[Programmers] Printer

https://programmers.co.kr/learn/courses/30/lessons/42587

Solution

You need to clearly understand when printing happens and increment a count for each print event.

Code 1

My approach. The branching conditions are a straightforward translation of the problem into code. One thing to watch out for: the very last line of solution must return count + 1. If execution reaches that line, it means all items in priorities have been popped, so the count hasn’t been incremented for the last one yet.

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def solution(priorities, location):
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    count = 0
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    while priorities:
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        front = priorities.pop(0)
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        if priorities:
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            if front < max(priorities):
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                priorities.append(front)
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                if location == 0:
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                    location = len(priorities) - 1
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                else:
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                    location -= 1
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            else:
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                count += 1
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                if location == 0:
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                    return count
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                else:
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                    location -= 1
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    return count + 1

Code 2

ref: https://programmers.co.kr/learn/courses/30/lessons/42587/solution_groups?language=python3

any: returns True if any element in the iterable is True, False otherwise.

I tracked the location’s position, but this solution pre-stores the original position using tuples. That makes the code much cleaner.

It also uses any for the priority comparison. Using max might look cleaner, but when you need more complex conditions rather than simple max/min, any can make the code much more concise.

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def solution(priorities, location):
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    queue =  [(i,p) for i,p in enumerate(priorities)]
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    answer = 0
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    while True:
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        cur = queue.pop(0)
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        if any(cur[1] < q[1] for q in queue):
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            queue.append(cur)
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        else:
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            answer += 1
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            if cur[0] == location:
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                return answer