CareerCup

private static void displayNumbersHasSum(int[] arr,int sum) {
		HashMap<Integer,Integer> ht=new HashMap<Integer,Integer>();
		for(int i=0;i<arr.length;i++)
		{
			int k=sum-arr[i];
			if(ht.containsKey(arr[i]))
			{
				System.out.println("Numbers are :- "+k+" "+arr[i]);
			}
			else
			{
				ht.put(arr[i], 1);
			}
		}
		
	}
Override the hashcode & equals method

I will use a HashMap to match a pair of two numbers such that sum is K.
Iterate the unsorted array and for each number a, calculate b = k - a
If number b exists in the map, pair up those a and b and return the pair
otherwise put number a into the map and go on to next number in the array.
Once you go through the array, you can find all pairs that sum up to K.

def solve(array, K):
    for indexn, n  in enumerate(array):
        for indexm, m  in enumerate(array):
            if indexm != indexn and (n+m) == K:
                print n,m

solve([1,2,3,4,5,6,7], 8)

I used two indices, s and e after sorting. Time complexity is O(n * log n). Space complexity is O(1).

static void printSum(int[] arr, int K) {
	Arrays.sort(arr);
	
	int s = 0;
	int e = arr.length - 1;
	while (s < e) {
		if (arr[s] + arr[e] == K) {
			System.out.println(arr[s] + ", " + arr[e]);
			s++;
		} else if (arr[s] + arr[e] > K) {
			e--;
		} else {
			s++;
		}
	}
}

Runtime O(n), Memory O(n)

public static int[] getSumPair(int[] arr, int sum){
    HashSet<Integer> contains = new HashSet<Integer>();
    for(int i : arr){
        int needed = sum - i;
        if(contains.contains(needed)){
            return new int[]{i, needed};
        }
        contained.add(i);
    }
    return null;
}