on above algo , there is no need to calculate sum of array so we can do it
in O(N) and not O(n).

On Fri, Dec 16, 2011 at 12:59 AM, Ankur Garg <ankurga...@gmail.com> wrote:

> Hi Topcoder.
>
> First of all you  posted  the wrong array .
>
> The array should be
>
> 4,      5,      10,     7,      12,     13
> a+b  a+c   a+d   b+c    b+d   c+d
>
> Now first calculate a+b+c+d which will be (sumofarray)/N-1
>
> So here a+b+c+d = 17
>
> Now take a[1] is a+c
> and a[N-1] =  b+c
> subtracting them gives b-a = 2
>                              a[0] is b+a=4
> that gives  b=3,a=1
> Now u have a and b calculate c as a[1]-a=4
> and d as9 . For this we traverse from a[1] to a[N-2]
> We calculate a and b because we know the order of sum of their
> elements(a+bis given and b's  addition with rest elements are there in
> array)
>
> This will work in Linear Time
>
> Now lets take an example with  8 elements to
> let  a=1,b=2,c=3,d=4,e=5,f=6,g=7,h=8
>
> then N=8 and array is
>  3  4  5  6 7 8 9 5 6 7 8 9 10 7 8 9 10 11 9 10 11 12 11 12 13 13 14 15
> Now by above logic  first
> a+b+c+d+e+f+g+h = (sum)/7 = 252/7 = 36
> Now a[1]=a+c = 4 and a[N-1] =a[7]=b+c=5
> a[8]-a[1]= b-a=1 and a+b=a[0]=3 gives b=2 and a =1
> Now we have a=1,b=2
> So we traverse from a[1] to a[N-2] to calculate values c to h
> c= a[1]-a=4-1=3
> d=a[2]-a=5-1=4
> e=a[3]-a=6-1=5
> similarly f=a[4]=6,g=a[5]=7 and h=a[6]=8
>
> This will work in O(n)
>
> Regards
> Ankur
>
> On Thu, Dec 15, 2011 at 12:42 PM, WgpShashank 
> <shashank7andr...@gmail.com>wrote:
>
>> @all , a Naive Approach with Quadratic Time will be for each i=1 to n ,
>> check if i and a[i]-i makes given sum , so for each each number we will do
>> the thus can achieve the solution ...i am just thinking about if we can do
>> it linear time ..will post if able to do it :)
>>
>>
>> Thanks
>> Shashank
>> CSe BIT Mesra
>>
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