02 '14 at 01:17

Here's a solution I came up with in Python (100/100 on codility); it's a little different than others I've seen on here so thought I'd share:

def solution(N, A):
    count = [0] * N
    max_counter = [i for i, a in enumerate(A) if a == N+1]
    if len(max_counter) == len(A):
        return count
    if max_counter:
        mode = 0
        for i, m in enumerate(max_counter):
            if m == 0 or m - max_counter[i-1] == 1:
                continue
            subcount = {}
            if i == 0:
                for k in A[:m]:
                    if k not in subcount:
                        subcount[k] = 1
                    else:
                        subcount[k] += 1
            else:
                for k in A[max_counter[i-1]+1:m]:
                    if k not in subcount:
                        subcount[k] = 1
                    else:
                        subcount[k] += 1
            mode += max(subcount.values())
        count = [mode] * N
        for k in A[max_counter[-1]+1:]:
            count[k-1] += 1
    else:
        for k in A:
            count[k-1] += 1
    return count

score 1 · Answer 5 · answered Sep 20 '13 at 04:56

Let's see...

public int[] Solution(int N, int[] A)
{
    int[] data = new int[N];
    int maxval = 0;
    int baseval = 0;
    for (int K = 0; K < A.length; K++)
    {
        int index = A[K] - 1;
        if (index < 0 || index > N)
            throw new InvalidOperationException();

        if (index < N)
            maxval = baseval + Math.Max(maxval, ++data[index]);
        else
        {
            baseval = maxval;
            data = new int[N];
        }
    }

    for (int K = 0; K < N; K++)
        data[K] += baseval;

    return data;
}

I think that's O(N+K). Depends on how you count the Order of re-initializing the array.

score 1 · Answer 6 · answered Jun 28 '15 at 17:09

Same principle as everybody scoring 100% really, it is just that I find this version easier to read (and it is probably only because I wrote it).

using System;
using System.Linq;

class Solution 
{
    public int[] solution(int N, int[] A) 
    {

        var currentMax = 0;
        var resetValue = 0;
        var counters = Enumerable.Range(1, N).ToDictionary(i => i, i => 0);

        foreach (var a in A)
        {
            if (a == N + 1) resetValue = currentMax;
            else
            {
                counters[a] = Math.Max(counters[a], resetValue) + 1;
                currentMax = Math.Max(currentMax, counters[a]);
            }
        }
        return counters.Values.Select(v => Math.Max(v,resetValue)).ToArray();
    }
}

score 1 · Answer 7 · answered Apr 19 '20 at 09:54

Swift solution 100%

public func solution(_ N : Int, _ A : inout [Int]) -> [Int] {
    // write your code in Swift 4.2.1 (Linux)

    var solution = Array.init(repeating: 0, count: N)

        var max = 0
        var actualMaxValue = 0

        for obj in A {

            if (obj <= N && obj >= 1 ) {

                if solution[obj-1] < actualMaxValue {

                    solution [obj-1] = actualMaxValue + 1
                } else {

                    solution[obj-1] += 1
                }

                if (solution[obj-1] > max) {

                    max = solution[obj-1]
                }
            }
            else if obj == N+1 {

              actualMaxValue = max
            }
        }

    for (index, value) in solution.enumerated() {


        if value < actualMaxValue {

            solution[index] = actualMaxValue
        }
    }


    return solution
}

Please don't post only code as an answer, but also provide an explanation what your code does and how it solves the problem of the question. Answers with an explanation are usually of higher quality, and are more likely to attract upvotes. — David Buck, Apr 19 '20 at 11:01

score 1 · Answer 8 · edited Nov 01 '20 at 00:22

Here is an implementation in PHP:

function solution($N, $A) {
    $output = array_fill(0, $N, 0);
    $maxCounter = 0;
    $minCounter = 0;
    foreach ($A as $number) {
        if($number === $N + 1) {
            $minCounter = $maxCounter;
        } else if($number <= $N) {
            $number--;
            if($minCounter > $output[$number]) {
                $output[$number] = $minCounter;
            }
            $output[$number]++;
            if($output[$number] > $maxCounter) $maxCounter = $output[$number];
        }
    }

    foreach ($output as $index => $number) {
        if($number < $minCounter) $output[$index] = $minCounter;
    }

//    var_dump($output);
    return $output;
}

score 0 · Answer 9 · answered Nov 09 '13 at 13:38

public int[] counters(int N, int[] A)
{
    int[] count = new int[N];
    int maxCount = 0;
    int setAll = 0;

    for (int i = 0; i < A.Length; i++)
    {
        if (A[i] == N + 1)
        {
            setAll += maxCount;
            maxCount = 0;
            count = new int[N];
        }
        else
        {
            count[A[i] - 1] += 1;


            if (count[A[i] - 1] > maxCount)
            {
                maxCount = count[A[i] - 1];
            }
        }
    }

    for (int j = 0; j < count.Length; j++)
    {
        count[j] += setAll;
    }

    return count;
}

I think this is O(N+K), but codility say its O(N*K)? Would appreciate if anyone could explain which is correct...

