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:

def 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:

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 Python

Time spent on task 1 minutes

Notes

not defined yet

Task timeline

Code: 18:31:21 UTC, py, 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

def solution(N, A):
    # Implement your solution here
    answer = [0] *N
    max_counter = N +1
    temp = 0
    curr_max = 0

    for i in A:
        if i < max_counter: #그냥 더하기
            if answer[i-1] < temp:
                #만약 max counter 발생 후에 temp에 저장된 값보다 작다면
                #max counter 발생 전에 최대값보다 작았다.
                answer[i-1] = temp +1
            else:
                answer[i-1] += 1
            if answer[i-1] > curr_max:
                #계속해서 현재 최대값 update
                curr_max = answer[i-1]
        else:
            # Max counter 발생!
            temp = curr_max #temp 변수에 현재 max_val 값을 저장 
        
    for idx in range(N):
        if answer[idx] < temp:
            answer[idx] =temp
    
    return answer

Analysis

▶

example
example test

✔

Code: 18:31:24 UTC, py, 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

def solution(N, A):
    # Implement your solution here
    answer = [0] *N
    max_counter = N +1
    temp = 0
    curr_max = 0

    for i in A:
        if i < max_counter: #그냥 더하기
            if answer[i-1] < temp:
                #만약 max counter 발생 후에 temp에 저장된 값보다 작다면
                #max counter 발생 전에 최대값보다 작았다.
                answer[i-1] = temp +1
            else:
                answer[i-1] += 1
            if answer[i-1] > curr_max:
                #계속해서 현재 최대값 update
                curr_max = answer[i-1]
        else:
            # Max counter 발생!
            temp = curr_max #temp 변수에 현재 max_val 값을 저장 
        
    for idx in range(N):
        if answer[idx] < temp:
            answer[idx] =temp
    
    return answer

Analysis

▶

example
example test

✔

Code: 18:31:27 UTC, py, 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

def solution(N, A):
    # Implement your solution here
    answer = [0] *N
    max_counter = N +1
    temp = 0
    curr_max = 0

    for i in A:
        if i < max_counter: #그냥 더하기
            if answer[i-1] < temp:
                #만약 max counter 발생 후에 temp에 저장된 값보다 작다면
                #max counter 발생 전에 최대값보다 작았다.
                answer[i-1] = temp +1
            else:
                answer[i-1] += 1
            if answer[i-1] > curr_max:
                #계속해서 현재 최대값 update
                curr_max = answer[i-1]
        else:
            # Max counter 발생!
            temp = curr_max #temp 변수에 현재 max_val 값을 저장 
        
    for idx in range(N):
        if answer[idx] < temp:
            answer[idx] =temp
    
    return answer

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

✔

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large_random2
large random test, 10000 max_counter operations

✔

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extreme_large
all max_counter operations

✔