The trivial solution would be as follows:
findAndUpdateMin(List Truck,value):
minDiff=MAX_INT
minIdx=-1
for i in Truck.size do:
if(Truck[i].FreeSpace>=value&& minDiff) then
minDiff=Truck[i].FreeSpace-value
minIdx=i
endif
done
if(minIdx==-1) return failiure
Truck[minIdx].FreeSpace=Truck[minIdx].FreeSpace-value
return minIdx
Saurabh Singh
B.Tech (Computer Science)
MNNIT
blog:geekinessthecoolway.blogspot.com
On Sat, Mar 10, 2012 at 10:43 PM, Bhavani Pradeep <[email protected]>wrote:
> yeah.. they can be removed too
>
>
> On Sat, Mar 10, 2012 at 10:41 PM, saurabh singh <[email protected]>wrote:
>
>> Is it possible items are removed from the truck too?
>> Saurabh Singh
>> B.Tech (Computer Science)
>> MNNIT
>> blog:geekinessthecoolway.blogspot.com
>>
>>
>>
>> On Sat, Mar 10, 2012 at 9:39 PM, shruthi sharma <
>> [email protected]> wrote:
>>
>>> I implemented it by using arraylist. I sorted the trucks according to
>>> the free space and when a load comes I iterated till the point, say j where
>>> I find a suitable space and then sorted it with the elements till j. I want
>>> to know if I can optimize it even further if possible. I want to know if
>>> there a best possible solution than what I implemented?
>>>
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>
>
> --
> Bhavani Pradeep Darbha
> Software Engineer
> IBM - Indian Software Lab
> Pune.
> Ph: +91-9049996768
>
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