Grokking the Coding Interview: Patterns for Coding Questions
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Problem Challenge 2: Maximum CPU Load (hard)

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A

Anonymous

· 2 years ago

from heapq import *


#class job:
#  def __init__(self, start, end, cpu_load):
#    self.start = start
#    self.end = end
#    self.cpuLoad = cpu_load

# You can override/modify __lt__ as per your need
#setattr(Job, "__lt__", lambda self, other: write logic here)

class Solution:
  def findMaxCPULoad(self, jobs):
    max_cpu_load = 0

    sum = 0 
    
    jobs.sort(key=lambda x: x.start)

    def overlap(a, b):
      return b.start < a.end

    for i in range(len(jobs)):
      
      if i == 0 or not overlap(jobs[i-1], jobs[i]):
        sum = jobs[i].cpuLoad
      else:
        sum += jobs[i].cpuLoad

      max_cpu_load = max(max_cpu_load, sum)
      
    return max_cpu_load
Show 1 reply
A

Anonymous

· 2 years ago

[[1, 9, 3], [2, 5, 4], [7, 9, 6]]

Expected: 9

My code submission passes, but it should fail on this case (my return is 13).

A

Anonymous

· 3 years ago

Hello,

I wrote a solution that passed the tests even though it was wrong (for some cases). One case where my solution fails is: [[1, 4, 3], [2, 5, 4], [7, 9, 6]]

Maybe you can add that as an additional test case?

A

Anonymous

· 4 years ago

Does anyone know what number this would be on leetcode?

A

Anonymous

· 4 years ago

Can't a simpler implementation of adding up the loads of an interval work.

need to sort the array

def max_cpu_load(jobs):

if len(jobs) < 2: return jobs[0][2]

start = jobs[0][0] end = jobs[0][1] load = jobs[0][2] max_load = load

for i in range(1, len(jobs)): j = jobs[i] if j[0] < end: load += j[2] max_load = max( load, max_load ) end = max( end, j[1]) else: load = j[2] max_load = max(load, max_load) return max_load

print( max_cpu_load([[1,4,3], [2,5,4], [7,9,6]]) ) print( max_cpu_load([[2,4,11], [6,7,10], [8,12,15]]) ) print( max_cpu_load([[1,4,2], [2,4,1], [3,6,5]]) )

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