Building a greedy task scheduler - Python algorithm









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Working on the following Leetcode problem: https://leetcode.com/problems/task-scheduler/




Given a char array representing tasks CPU need to do. It contains
capital letters A to Z where different letters represent different
tasks.Tasks could be done without original order. Each task could be
done in one interval. For each interval, CPU could finish one task or
just be idle.



However, there is a non-negative cooling interval n that means between
two same tasks, there must be at least n intervals that CPU are doing
different tasks or just be idle.



You need to return the least number of intervals the CPU will take to
finish all the given tasks.




Example:



Input: tasks = ["A","A","A","B","B","B"], n = 2
Output: 8
Explanation: A -> B -> idle -> A -> B -> idle -> A -> B.


I've written code that passes the majority of the Leetcode tests cases, but is failing on a very large input. Here's my code:



import heapq
from collections import Counter

class Solution(object):
def leastInterval(self, tasks, n):
CLOCK = 0
if not tasks:
return len(tasks)

counts = Counter(tasks)
unvisited_tasks = counts.most_common()[::-1]
starting_task, _ = unvisited_tasks.pop()
queue = [[0, starting_task]]

while queue or unvisited_tasks:
while queue and CLOCK >= queue[0][0]:
_, task = heapq.heappop(queue)
counts[task] -= 1
if counts[task] > 0:
heapq.heappush(queue, [CLOCK + 1 + n, task])
CLOCK += 1

if unvisited_tasks:
t, _ = unvisited_tasks.pop()
heapq.heappush(queue, [0, t])
else:
# must go idle
if queue:
CLOCK += 1

return CLOCK


Here's the (large) input case:



tasks = ["G","C","A","H","A","G","G","F","G","J","H","C","A","G","E","A","H","E","F","D","B","D","H","H","E","G","F","B","C","G","F","H","J","F","A","C","G","D","I","J","A","G","D","F","B","F","H","I","G","J","G","H","F","E","H","J","C","E","H","F","C","E","F","H","H","I","G","A","G","D","C","B","I","D","B","C","J","I","B","G","C","H","D","I","A","B","A","J","C","E","B","F","B","J","J","D","D","H","I","I","B","A","E","H","J","J","A","J","E","H","G","B","F","C","H","C","B","J","B","A","H","B","D","I","F","A","E","J","H","C","E","G","F","G","B","G","C","G","A","H","E","F","H","F","C","G","B","I","E","B","J","D","B","B","G","C","A","J","B","J","J","F","J","C","A","G","J","E","G","J","C","D","D","A","I","A","J","F","H","J","D","D","D","C","E","D","D","F","B","A","J","D","I","H","B","A","F","E","B","J","A","H","D","E","I","B","H","C","C","C","G","C","B","E","A","G","H","H","A","I","A","B","A","D","A","I","E","C","C","D","A","B","H","D","E","C","A","H","B","I","A","B","E","H","C","B","A","D","H","E","J","B","J","A","B","G","J","J","F","F","H","I","A","H","F","C","H","D","H","C","C","E","I","G","J","H","D","E","I","J","C","C","H","J","C","G","I","E","D","E","H","J","A","H","D","A","B","F","I","F","J","J","H","D","I","C","G","J","C","C","D","B","E","B","E","B","G","B","A","C","F","E","H","B","D","C","H","F","A","I","A","E","J","F","A","E","B","I","G","H","D","B","F","D","B","I","B","E","D","I","D","F","A","E","H","B","I","G","F","D","E","B","E","C","C","C","J","J","C","H","I","B","H","F","H","F","D","J","D","D","H","H","C","D","A","J","D","F","D","G","B","I","F","J","J","C","C","I","F","G","F","C","E","G","E","F","D","A","I","I","H","G","H","H","A","J","D","J","G","F","G","E","E","A","H","B","G","A","J","J","E","I","H","A","G","E","C","D","I","B","E","A","G","A","C","E","B","J","C","B","A","D","J","E","J","I","F","F","C","B","I","H","C","F","B","C","G","D","A","A","B","F","C","D","B","I","I","H","H","J","A","F","J","F","J","F","H","G","F","D","J","G","I","E","B","C","G","I","F","F","J","H","H","G","A","A","J","C","G","F","B","A","A","E","E","A","E","I","G","F","D","B","I","F","A","B","J","F","F","J","B","F","J","F","J","F","I","E","J","H","D","G","G","D","F","G","B","J","F","J","A","J","E","G","H","I","E","G","D","I","B","D","J","A","A","G","A","I","I","A","A","I","I","H","E","C","A","G","I","F","F","C","D","J","J","I","A","A","F","C","J","G","C","C","H","E","A","H","F","B","J","G","I","A","A","H","G","B","E","G","D","I","C","G","J","C","C","I","H","B","D","J","H","B","J","H","B","F","J","E","J","A","G","H","B","E","H","B","F","F","H","E","B","E","G","H","J","G","J","B","H","C","H","A","A","B","E","I","H","B","I","D","J","J","C","D","G","I","J","G","J","D","F","J","E","F","D","E","B","D","B","C","B","B","C","C","I","F","D","E","I","G","G","I","B","H","G","J","A","A","H","I","I","H","A","I","F","C","D","A","C","G","E","G","E","E","H","D","C","G","D","I","A","G","G","D","A","H","H","I","F","E","I","A","D","H","B","B","G","I","C","G","B","I","I","D","F","F","C","C","A","I","E","A","E","J","A","H","C","D","A","C","B","G","H","G","J","G","I","H","B","A","C","H","I","D","D","C","F","G","B","H","E","B","B","H","C","B","G","G","C","F","B","E","J","B","B","I","D","H","D","I","I","A","A","H","G","F","B","J","F","D","E","G","F","A","G","G","D","A","B","B","B","J","A","F","H","H","D","C","J","I","A","H","G","C","J","I","F","J","C","A","E","C","H","J","H","H","F","G","E","A","C","F","J","H","D","G","G","D","D","C","B","H","B","C","E","F","B","D","J","H","J","J","J","A","F","F","D","E","F","C","I","B","H","H","D","E","A","I","A","B","F","G","F","F","I","E","E","G","A","I","D","F","C","H","E","C","G","H","F","F","H","J","H","G","A","E","H","B","G","G","D","D","D","F","I","A","F","F","D","E","H","J","E","D","D","A","J","F","E","E","E","F","I","D","A","F","F","J","E","I","J","D","D","G","A","C","G","G","I","E","G","E","H","E","D","E","J","B","G","I","J","C","H","C","C","A","A","B","C","G","B","D","I","D","E","H","J","J","B","F","E","J","H","H","I","G","B","D"]
n = 1


