CareerCup

This was posted by someone before for a very similar question:
This will give all existing pairs such that
A[i] + A[j] = k
Method 1: Sort the array and then do the following (O(n log n)):

p=0,q=n-1;
while(p<q)
{
  if(a[p]+a[q]==k)
  {
      cout<<a[p]<<"\t"<<a[q];
      p++;
      q--;
   }
   else if(a[p]+a[q]>k)
      q--;
   else 
      p++;
}

Method 2: We can use hash table (O(n)):

for(i=0;i<n;i++)
{
   Complement=K-a[i];
   if(SearchHashTable(Complement)!=-1)
     cout<<"Indices:\t"<<HashTable[Complement]<<"\t"<<i<<endl;
   HashTable[a[i]]=i;
}

Assuming that the array is sorted:

class Problem1
{
	public static void main(String[] args)
	{
		int[] iArray = {1,3,5,6,7,9};
		int len = iArray.length;
		int first = 0;
		int last = len - 1;
		int targetSum = 12;
		while(first<last)
		{
			if(iArray[first] + iArray[last] == targetSum)
			{
				System.out.println(first + "," + last);
				first++;
				last--;
			}
			else if((iArray[first] + iArray[last]) > targetSum)
				last--;
			else if((iArray[first] + iArray[last]) < targetSum)
				first++;
		}
	}
}