[PS] n-th system game

Programmers: n-th system game

Conditions

The base n, the numbers to be obtained in advance t, the number of participants in the game m, and the sequence p are given.
The number t that must be spoken is displayed in sequence without spaces.
10~15 are printed as capital letters A~F respectively.
Changes from 0 to n base and creates a string that continues with the value.
Since t is required, if the length of the game string is longer that $t \times m + p$ , the process of converting to n base stops.
The $i \times m + p$ -th character is concatenated until the length of answer becomes t.

Brute force

def solution(n, t, m, p):
    answer = ''
    num = 0
    game = ""

    while True:
        numToN = convertN(num, n)
        game += numToN

        if len(game) > (t * m + p):
            break

        num += 1

    for i in range(len(game)):
        if i % m == p - 1:
            answer += game[i]

        if len(answer) == t:
            break

    return answer

def convertN(num, n):
    if num == 0:
        return '0'

    digit = "0123456789ABCDEF"
    temp = ""

    while num:
        q, r = divmod(num, n)
        temp = digit[r] + temp
        num = q

    return temp

Max: Test 16 〉 Pass (37.41ms, 10.7MB)

Big-O notation

In convertN(), the while statement is repeated $log_n\ num$ times. Expanding this, convertN() is $O(log\ num)$ .
However, the actual time required for convertN() is 1 when num is 0, 1 when 1, 1 when 2, 2 when 3, 2 when 4 ... increases in logscale in this way. Therefore, the time taken for the while statement is the sum of these values. It will be less than $1+2+3+4+...+n = \frac{n(n+1)}{2}$ , so it will be roughly close to $O(NlogN)$ .
The for statement that makes answer works as much as $O(t)$ .
As a result, it operates at $O(NlogN)$ time.

Optimization

def convertN(num, n):
    if num == 0:
        return '0'

    digit = "0123456789ABCDEF"
    temp = ""

    while num:
        num, r = divmod(num, n)
        temp = digit[r] + temp

    return temp

It's a small difference, but don't have to use q.

def solution(n, t, m, p):
    answer = ''
    num = 0
    game = ""
    targetLen = t * m + p

    while len(game) < targetLen:
        game += convertN(num, n)
        num += 1

    for i in range(len(game)):
        if i % m == p - 1:
            answer += game[i]

        if len(answer) == t:
            break

    return answer

Removed numToN because it is not necessary to use it.
Even $t \times m + p$ does not need to be calculated every time.
If you change the condition of the while statement, you can also remove the if statement.

Max: Test 15 〉 Pass (36.15ms, 10.9MB)

Remember

It would be good to remember how to convert the base system.
In divmod(a, b), when a is 0, q was not returned. Caution.