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:
- Build a mapping of character to number of occurrences
- Start with the task that occurs the largest number of times.
- 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.
- If can't visit an already-visited task, enqueue a new one, and do so without incrementing the clock.
- 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
add a comment |
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:
- Build a mapping of character to number of occurrences
- Start with the task that occurs the largest number of times.
- 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.
- If can't visit an already-visited task, enqueue a new one, and do so without incrementing the clock.
- 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
add a comment |
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:
- Build a mapping of character to number of occurrences
- Start with the task that occurs the largest number of times.
- 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.
- If can't visit an already-visited task, enqueue a new one, and do so without incrementing the clock.
- 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
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:
- Build a mapping of character to number of occurrences
- Start with the task that occurs the largest number of times.
- 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.
- If can't visit an already-visited task, enqueue a new one, and do so without incrementing the clock.
- 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
python algorithm greedy
asked Nov 11 at 20:33
segue_segway
6391924
6391924
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
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.
Preliminary solution to the question: theleastInterval
function should depend on the number of task typesN
vs. the wait timen
. IfN>n
, then theleastInterval
is just the length of the list.
– Carol Ng
Nov 12 at 0:48
IfN<n
, then you'd want to count how manyidling
cycles you'd need OR whether or not your "bottomn
jobs" can fill the idle gaps of the "topn
jobs". A preliminary formula for the idle cycles is probably something liken*(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
add a comment |
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1 Answer
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1 Answer
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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.
Preliminary solution to the question: theleastInterval
function should depend on the number of task typesN
vs. the wait timen
. IfN>n
, then theleastInterval
is just the length of the list.
– Carol Ng
Nov 12 at 0:48
IfN<n
, then you'd want to count how manyidling
cycles you'd need OR whether or not your "bottomn
jobs" can fill the idle gaps of the "topn
jobs". A preliminary formula for the idle cycles is probably something liken*(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
add a comment |
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.
Preliminary solution to the question: theleastInterval
function should depend on the number of task typesN
vs. the wait timen
. IfN>n
, then theleastInterval
is just the length of the list.
– Carol Ng
Nov 12 at 0:48
IfN<n
, then you'd want to count how manyidling
cycles you'd need OR whether or not your "bottomn
jobs" can fill the idle gaps of the "topn
jobs". A preliminary formula for the idle cycles is probably something liken*(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
add a comment |
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.
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.
edited Nov 12 at 0:44
answered Nov 12 at 0:36
Carol Ng
39019
39019
Preliminary solution to the question: theleastInterval
function should depend on the number of task typesN
vs. the wait timen
. IfN>n
, then theleastInterval
is just the length of the list.
– Carol Ng
Nov 12 at 0:48
IfN<n
, then you'd want to count how manyidling
cycles you'd need OR whether or not your "bottomn
jobs" can fill the idle gaps of the "topn
jobs". A preliminary formula for the idle cycles is probably something liken*(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
add a comment |
Preliminary solution to the question: theleastInterval
function should depend on the number of task typesN
vs. the wait timen
. IfN>n
, then theleastInterval
is just the length of the list.
– Carol Ng
Nov 12 at 0:48
IfN<n
, then you'd want to count how manyidling
cycles you'd need OR whether or not your "bottomn
jobs" can fill the idle gaps of the "topn
jobs". A preliminary formula for the idle cycles is probably something liken*(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
add a comment |
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