My code is outputting an interval count of 1002, and the correct answer is 1000. Because the input size is so large, I'm having trouble debugging by hand on where this is going wrong.



My algorithm essentially does the following:



  1. Build a mapping of character to number of occurrences

  2. Start with the task that occurs the largest number of times.

  3. When you visit a task, enqueue the next task to be CLOCK + interval iterations later, because my premise is that you want to visit a task as soon as you're able to do so.

  4. If can't visit an already-visited task, enqueue a new one, and do so without incrementing the clock.

  5. If you have elements in the queue, but not enough time has passed, increment the clock.

At the end, the CLOCK variable describes how long (in other words, how many "intervals") passed before you're able to run all tasks.



Can someone spot the bug in my logic?












share|improve this question

























    up vote
    0
    down vote

    favorite












    Working on the following Leetcode problem: https://leetcode.com/problems/task-scheduler/




    Given a char array representing tasks CPU need to do. It contains
    capital letters A to Z where different letters represent different
    tasks.Tasks could be done without original order. Each task could be
    done in one interval. For each interval, CPU could finish one task or
    just be idle.



    However, there is a non-negative cooling interval n that means between
    two same tasks, there must be at least n intervals that CPU are doing
    different tasks or just be idle.



    You need to return the least number of intervals the CPU will take to
    finish all the given tasks.




    Example:



    Input: tasks = ["A","A","A","B","B","B"], n = 2
    Output: 8
    Explanation: A -> B -> idle -> A -> B -> idle -> A -> B.


    I've written code that passes the majority of the Leetcode tests cases, but is failing on a very large input. Here's my code:



    import heapq
    from collections import Counter

    class Solution(object):
    def leastInterval(self, tasks, n):
    CLOCK = 0
    if not tasks:
    return len(tasks)

    counts = Counter(tasks)
    unvisited_tasks = counts.most_common()[::-1]
    starting_task, _ = unvisited_tasks.pop()
    queue = [[0, starting_task]]

    while queue or unvisited_tasks:
    while queue and CLOCK >= queue[0][0]:
    _, task = heapq.heappop(queue)
    counts[task] -= 1
    if counts[task] > 0:
    heapq.heappush(queue, [CLOCK + 1 + n, task])
    CLOCK += 1

    if unvisited_tasks:
    t, _ = unvisited_tasks.pop()
    heapq.heappush(queue, [0, t])
    else:
    # must go idle
    if queue:
    CLOCK += 1

    return CLOCK


    Here's the (large) input case:



