in above case we need to do some checking like in case when next element is
10;

next elements is 10
10-4 = 6 , 6 > 4

add(1,3,4 ) = 8

8 < 10(required_element) , so add 1 to 8 = 9 and do same as mentioned

but if sum( number_found_so_far ) > required_num then add 1 to difference
of required_num - least number .


for example :-

suppose in the same example if i add 15 in the end of given input

now next elem 15 - 4 = 11

sum(1,3,4,9) = 17

15 < 17 , so add 1 to 11 = 12 and use this number(i.e 12 ) and
element_found_so_far(i.e 1,3,4,9) to form 15 ( 12 + 3 )  : yes , so 12
should be there

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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