score 0 · Answer 10 · answered Feb 17 '14 at 11:35

A 100/100 solution in php

function solution($N, $A){
    $cond     = $N + 1;
    $cur_max  = 0;
    $last_upd = 0;
    $cnt_arr  = array();
    $cnt      = count($A);
    for($i = 0; $i < $cnt; $i++){
        $cur = $A[$i];
        if($cur == $cond){
            $last_upd = $cur_max;
        }
        else{
            $pos = $cur - 1;
            if(!isset($cnt_arr[$pos])){
                $cnt_arr[$pos] = 0;
            }
            if($cnt_arr[$pos] < $last_upd){
                $cnt_arr[$pos] = $last_upd + 1;
            }
            else{
                $cnt_arr[$pos] ++;
            }
            if($cnt_arr[$pos] > $cur_max){
                $cur_max = $cnt_arr[$pos];
            }
        }
    }
    for($i = 0; $i < $N; $i++){
        if(!isset($cnt_arr[$i])){
            $cnt_arr[$i] = 0;
        }
        if($cnt_arr[$i] < $last_upd){
            $cnt_arr[$i] = $last_upd;
        }
    }
    return $cnt_arr;
}

user3648624 · Answer 11 · 2014-05-17T23:31:07.673

Rue, I just ran this locally. Watched the counters myself. I used this algorithm:

    public int[] solution(int N, int[] A)
    {
        int[] result = new int[N];
        int maximum = 0;
        int resetlimit = 0;

        for (int K = 0; K < A.Length; K++)
        {
            if (A[K] < 1 || A[K] > N + 1)
            {
                throw new InvalidOperationException();
            }

            if (A[K] >= 1 && A[K] <= N)
            {
                if (result[A[K] - 1] < resetlimit)
                {
                    result[A[K] - 1] = resetlimit + 1;
                }
                else
                {
                    result[A[K] - 1]++;
                }

                if (result[A[K] - 1] > maximum)
                {
                    maximum = result[A[K] - 1];
                }
            }
            else
            {
                resetlimit = maximum;
                result = Enumerable.Repeat(maximum, result.Length).ToArray();
            }
        }

        //for (int i = 0; i < result.Length; i++)
        //{
        //    result[i] = Math.Max(resetlimit, result[i]);
        //}

        return result;
    }
}

looking at the problem and the result sets, you must include in the inefficient for loop in the else statement. The for loop outside does not replicate the 2nd operation
•if A[K] = N + 1 then operation K is max_counter.

In order for iteration A[3] = 6 to set result[] all to '2' you must load the result array with the maximum counter. Otherwise your return will never have (2,2,2,2,2) as the 4th example array shows.

I too must take a test to get my dream job so the little inefficiency here is important;

the statement

  result = Enumerable.Repeat(maximum, result.Length).ToArray();

loads the array all in one shot so no inner loop and no inner efficiency. I think that is pretty close to the correct result sets. I'm surprised they didn't ask to return like a jagged array of the total return. Still, this codility test scares me a lot.

score 0 · Answer 12 · answered Jul 23 '15 at 05:06

C++11 code:

#include <algorithm>

vector<int> solution(int N, vector<int> &A) {

    vector<int> hist(N, 0);

    int last_update = 0;
    int max_value = 0;

    for (auto i : A){

        if (1 <= i && i <= N){
            int& val = hist[i - 1];

            if (val < last_update)
                val = last_update + 1;
            else
                val++;

            if (max_value < val)
                max_value = val;
        }

        if (i == (N+1)){
            last_update = max_value;
        }

    }

    replace_if(hist.begin(), hist.end(), [last_update](int x){return x < last_update;}, last_update);

    return hist;
}

score 0 · Answer 13 · answered Sep 09 '15 at 12:44

The key is that [0] * N is an N operation. If that exists in a for loop it will become N*M. Tested in Codility 100%

            # you can write to stdout for debugging purposes, e.g.
            # print "this is a debug message"

            def solution(N, A):
                # write your code in Python 2.7
                count = [0] * N
                maxCounter = 0
                minCounter = 0
                for x in A:
                    if x <= N and x >= 1:
                        count[x-1] = max(count[x-1], minCounter) + 1
                        if maxCounter < count[x-1]:
                            maxCounter = count[x-1]
                    if x == N + 1:
                        minCounter = maxCounter                
                for i in xrange(N):
                    count[i] = max(count[i], minValue)
                return count

score 0 · Answer 14 · answered Oct 26 '15 at 17:06

Here is Scala version, 100% on codility:

import java.util

def solution(N: Int, A: Array[Int]): Array[Int] = {

    var counters = new Array[Int](N)

    var maxCounter = 0

    for(ind <- 0 to A.length-1){


      if(A(ind) == (N+1)){

        //all to max
        util.Arrays.fill(counters,maxCounter)

