CareerCup

Use two indexes, one at the beginning and one at the end.

while(start < end) {
 if (a[start] + a[end] > number)
  end--;
 else
  if(a[start] + a[end] < number)
   start++;
  else
   break; // found 
}

Thanks S for the answer

@Paras: Given a sorted array, your approach is definitely higher in complexity. However if we are given an unsorted array, the sorting itself will take O (n lgn). Then your approach will also be O (n lgn). No loss no gain. However S's approach is recommended even in that case!

We can use Hashtable to achieve better time-complexity.
Just Iterate through an array i.e. remainder= NumberGiven - A[i] and search the hashTable for remainder.
I hope, it does work.
or Just check all possibilities.

1) Find the index (k) of the number in array.
2) Parse a[] up to k-1 by with n/2(n/2-1) iterations to verify. The complexity will be O(n2/2)

Just another idea (though the complexity will be high)
for each element a[i] in the array
do a binary search on the array for number-a[i].
if found then these two numbers are the pair.