    tasks = ["G","C","A","H","A","G","G","F","G","J","H","C","A","G","E","A","H","E","F","D","B","D","H","H","E","G","F","B","C","G","F","H","J","F","A","C","G","D","I","J","A","G","D","F","B","F","H","I","G","J","G","H","F","E","H","J","C","E","H","F","C","E","F","H","H","I","G","A","G","D","C","B","I","D","B","C","J","I","B","G","C","H","D","I","A","B","A","J","C","E","B","F","B","J","J","D","D","H","I","I","B","A","E","H","J","J","A","J","E","H","G","B","F","C","H","C","B","J","B","A","H","B","D","I","F","A","E","J","H","C","E","G","F","G","B","G","C","G","A","H","E","F","H","F","C","G","B","I","E","B","J","D","B","B","G","C","A","J","B","J","J","F","J","C","A","G","J","E","G","J","C","D","D","A","I","A","J","F","H","J","D","D","D","C","E","D","D","F","B","A","J","D","I","H","B","A","F","E","B","J","A","H","D","E","I","B","H","C","C","C","G","C","B","E","A","G","H","H","A","I","A","B","A","D","A","I","E","C","C","D","A","B","H","D","E","C","A","H","B","I","A","B","E","H","C","B","A","D","H","E","J","B","J","A","B","G","J","J","F","F","H","I","A","H","F","C","H","D","H","C","C","E","I","G","J","H","D","E","I","J","C","C","H","J","C","G","I","E","D","E","H","J","A","H","D","A","B","F","I","F","J","J","H","D","I","C","G","J","C","C","D","B","E","B","E","B","G","B","A","C","F","E","H","B","D","C","H","F","A","I","A","E","J","F","A","E","B","I","G","H","D","B","F","D","B","I","B","E","D","I","D","F","A","E","H","B","I","G","F","D","E","B","E","C","C","C","J","J","C","H","I","B","H","F","H","F","D","J","D","D","H","H","C","D","A","J","D","F","D","G","B","I","F","J","J","C","C","I","F","G","F","C","E","G","E","F","D","A","I","I","H","G","H","H","A","J","D","J","G","F","G","E","E","A","H","B","G","A","J","J","E","I","H","A","G","E","C","D","I","B","E","A","G","A","C","E","B","J","C","B","A","D","J","E","J","I","F","F","C","B","I","H","C","F","B","C","G","D","A","A","B","F","C","D","B","I","I","H","H","J","A","F","J","F","J","F","H","G","F","D","J","G","I","E","B","C","G","I","F","F","J","H","H","G","A","A","J","C","G","F","B","A","A","E","E","A","E","I","G","F","D","B","I","F","A","B","J","F","F","J","B","F","J","F","J","F","I","E","J","H","D","G","G","D","F","G","B","J","F","J","A","J","E","G","H","I","E","G","D","I","B","D","J","A","A","G","A","I","I","A","A","I","I","H","E","C","A","G","I","F","F","C","D","J","J","I","A","A","F","C","J","G","C","C","H","E","A","H","F","B","J","G","I","A","A","H","G","B","E","G","D","I","C","G","J","C","C","I","H","B","D","J","H","B","J","H","B","F","J","E","J","A","G","H","B","E","H","B","F","F","H","E","B","E","G","H","J","G","J","B","H","C","H","A","A","B","E","I","H","B","I","D","J","J","C","D","G","I","J","G","J","D","F","J","E","F","D","E","B","D","B","C","B","B","C","C","I","F","D","E","I","G","G","I","B","H","G","J","A","A","H","I","I","H","A","I","F","C","D","A","C","G","E","G","E","E","H","D","C","G","D","I","A","G","G","D","A","H","H","I","F","E","I","A","D","H","B","B","G","I","C","G","B","I","I","D","F","F","C","C","A","I","E","A","E","J","A","H","C","D","A","C","B","G","H","G","J","G","I","H","B","A","C","H","I","D","D","C","F","G","B","H","E","B","B","H","C","B","G","G","C","F","B","E","J","B","B","I","D","H","D","I","I","A","A","H","G","F","B","J","F","D","E","G","F","A","G","G","D","A","B","B","B","J","A","F","H","H","D","C","J","I","A","H","G","C","J","I","F","J","C","A","E","C","H","J","H","H","F","G","E","A","C","F","J","H","D","G","G","D","D","C","B","H","B","C","E","F","B","D","J","H","J","J","J","A","F","F","D","E","F","C","I","B","H","H","D","E","A","I","A","B","F","G","F","F","I","E","E","G","A","I","D","F","C","H","E","C","G","H","F","F","H","J","H","G","A","E","H","B","G","G","D","D","D","F","I","A","F","F","D","E","H","J","E","D","D","A","J","F","E","E","E","F","I","D","A","F","F","J","E","I","J","D","D","G","A","C","G","G","I","E","G","E","H","E","D","E","J","B","G","I","J","C","H","C","C","A","A","B","C","G","B","D","I","D","E","H","J","J","B","F","E","J","H","H","I","G","B","D"]
    n = 1


    My code is outputting an interval count of 1002, and the correct answer is 1000. Because the input size is so large, I'm having trouble debugging by hand on where this is going wrong.



