use map<int,bool> S
when you encounter a particular number check S[x-num] is true, if so return
that pair. else
S[num]=true

On Sat, Sep 11, 2010 at 1:09 PM, topojoy biswas <[email protected]>wrote:

> put all the numbers in a hash table( which demands O(n) space) and then
> pick each number in the array and check in the hashtable whether x-chosen
> number is present. this would take O(1) per search. Do it for all the
> numbers in the array. It wold take O(n).
>
>
> On Fri, Sep 10, 2010 at 11:00 PM, jagadish <[email protected]> wrote:
>
>> Hi,
>> O(n) solution is possible if the array is presorted!
>> Else i think it is NOT.
>> 1.Have two ptrs (one from first and other to track the array from
>> last)
>> 2.s=a[left]+a[right]
>> 3.if(s<sum) left++;
>>  else if(s>sum) right--;
>>  else
>>  print "sum found"
>>
>> On Sep 10, 10:19 pm, bittu <[email protected]> wrote:
>> > Q Given an array, find out if there exists a pair of nos. that adds up
>> > to a sum 'x'
>> >
>> > in O(n) ???? possible,,
>>
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>
>
> --
> Topo.
>
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