I can think of this solution

Put first elements in a max heap ..O(10)

Now take 11 th element ..compare it with root of this max heap ..If less
than root ignore ..else remove root and call max-heapify and put this
element in right place

Similarly do this for rest

Complexity will be  n+nlogk where k is 10

Please put ur comments and correct if wrong !!

Regards

Ankur

On Mon, Aug 8, 2011 at 4:36 PM, shady <[email protected]> wrote:

> @DK can you please explain your solution, i couldn't understand...
>
> @all any alternate solution ?
>
>
> On Sun, Aug 7, 2011 at 11:02 PM, DK <[email protected]> wrote:
>
>> This is a simple problem:
>>
>> 1. Understand that if you are selecting 2 petrol pumps, say P and Q with
>> say another petrol pump R between them, then it is optimal to include R into
>> the set of selected petrol pumps as the max-distance for the chosen pumps
>> remains PQ but number of included petrol pumps is reduced.
>>
>> Diagram: P --- R --- Q
>>
>> Therefore, all the selected pumps in the final solution are going to be
>> contiguous.
>>
>> The simplest solution thus is to sort the points along the line AB. O(n
>> log n)
>> Then the answer is min{ai+10 - ai} for i = 0 to N-11 - O(n)
>>
>> Therefore net complexity is O(n log n).
>> If the petrol pumps are given in sorted order from distance from A, then
>> the complexity is simply O(n).
>>
>> --
>> DK
>>
>> http://gplus.to/divyekapoor
>> http://www.divye.in
>> http://twitter.com/divyekapoor
>>
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