    My algorithm essentially does the following:



    1. Build a mapping of character to number of occurrences

    2. Start with the task that occurs the largest number of times.

    3. When you visit a task, enqueue the next task to be CLOCK + interval iterations later, because my premise is that you want to visit a task as soon as you're able to do so.

    4. If can't visit an already-visited task, enqueue a new one, and do so without incrementing the clock.

    5. If you have elements in the queue, but not enough time has passed, increment the clock.

    At the end, the CLOCK variable describes how long (in other words, how many "intervals") passed before you're able to run all tasks.



    Can someone spot the bug in my logic?












    share|improve this question























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Working on the following Leetcode problem: https://leetcode.com/problems/task-scheduler/




      Given a char array representing tasks CPU need to do. It contains
      capital letters A to Z where different letters represent different
      tasks.Tasks could be done without original order. Each task could be
      done in one interval. For each interval, CPU could finish one task or
      just be idle.



      However, there is a non-negative cooling interval n that means between
      two same tasks, there must be at least n intervals that CPU are doing
      different tasks or just be idle.



      You need to return the least number of intervals the CPU will take to
      finish all the given tasks.




      Example:



      Input: tasks = ["A","A","A","B","B","B"], n = 2
      Output: 8
      Explanation: A -> B -> idle -> A -> B -> idle -> A -> B.


      I've written code that passes the majority of the Leetcode tests cases, but is failing on a very large input. Here's my code:



      import heapq
      from collections import Counter

      class Solution(object):
      def leastInterval(self, tasks, n):
      CLOCK = 0
      if not tasks:
      return len(tasks)

      counts = Counter(tasks)
      unvisited_tasks = counts.most_common()[::-1]
      starting_task, _ = unvisited_tasks.pop()
      queue = [[0, starting_task]]

      while queue or unvisited_tasks:
      while queue and CLOCK >= queue[0][0]:
      _, task = heapq.heappop(queue)
      counts[task] -= 1
      if counts[task] > 0:
      heapq.heappush(queue, [CLOCK + 1 + n, task])
      CLOCK += 1

      if unvisited_tasks:
      t, _ = unvisited_tasks.pop()
      heapq.heappush(queue, [0, t])
      else:
      # must go idle
      if queue:
      CLOCK += 1

      return CLOCK


      Here's the (large) input case:



