CareerCup

1) sort the array
2) then start with one pointer i=0 and another j=n(array length)
3) calculate sum if greater then k then j-- and if less then k than i++
4) repeat 3 step till you find the sum or i>j

1) Hash the values in a table 'Table'.
2) For all values 'e_i' in the series
a) if 'K - e_i' is found in the table at position j, return (i, j)
3) return null

or sort in (nlogn) then do a binary search .

Here's a solution using a hashmap to make the time complexity O(n)

private static void printNumsThatAddUpTo(int[] arr, int target) {
	Map<Integer, Integer> storage = new HashMap<Integer,Integer>();
	for (int i = 0; i < arr.length;i++){
		if ( storage.containsKey(target - arr[i])){
			if (!storage.containsKey(arr[i])){
				System.out.println(arr[i]+ " "+(target - arr[i]));
				storage.put(arr[i], arr[i]);
			}
		}else if (!storage.containsKey(arr[i])){
			storage.put(arr[i], arr[i]);
		}
	}

}