Test results - Codility

Tasks Details

medium

1. MaxCounters

Calculate the values of counters after applying all alternating operations: increase counter by 1; set value of all counters to current maximum.

88%

Not assessed

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:

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

Programming language used Java 8

Time spent on task 4 minutes

Notes

not defined yet

Task timeline

Code: 12:19:07 UTC, java, verify, result: Passed

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

// you can also use imports, for example:

import java.util.Arrays;

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

        final int condition = N + 1;
        int currentMax = 0;
        int countersArray[] = new int[N];

        for (int iii = 0; iii < A.length; iii++) {
            int currentValue = A[iii];
            if (currentValue == condition) {
                Arrays.fill(countersArray, currentMax);
            } else {
                int position = currentValue - 1;
                int localValue = countersArray[position] + 1;
                countersArray[position] = localValue;
                
                if (localValue > currentMax) {
                    currentMax = localValue;
                }
            }

        }

        return countersArray;
    }
}

Analysis

▶

example
example test

✔

Code: 12:19:20 UTC, java, verify, result: Passed

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

// you can also use imports, for example:

import java.util.Arrays;

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

        final int condition = N + 1;
        int currentMax = 0;
        int countersArray[] = new int[N];

        for (int iii = 0; iii < A.length; iii++) {
            int currentValue = A[iii];
            if (currentValue == condition) {
                Arrays.fill(countersArray, currentMax);
            } else {
                int position = currentValue - 1;
                int localValue = countersArray[position] + 1;
                countersArray[position] = localValue;
                
                if (localValue > currentMax) {
                    currentMax = localValue;
                }
            }

        }

        return countersArray;
    }
}

Analysis

▶

example
example test

✔

Code: 12:19:25 UTC, java, final, score: 88

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

// you can also use imports, for example:

import java.util.Arrays;

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

        final int condition = N + 1;
        int currentMax = 0;
        int countersArray[] = new int[N];

        for (int iii = 0; iii < A.length; iii++) {
            int currentValue = A[iii];
            if (currentValue == condition) {
                Arrays.fill(countersArray, currentMax);
            } else {
                int position = currentValue - 1;
                int localValue = countersArray[position] + 1;
                countersArray[position] = localValue;
                
                if (localValue > currentMax) {
                    currentMax = localValue;
                }
            }

        }

        return countersArray;
    }
}

Analysis summary

The following issues have been detected: timeout errors.

Analysis

Detected time complexity:

O(N + M)

▶

example
example test

✔

▶

extreme_small
all max_counter operations

✔

▶

single
only one counter

✔

▶

small_random1
small random test, 6 max_counter operations

✔

▶

small_random2
small random test, 10 max_counter operations

✔

▶

medium_random1
medium random test, 50 max_counter operations

✔

▶

medium_random2
medium random test, 500 max_counter operations

✔

▶

large_random1
large random test, 2120 max_counter operations

✔

▶

large_random2
large random test, 10000 max_counter operations

✔

▶

extreme_large
all max_counter operations

✘

TIMEOUT ERROR
running time: >2.18 sec., time limit: 1.38 sec.