힙 02 - 디스크컨트롤러, 프로그래머스, 알고리즘, Programmers, Stack, Queue, Hash, 코딩테스트, Algorithm, 풀이과정, Leetcode, 릿코드, 코딩테스트, Tech interview, Heap, 힙

 

# Process SJF(Short Job First)

# 1. Input tasks [[input_time, task_time], ..]

# 2. Sort using input_time

# 3. Iterate till tasks is remained

#  3.1. Search shortest task which input time is under time spent

#  3.2. Do task and ++time

# 4. Return total time / initial length of tasks

 

import heapq

import math



def solution(tasks):

    

    whole_task_time = 0

    time_spent = 0

    len_tasks = len(tasks)

    

    # 2.

    tasks_sorted = sorted(tasks, key=lambda k: (k[0]), reverse=False)

        

    if len_tasks > 0:

        # 3.

        while len(tasks_sorted) > 0:

            # 3.1.

            current_input_time = tasks_sorted[0][0]

            shortest_task_time = tasks_sorted[0][1]

            index = 0

            i = 1

            if time_spent <current_input_time:

                time_spent = current_input_time

            while i < len(tasks_sorted) and tasks_sorted[i][0] <= time_spent:

                if tasks_sorted[i][1] <= shortest_task_time:

                    index = i

                    shortest_task_time = tasks_sorted[i][1]

                i += 1

            working_task = tasks_sorted.pop(index)

            

            # 3.2.

            time_spent += working_task[1]

            

            whole_task_time += (time_spent - (working_task[0]))

            

    return math.floor(whole_task_time / len_tasks)

 

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