how is the case taken of when 2 pairs add to the same sum?...

On Tue, Dec 13, 2011 at 11:35 AM, atul anand <[email protected]>wrote:

> hmmm i guess i screwed by taking least element as a part of the output set
> directly.
>
>
>
>
> On Wed, Dec 14, 2011 at 12:57 AM, sayan nayak <[email protected]>wrote:
>
>> @atul:
>> Suppose the input is :(7,8,9)
>>
>> So output should be (3,4,5)
>>
>> then ur approach is not giving the answers..
>>
>> Regards,
>> Sayan
>>
>> On Tue, Dec 13, 2011 at 11:51 PM, atul anand <[email protected]>wrote:
>>
>>> i am not sure , but this came to me when i first read it
>>>
>>> here is the idea:-
>>> given : 4 5 7 10 12 13
>>>
>>> 4 should be there boz it is the least.
>>>
>>> next is 5 , 5-4 =1 which is less that 4 , so 1 should be there
>>>
>>> next is 7 , 7-4 = 3 which is less than 4 , so 3 should be there
>>>
>>> next is 10 , 10-4 = 6 which is greater then 4 , so add previous found
>>> elements
>>> i.e 1,3,4 add them 1+3+4 = 8 , add 1 to 8 = 9
>>>
>>> now check ,can we use this number(i.e 9 ) and previous found elements (
>>> 1,3,4) to
>>> form 10 ( 9 +1) : yes -> so 9 should be there
>>>
>>> next is 12 , 12-4 = 8 ( but now greatest element among 1,3,4,9 is 9) and
>>> 8 < 9 ,so skip;
>>>
>>> next is 13 , 13-4 = 9 same reason for skipping as for number 12 before.
>>>
>>>
>>>
>>> On Tue, Dec 13, 2011 at 10:28 PM, top coder <[email protected]>wrote:
>>>
>>>> If pairwise sums of 'n' numbers are given in non-decreasing order
>>>> identify the individual numbers. If the sum is corrupted print -1
>>>> Example:
>>>> i/p:
>>>> 4
>>>> 4 5 7 10 12 13
>>>>
>>>> o/p:
>>>> 1 3 4 9
>>>>
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