      }else{
        //ind -1  cause array index start with 0 which is marked as 1 in the input data
        counters(A(ind)-1) += 1

        //update max counter if necessary
        if(maxCounter< (counters(A(ind)-1))) maxCounter = (counters(A(ind)-1))

      }

    }

    return counters
  }

Performance: https://codility.com/demo/results/trainingKJT6Y3-74G/

score 0 · Answer 15 · answered Feb 08 '16 at 07:17

0

Ruby Codility Code that got 100/100

def solution(a)

  if a.length < 3
      0
  end
  a.sort!
  for i in 2..a.length - 1
    if (a[i-2] + a[i-1]) > a[i]
      return 1
    end
  end
 0
end

answered Feb 08 '16 at 07:17

Ankit Agarwal

11
2

score 0 · Answer 16 · answered Oct 29 '18 at 05:14

def solution(N, A):
    res = [0] * N
    maxV, minV = 0, 0
    for x in A:
        if 1 <= x <= N:
            res[x-1] = max(res[x-1], minV) + 1
            maxV = max(maxV, res[x-1])
        else: minV = maxV
    for i in range(N):
        res[i] = max(res[i], minV)
    return res

score 0 · Answer 17 · answered Jun 06 '19 at 00:09

0

Ruby 100%


def solution(n, a)
  max = 0
  offsets = a.inject(Hash.new(max)) do |acc, el|
    next Hash.new(max) if el == n+1
    acc[el] +=1
    max = acc[el] if max < acc[el]
    acc
  end
  (1..n).map{|i| offsets[i]}
end

answered Jun 06 '19 at 00:09

Mike Belyakov

1,355
1
13
24

score 0 · Answer 18 · answered Jan 20 '20 at 20:31

Here is the kotlin version, 100% on codility

fun solutionMissingInteger(N: Int, A: IntArray): IntArray {
    val counters = IntArray(N)
    var max = 0
    var lastUpdate = 0
    for (index in A.indices) {
        val element = A[index]
        if (element == N + 1) {
            lastUpdate = max
        } else {
            val position = element - 1
            if (counters[position] < lastUpdate) {
                counters[position] = lastUpdate + 1
            } else {
                counters[position] = counters[position] +1
            }

            if (counters[position] > max) {
                max = counters[position]
            }
        }
    }
    setAllCountersToMaxValue(lastUpdate, counters)
    return counters
}

private fun setAllCountersToMaxValue(lastUpdate: Int, counters: IntArray) {
    for (index in counters.indices) {
        if (counters[index] < lastUpdate)
            counters[index] = lastUpdate
    }
}

score 0 · Answer 19 · answered Jan 21 '20 at 05:26

with js max score that I can get is 77%

any improvement?

function solution(N, A) {

  let counters = [];

  //fill counter with 0
  for (let i = 0; i < N; i += 1) {
    counters[i] = 0;
  }

  //loop array and set counters
  for (let i = 0; i < A.length; i += 1) {

    //0 index fix
    let position = A[i] - 1;

    if (A[i] <= N) {
      counters[position] += 1;
    } else {
      let max = Math.max(...counters);
      counters.fill(max)
    }

  }

  return counters;

}

score 0 · Answer 20 · answered Mar 02 '20 at 09:51

Java, 100%/100%


public int[] solution(int N, int[] A) {

    int[] counters = new int[N];

    int currentMax = 0;
    int sumOfMaxCounters = 0;
    boolean justDoneMaxCounter = false;

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

        if (A[i]  <= N) {

            justDoneMaxCounter = false;
            counters[A[i]-1]++;
            currentMax = currentMax < counters[A[i]-1] ? counters[A[i]-1] : currentMax;

        }else if (!justDoneMaxCounter){

            sumOfMaxCounters += currentMax;     
            currentMax = 0;
            counters = new int[N];
            justDoneMaxCounter = true;

        }
    }


    for (int j = 0; j < counters.length; j++) {
        counters[j] = counters[j] + sumOfMaxCounters;
    }

    return counters;
}

score 0 · Answer 21 · answered May 27 '20 at 15:27

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
}

score 0 · Answer 22 · answered Oct 18 '20 at 20:05

ES6

const solution = (n, a) => {
    // Initialize to zero
    let counter =  new Array(n);
    for(let i = 0  ; i < n ; i++ ){
        counter[i] = 0;
    }
    let max = 0;
    for(let j = 0  ; j < a.length ; j++ ){
        const item = a[j];
        if( item > n) {
            for(let i = 0  ; i < n ; i++ ){
                counter[i] = max;
            }
        }
        else{
            counter[item-1]++; 
            if(max < counter[item-1])
            {
                max = counter[item-1];
            }
        }
    }
    return counter;
};

Algorithm: Max Counters

22 Answers22

O(N*M) Correctness 100% / Performance 60% (high redability)

O(N+M) Correctness 100% / Performance 100% (not so high redability)

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