Thanks for the link, today I learned sth new :)

For that case, there is an info in the description:

Every letter has to start with an increasing index.. "ONENO" results to 11, because the E can be used two times. Endless loops are not possible!

111 wouldn't be possible, since the index is increasing and the strings "ONE", "NEN" and "ENO" would be used.

Maybe still not efficient enough?

I'm agree. There is something unclear with instruction and/or tests.

Thank you, I added examples.
Please check=)

eg: queue_time([2,10,3,4],2)

You have 2 box.

first , one box takes 2,and other one takes 10. 2 ,10

second, the box taken 2 will plus 3,the other box also 10. 2+3=5 ,10

because, 5<10,so the '10' box cannot take the number '4' 5+4=9 ,10

At last, you cann't take any number ,then you compare the boxs,return the max 10