      tasks = ["G","C","A","H","A","G","G","F","G","J","H","C","A","G","E","A","H","E","F","D","B","D","H","H","E","G","F","B","C","G","F","H","J","F","A","C","G","D","I","J","A","G","D","F","B","F","H","I","G","J","G","H","F","E","H","J","C","E","H","F","C","E","F","H","H","I","G","A","G","D","C","B","I","D","B","C","J","I","B","G","C","H","D","I","A","B","A","J","C","E","B","F","B","J","J","D","D","H","I","I","B","A","E","H","J","J","A","J","E","H","G","B","F","C","H","C","B","J","B","A","H","B","D","I","F","A","E","J","H","C","E","G","F","G","B","G","C","G","A","H","E","F","H","F","C","G","B","I","E","B","J","D","B","B","G","C","A","J","B","J","J","F","J","C","A","G","J","E","G","J","C","D","D","A","I","A","J","F","H","J","D","D","D","C","E","D","D","F","B","A","J","D","I","H","B","A","F","E","B","J","A","H","D","E","I","B","H","C","C","C","G","C","B","E","A","G","H","H","A","I","A","B","A","D","A","I","E","C","C","D","A","B","H","D","E","C","A","H","B","I","A","B","E","H","C","B","A","D","H","E","J","B","J","A","B","G","J","J","F","F","H","I","A","H","F","C","H","D","H","C","C","E","I","G","J","H","D","E","I","J","C","C","H","J","C","G","I","E","D","E","H","J","A","H","D","A","B","F","I","F","J","J","H","D","I","C","G","J","C","C","D","B","E","B","E","B","G","B","A","C","F","E","H","B","D","C","H","F","A","I","A","E","J","F","A","E","B","I","G","H","D","B","F","D","B","I","B","E","D","I","D","F","A","E","H","B","I","G","F","D","E","B","E","C","C","C","J","J","C","H","I","B","H","F","H","F","D","J","D","D","H","H","C","D","A","J","D","F","D","G","B","I","F","J","J","C","C","I","F","G","F","C","E","G","E","F","D","A","I","I","H","G","H","H","A","J","D","J","G","F","G","E","E","A","H","B","G","A","J","J","E","I","H","A","G","E","C","D","I","B","E","A","G","A","C","E","B","J","C","B","A","D","J","E","J","I","F","F","C","B","I","H","C","F","B","C","G","D","A","A","B","F","C","D","B","I","I","H","H","J","A","F","J","F","J","F","H","G","F","D","J","G","I","E","B","C","G","I","F","F","J","H","H","G","A","A","J","C","G","F","B","A","A","E","E","A","E","I","G","F","D","B","I","F","A","B","J","F","F","J","B","F","J","F","J","F","I","E","J","H","D","G","G","D","F","G","B","J","F","J","A","J","E","G","H","I","E","G","D","I","B","D","J","A","A","G","A","I","I","A","A","I","I","H","E","C","A","G","I","F","F","C","D","J","J","I","A","A","F","C","J","G","C","C","H","E","A","H","F","B","J","G","I","A","A","H","G","B","E","G","D","I","C","G","J","C","C","I","H","B","D","J","H","B","J","H","B","F","J","E","J","A","G","H","B","E","H","B","F","F","H","E","B","E","G","H","J","G","J","B","H","C","H","A","A","B","E","I","H","B","I","D","J","J","C","D","G","I","J","G","J","D","F","J","E","F","D","E","B","D","B","C","B","B","C","C","I","F","D","E","I","G","G","I","B","H","G","J","A","A","H","I","I","H","A","I","F","C","D","A","C","G","E","G","E","E","H","D","C","G","D","I","A","G","G","D","A","H","H","I","F","E","I","A","D","H","B","B","G","I","C","G","B","I","I","D","F","F","C","C","A","I","E","A","E","J","A","H","C","D","A","C","B","G","H","G","J","G","I","H","B","A","C","H","I","D","D","C","F","G","B","H","E","B","B","H","C","B","G","G","C","F","B","E","J","B","B","I","D","H","D","I","I","A","A","H","G","F","B","J","F","D","E","G","F","A","G","G","D","A","B","B","B","J","A","F","H","H","D","C","J","I","A","H","G","C","J","I","F","J","C","A","E","C","H","J","H","H","F","G","E","A","C","F","J","H","D","G","G","D","D","C","B","H","B","C","E","F","B","D","J","H","J","J","J","A","F","F","D","E","F","C","I","B","H","H","D","E","A","I","A","B","F","G","F","F","I","E","E","G","A","I","D","F","C","H","E","C","G","H","F","F","H","J","H","G","A","E","H","B","G","G","D","D","D","F","I","A","F","F","D","E","H","J","E","D","D","A","J","F","E","E","E","F","I","D","A","F","F","J","E","I","J","D","D","G","A","C","G","G","I","E","G","E","H","E","D","E","J","B","G","I","J","C","H","C","C","A","A","B","C","G","B","D","I","D","E","H","J","J","B","F","E","J","H","H","I","G","B","D"]
      n = 1


      My code is outputting an interval count of 1002, and the correct answer is 1000. Because the input size is so large, I'm having trouble debugging by hand on where this is going wrong.



      My algorithm essentially does the following:



      1. Build a mapping of character to number of occurrences

      2. Start with the task that occurs the largest number of times.

      3. When you visit a task, enqueue the next task to be CLOCK + interval iterations later, because my premise is that you want to visit a task as soon as you're able to do so.

      4. If can't visit an already-visited task, enqueue a new one, and do so without incrementing the clock.

      5. If you have elements in the queue, but not enough time has passed, increment the clock.

      At the end, the CLOCK variable describes how long (in other words, how many "intervals") passed before you're able to run all tasks.



      Can someone spot the bug in my logic?












      share|improve this question













      Working on the following Leetcode problem: https://leetcode.com/problems/task-scheduler/




      Given a char array representing tasks CPU need to do. It contains
      capital letters A to Z where different letters represent different
      tasks.Tasks could be done without original order. Each task could be
      done in one interval. For each interval, CPU could finish one task or
      just be idle.



      However, there is a non-negative cooling interval n that means between
      two same tasks, there must be at least n intervals that CPU are doing
      different tasks or just be idle.



      You need to return the least number of intervals the CPU will take to
      finish all the given tasks.




      Example:



      Input: tasks = ["A","A","A","B","B","B"], n = 2
      Output: 8
      Explanation: A -> B -> idle -> A -> B -> idle -> A -> B.


