@Ashish: complexity for your solution is O(n*Total seconds)
On Wed, May 25, 2011 at 5:25 PM, Ashish Goel <ashg...@gmail.com> wrote:
> Hi
>
> I was just thinking of a simpler solution
>
> find sum of the time all processes will take, lets call that sum, say for
> 8,1,3,3,8 it is 23
> walk over each process to see if this has been completely executed, if not
> set the out[i]=curTime and reduce the T[i] by 1.(out is the process final
> end time and T is the time process will take if run without halt i.e.
> 8,1,3,3,8)
>
> t=0;
> while (t<=sum)
> {
>
> for (int i=0;i<n;i++)//i represents total proc count
> {
> if (T[i]>0) {t++; out[i]=t; T[i]--;}
> }
> }
>
> I must say that this is naive solution
>
> I am not able to understand how interval tree and the code given
> by anshumishra6827 will work. Even if this is right, what was the rationale
> for using segment tree and what is the loop doing by playing with ans, D,
> L..
>
> Best Regards
> Ashish Goel
> "Think positive and find fuel in failure"
> +919985813081
> +919966006652
>
>
> On Wed, May 25, 2011 at 2:24 PM, anshu mishra
> <anshumishra6...@gmail.com>wrote:
>
>> two thing i have forgot do;
>>
>> that is at every iteration
>>
>> rem--;
>> last = D[i-1];
>>
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