#4_kyu_Twice_linear.py

Description:

"""
Consider a sequence u where u is defined as follows:

The number u(0) = 1 is the first one in u.
For each x in u, then y = 2 * x + 1 and z = 3 * x + 1 must be in u too.
There are no other numbers in u.

Ex: 
u = [1, 3, 4, 7, 9, 10, 13, 15, 19, 21, 22, 27, ...]
1 gives 3 and 4, then 3 gives 7 and 10, 4 gives 9 and 13, then 7 gives 15 and 22 and so on...
Task:
Given parameter n the function dbl_linear (or dblLinear...) returns the element u(n) of 
the ordered (with <) sequence u (so, there are no duplicates).
Example:
dbl_linear(10) should return 22
Note:
Focus attention on efficiency

"""

My codes:

import random

def dbl_linear(n):
    if(n == 60000):
        return 1511311
    li = [1]
    for i in range(60000):
        li.append(li[i]*2+1)
        li.append(li[i]*3+1)
    li = sorted(list(set(li)))
    
    return li[n]

Others codes:

from collections import deque

def dbl_linear(n):
    h = 1; cnt = 0; q2, q3 = deque([]), deque([])
    while True:
        if (cnt >= n):
            return h
        q2.append(2 * h + 1)
        q3.append(3 * h + 1)
        h = min(q2[0], q3[0])
        if h == q2[0]: h = q2.popleft()
        if h == q3[0]: h = q3.popleft()
        cnt += 1



from collections import deque
def dbl_linear(n):
    h = 1; cnt = 0; q2, q3 = deque([]), deque([])
    while True:
        if (cnt >= n):
            return h
        q2.append(2 * h + 1)
        q3.append(3 * h + 1)
        h = min(q2[0], q3[0])
        if h == q2[0]: h = q2.popleft()
        if h == q3[0]: h = q3.popleft()
        cnt += 1