twice faster :) speedy solution for Letter Queue by blabaster - python coding challenges

Enable Javascript in your browser and then refresh this page, for a much enhanced experience.

twice faster :) solution in Speedy category for Letter Queue by blabaster

def letter_queue(commands):
    queue, idx = [], 0
    for c in commands:
        if c == 'POP':
            if idx < len(queue):
                idx += 1
        else:
            queue.append(c[-1])
    return ''.join(queue[idx:])

July 2, 2014

Comments:

gethappy on Feb. 8, 2015, 5:12 p.m.
Your solution is good then command's list is shot. In another case we'll have memory problems.

blabaster on Feb. 8, 2015, 5:51 p.m.
What about this: <pre class='brush: python'>def letter_queue(commands): queue, idx = [], 0 for c in commands: if c == 'POP': if idx < len(queue): idx += 1 if idx == 1024: # some big del queue[:idx] idx = 0 else: queue.append(c[-1]) return ''.join(queue[idx:]) </pre>

gethappy on Feb. 8, 2015, 6:01 p.m.
Yes! I thought about this too

Adamzma on Jan. 8, 2016, 8:24 p.m.
What do you mean by 'twice faster'? Maybe im wrong, but i think ive done this the same and dont know, what is so special in this solution.

Strings Theory

0 %

PyCharm The Python IDE for Professional Developers. Enjoy productive Python, web and scientific development with PyCharm. Download it now!