      I've written code that passes the majority of the Leetcode tests cases, but is failing on a very large input. Here's my code:



      import heapq
      from collections import Counter

      class Solution(object):
      def leastInterval(self, tasks, n):
      CLOCK = 0
      if not tasks:
      return len(tasks)

      counts = Counter(tasks)
      unvisited_tasks = counts.most_common()[::-1]
      starting_task, _ = unvisited_tasks.pop()
      queue = [[0, starting_task]]

      while queue or unvisited_tasks:
      while queue and CLOCK >= queue[0][0]:
      _, task = heapq.heappop(queue)
      counts[task] -= 1
      if counts[task] > 0:
      heapq.heappush(queue, [CLOCK + 1 + n, task])
      CLOCK += 1

      if unvisited_tasks:
      t, _ = unvisited_tasks.pop()
      heapq.heappush(queue, [0, t])
      else:
      # must go idle
      if queue:
      CLOCK += 1

      return CLOCK


      Here's the (large) input case:



      tasks = ["G","C","A","H","A","G","G","F","G","J","H","C","A","G","E","A","H","E","F","D","B","D","H","H","E","G","F","B","C","G","F","H","J","F","A","C","G","D","I","J","A","G","D","F","B","F","H","I","G","J","G","H","F","E","H","J","C","E","H","F","C","E","F","H","H","I","G","A","G","D","C","B","I","D","B","C","J","I","B","G","C","H","D","I","A","B","A","J","C","E","B","F","B","J","J","D","D","H","I","I","B","A","E","H","J","J","A","J","E","H","G","B","F","C","H","C","B","J","B","A","H","B","D","I","F","A","E","J","H","C","E","G","F","G","B","G","C","G","A","H","E","F","H","F","C","G","B","I","E","B","J","D","B","B","G","C","A","J","B","J","J","F","J","C","A","G","J","E","G","J","C","D","D","A","I","A","J","F","H","J","D","D","D","C","E","D","D","F","B","A","J","D","I","H","B","A","F","E","B","J","A","H","D","E","I","B","H","C","C","C","G","C","B","E","A","G","H","H","A","I","A","B","A","D","A","I","E","C","C","D","A","B","H","D","E","C","A","H","B","I","A","B","E","H","C","B","A","D","H","E","J","B","J","A","B","G","J","J","F","F","H","I","A","H","F","C","H","D","H","C","C","E","I","G","J","H","D","E","I","J","C","C","H","J","C","G","I","E","D","E","H","J","A","H","D","A","B","F","I","F","J","J","H","D","I","C","G","J","C","C","D","B","E","B","E","B","G","B","A","C","F","E","H","B","D","C","H","F","A","I","A","E","J","F","A","E","B","I","G","H","D","B","F","D","B","I","B","E","D","I","D","F","A","E","H","B","I","G","F","D","E","B","E","C","C","C","J","J","C","H","I","B","H","F","H","F","D","J","D","D","H","H","C","D","A","J","D","F","D","G","B","I","F","J","J","C","C","I","F","G","F","C","E","G","E","F","D","A","I","I","H","G","H","H","A","J","D","J","G","F","G","E","E","A","H","B","G","A","J","J","E","I","H","A","G","E","C","D","I","B","E","A","G","A","C","E","B","J","C","B","A","D","J","E","J","I","F","F","C","B","I","H","C","F","B","C","G","D","A","A","B","F","C","D","B","I","I","H","H","J","A","F","J","F","J","F","H","G","F","D","J","G","I","E","B","C","G","I","F","F","J","H","H","G","A","A","J","C","G","F","B","A","A","E","E","A","E","I","G","F","D","B","I","F","A","B","J","F","F","J","B","F","J","F","J","F","I","E","J","H","D","G","G","D","F","G","B","J","F","J","A","J","E","G","H","I","E","G","D","I","B","D","J","A","A","G","A","I","I","A","A","I","I","H","E","C","A","G","I","F","F","C","D","J","J","I","A","A","F","C","J","G","C","C","H","E","A","H","F","B","J","G","I","A","A","H","G","B","E","G","D","I","C","G","J","C","C","I","H","B","D","J","H","B","J","H","B","F","J","E","J","A","G","H","B","E","H","B","F","F","H","E","B","E","G","H","J","G","J","B","H","C","H","A","A","B","E","I","H","B","I","D","J","J","C","D","G","I","J","G","J","D","F","J","E","F","D","E","B","D","B","C","B","B","C","C","I","F","D","E","I","G","G","I","B","H","G","J","A","A","H","I","I","H","A","I","F","C","D","A","C","G","E","G","E","E","H","D","C","G","D","I","A","G","G","D","A","H","H","I","F","E","I","A","D","H","B","B","G","I","C","G","B","I","I","D","F","F","C","C","A","I","E","A","E","J","A","H","C","D","A","C","B","G","H","G","J","G","I","H","B","A","C","H","I","D","D","C","F","G","B","H","E","B","B","H","C","B","G","G","C","F","B","E","J","B","B","I","D","H","D","I","I","A","A","H","G","F","B","J","F","D","E","G","F","A","G","G","D","A","B","B","B","J","A","F","H","H","D","C","J","I","A","H","G","C","J","I","F","J","C","A","E","C","H","J","H","H","F","G","E","A","C","F","J","H","D","G","G","D","D","C","B","H","B","C","E","F","B","D","J","H","J","J","J","A","F","F","D","E","F","C","I","B","H","H","D","E","A","I","A","B","F","G","F","F","I","E","E","G","A","I","D","F","C","H","E","C","G","H","F","F","H","J","H","G","A","E","H","B","G","G","D","D","D","F","I","A","F","F","D","E","H","J","E","D","D","A","J","F","E","E","E","F","I","D","A","F","F","J","E","I","J","D","D","G","A","C","G","G","I","E","G","E","H","E","D","E","J","B","G","I","J","C","H","C","C","A","A","B","C","G","B","D","I","D","E","H","J","J","B","F","E","J","H","H","I","G","B","D"]
      n = 1


