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.

100%

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: 01:00:47 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 31 32 33 34 35

// you can also use imports, for example:
// import java.util.*;

// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");

class Solution {
    public int[] solution(int N, int[] A) {
        // write your code in Java SE 8
        int size = A.length;
        int max = 0, maxtmp = 0;
        int []res = new int[N];
        boolean allmax = false;
        for(int i=0; i<size; i++){
            if(A[i] <= N) {
                if(res[A[i]-1] > max) {
                    res[A[i]-1] +=1;
                } else {
                    res[A[i]-1] = max + 1;
                }
                if(res[A[i]-1] > maxtmp) {
                    maxtmp = res[A[i]-1];
                }
            }else if(A[i] == N+1) {
                max = maxtmp;
            }
        }
        for(int j=0; j<N; j++){
            if(res[j] < max) {
                res[j] = max;
            }
        }
        return res;
    }
}

Analysis

▶

example
example test

✔

Code: 01:01:28 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 31 32 33 34 35

// you can also use imports, for example:
// import java.util.*;

// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");

class Solution {
    public int[] solution(int N, int[] A) {
        // write your code in Java SE 8
        int size = A.length;
        int max = 0, maxtmp = 0;
        int []res = new int[N];
        boolean allmax = false;
        for(int i=0; i<size; i++){
            if(A[i] <= N) {
                if(res[A[i]-1] > max) {
                    res[A[i]-1] +=1;
                } else {
                    res[A[i]-1] = max + 1;
                }
                if(res[A[i]-1] > maxtmp) {
                    maxtmp = res[A[i]-1];
                }
            }else if(A[i] == N+1) {
                max = maxtmp;
            }
        }
        for(int j=0; j<N; j++){
            if(res[j] < max) {
                res[j] = max;
            }
        }
        return res;
    }
}

Analysis

▶

example
example test

✔

Code: 01:01:37 UTC, java, final, score: 100

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 31 32 33 34 35

// you can also use imports, for example:
// import java.util.*;

// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");

class Solution {
    public int[] solution(int N, int[] A) {
        // write your code in Java SE 8
        int size = A.length;
        int max = 0, maxtmp = 0;
        int []res = new int[N];
        boolean allmax = false;
        for(int i=0; i<size; i++){
            if(A[i] <= N) {
                if(res[A[i]-1] > max) {
                    res[A[i]-1] +=1;
                } else {
                    res[A[i]-1] = max + 1;
                }
                if(res[A[i]-1] > maxtmp) {
                    maxtmp = res[A[i]-1];
                }
            }else if(A[i] == N+1) {
                max = maxtmp;
            }
        }
        for(int j=0; j<N; j++){
            if(res[j] < max) {
                res[j] = max;
            }
        }
        return res;
    }
}

Analysis summary

The solution obtained perfect score.

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

✔