WELL my answer is complexity is between   n^2 to n

SOLUTION LIES WITHIN THE SOLUTION Array  AS:

A= {3  4 5 15 19 20 25}


first  we have in B={ 3,4}    now we have   PTR1 @ 5    and    PTR2  always
one less thab PTR1   means @ 4

now from B[0] upto  the same element ie 4  add      ptr2  and check
 3+4>5   so we hav

B={3,4,5}            now ptr1 @15   and ptr2  @ 5   checking   with B[0]
 3+5<15 &    4+5 <15
so B={3,4,5,8,9}
Now we missed one thing that is 12
so we need one more check that is while we move to   to each node we can
remeber sum   that is  while on 4 we have 7


 on 5 we have 12
so we need to check also we SUM <15        we add 12  so B={3,4,5,8,9,12}

""""""""""""""""""""""IMP NOTE : we check sum only if   pointer moving from
B[0] to reaches the no. itself """"""""""""""""

next we hav ptr1 @ 19 ptr2 @ 15    so          15+b[0] =15+3<19   & 15+4=19
so   we insert B={3,4,5,8,9,12,15,18}
& sum=3+4+5+8 =12+8=20 as sum>19   so no change

3,4,5,8,9,12,15,18,19,20,     20+B[0]  =23 ,20+b[1]=24   ,25 sooooooooooooo
onnnnnnnnn




On Fri, Apr 8, 2011 at 11:44 PM, Akash Agrawal <[email protected]>wrote:

> what do u mean by* top k values?*
> can't the answer  be  3,4,5,15,3+15 etc
>
> Regards,
> Akash Agrawal
> http://tech-queries.blogspot.com/
>
>
>
> On Fri, Apr 8, 2011 at 7:22 PM, saurabh agrawal <[email protected]>wrote:
>
>>  ou are given an array A of k values which contain int values in sorted
>> (asec) order. Find top k values (asec) which can either be the number from
>> the array A, or sum of any two numbers from A or sum of any three numbers
>> from A. So, if A's values are represented as : a1,a2,...,ak , the possible
>> numbers can be: a(i), a(i)+a(j), a(i)+a(j)+a(l) for any i,j,l < k
>>
>> Ex: A[7] = {3,4,5,15,19,20,25}
>> output B[7] = {3,4,5,(3+4),(3+5),(4+5),(3+4+5)}
>>
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-- 
Arpit Bhatnagar
(MNIT JAIPUR)

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