      My code is outputting an interval count of 1002, and the correct answer is 1000. Because the input size is so large, I'm having trouble debugging by hand on where this is going wrong.



      My algorithm essentially does the following:



      1. Build a mapping of character to number of occurrences

      2. Start with the task that occurs the largest number of times.

      3. When you visit a task, enqueue the next task to be CLOCK + interval iterations later, because my premise is that you want to visit a task as soon as you're able to do so.

      4. If can't visit an already-visited task, enqueue a new one, and do so without incrementing the clock.

      5. If you have elements in the queue, but not enough time has passed, increment the clock.

      At the end, the CLOCK variable describes how long (in other words, how many "intervals") passed before you're able to run all tasks.



      Can someone spot the bug in my logic?









      python algorithm greedy






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Nov 11 at 20:33









      segue_segway

      6391924




      6391924






















          1 Answer
          1






          active

          oldest

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          up vote
          0
          down vote













          Consider a case where the delay n=1, and you have a task distribution like so, for which the least number of cycles is just the length of the list (the tasks could be run like "ABCABC...D"):



          "A": 100, "B": 100, "C": 99, "D": 1 # { "task": <# of occurrences>, ...



          Using your algorithm, you would process all the cases of "A" and "B" first, since you want to move onto the next task in the same type as soon as possible, without considering other task types. After processing those two, you're left with:



          "C": 99, "D": 1



          which results in at least 96 idle cycles.



          To fix this, the ideal task configuration would be something like a round robin of sorts.






          share|improve this answer






















          • Preliminary solution to the question: the leastInterval function should depend on the number of task types N vs. the wait time n. If N>n, then the leastInterval is just the length of the list.
            – Carol Ng
            Nov 12 at 0:48











          • If N<n, then you'd want to count how many idling cycles you'd need OR whether or not your "bottom n jobs" can fill the idle gaps of the "top n jobs". A preliminary formula for the idle cycles is probably something like n*(most common job #) - (length of the list) for a heavily imbalanced task distribution, with a correction term for the terminating sequence.
            – Carol Ng
            Nov 12 at 0:51










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          1 Answer
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          active

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          up vote
          0
          down vote













          Consider a case where the delay n=1, and you have a task distribution like so, for which the least number of cycles is just the length of the list (the tasks could be run like "ABCABC...D"):



          "A": 100, "B": 100, "C": 99, "D": 1 # { "task": <# of occurrences>, ...



          Using your algorithm, you would process all the cases of "A" and "B" first, since you want to move onto the next task in the same type as soon as possible, without considering other task types. After processing those two, you're left with:



          "C": 99, "D": 1



          which results in at least 96 idle cycles.



          To fix this, the ideal task configuration would be something like a round robin of sorts.






          share|improve this answer






















          • Preliminary solution to the question: the leastInterval function should depend on the number of task types N vs. the wait time n. If N>n, then the leastInterval is just the length of the list.
            – Carol Ng
            Nov 12 at 0:48











          • If N<n, then you'd want to count how many idling cycles you'd need OR whether or not your "bottom n jobs" can fill the idle gaps of the "top n jobs". A preliminary formula for the idle cycles is probably something like n*(most common job #) - (length of the list) for a heavily imbalanced task distribution, with a correction term for the terminating sequence.
            – Carol Ng
            Nov 12 at 0:51














          up vote
          0
          down vote













          Consider a case where the delay n=1, and you have a task distribution like so, for which the least number of cycles is just the length of the list (the tasks could be run like "ABCABC...D"):



          "A": 100, "B": 100, "C": 99, "D": 1 # { "task": <# of occurrences>, ...



          Using your algorithm, you would process all the cases of "A" and "B" first, since you want to move onto the next task in the same type as soon as possible, without considering other task types. After processing those two, you're left with:



          "C": 99, "D": 1



          which results in at least 96 idle cycles.



          To fix this, the ideal task configuration would be something like a round robin of sorts.






          share|improve this answer






















          • Preliminary solution to the question: the leastInterval function should depend on the number of task types N vs. the wait time n. If N>n, then the leastInterval is just the length of the list.
            – Carol Ng
            Nov 12 at 0:48











          • If N<n, then you'd want to count how many idling cycles you'd need OR whether or not your "bottom n jobs" can fill the idle gaps of the "top n jobs". A preliminary formula for the idle cycles is probably something like n*(most common job #) - (length of the list) for a heavily imbalanced task distribution, with a correction term for the terminating sequence.
            – Carol Ng
            Nov 12 at 0:51












          up vote
          0
          down vote










          up vote
          0
          down vote









          Consider a case where the delay n=1, and you have a task distribution like so, for which the least number of cycles is just the length of the list (the tasks could be run like "ABCABC...D"):



          "A": 100, "B": 100, "C": 99, "D": 1 # { "task": <# of occurrences>, ...



          Using your algorithm, you would process all the cases of "A" and "B" first, since you want to move onto the next task in the same type as soon as possible, without considering other task types. After processing those two, you're left with:



          "C": 99, "D": 1



          which results in at least 96 idle cycles.



          To fix this, the ideal task configuration would be something like a round robin of sorts.






          share|improve this answer














          Consider a case where the delay n=1, and you have a task distribution like so, for which the least number of cycles is just the length of the list (the tasks could be run like "ABCABC...D"):



          "A": 100, "B": 100, "C": 99, "D": 1 # { "task": <# of occurrences>, ...



          Using your algorithm, you would process all the cases of "A" and "B" first, since you want to move onto the next task in the same type as soon as possible, without considering other task types. After processing those two, you're left with:



          "C": 99, "D": 1



          which results in at least 96 idle cycles.



          To fix this, the ideal task configuration would be something like a round robin of sorts.







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Nov 12 at 0:44

























          answered Nov 12 at 0:36









          Carol Ng

          39019




          39019











          • Preliminary solution to the question: the leastInterval function should depend on the number of task types N vs. the wait time n. If N>n, then the leastInterval is just the length of the list.
            – Carol Ng
            Nov 12 at 0:48











          • If N<n, then you'd want to count how many idling cycles you'd need OR whether or not your "bottom n jobs" can fill the idle gaps of the "top n jobs". A preliminary formula for the idle cycles is probably something like n*(most common job #) - (length of the list) for a heavily imbalanced task distribution, with a correction term for the terminating sequence.
            – Carol Ng
            Nov 12 at 0:51
















          • Preliminary solution to the question: the leastInterval function should depend on the number of task types N vs. the wait time n. If N>n, then the leastInterval is just the length of the list.
            – Carol Ng
            Nov 12 at 0:48











          • If N<n, then you'd want to count how many idling cycles you'd need OR whether or not your "bottom n jobs" can fill the idle gaps of the "top n jobs". A preliminary formula for the idle cycles is probably something like n*(most common job #) - (length of the list) for a heavily imbalanced task distribution, with a correction term for the terminating sequence.
            – Carol Ng
            Nov 12 at 0:51















          Preliminary solution to the question: the leastInterval function should depend on the number of task types N vs. the wait time n. If N>n, then the leastInterval is just the length of the list.
          – Carol Ng
          Nov 12 at 0:48





          Preliminary solution to the question: the leastInterval function should depend on the number of task types N vs. the wait time n. If N>n, then the leastInterval is just the length of the list.
          – Carol Ng
          Nov 12 at 0:48













          If N<n, then you'd want to count how many idling cycles you'd need OR whether or not your "bottom n jobs" can fill the idle gaps of the "top n jobs". A preliminary formula for the idle cycles is probably something like n*(most common job #) - (length of the list) for a heavily imbalanced task distribution, with a correction term for the terminating sequence.
          – Carol Ng
          Nov 12 at 0:51




          If N<n, then you'd want to count how many idling cycles you'd need OR whether or not your "bottom n jobs" can fill the idle gaps of the "top n jobs". A preliminary formula for the idle cycles is probably something like n*(most common job #) - (length of the list) for a heavily imbalanced task distribution, with a correction term for the terminating sequence.
          – Carol Ng
          Nov 12 at 0:51

















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