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solution 4.1

public TreeNode solve(TreeNode root) {
        TreeNode[] res = treetoDLL(root);
        if(res == null) return null;
        res[0].left = res[1];  //make linked list circular
        res[1].right = res[0];
        return res[0];  //return a node in the chain
    }

    public TreeNode[] treetoDLL(TreeNode root) { //recursion to create linked list in place    

        if(root == null) return null;
    

        TreeNode[] prev = treetoDLL(root.left);

        TreeNode[] next = treetoDLL(root.right);


        TreeNode[] res = new TreeNode[] {root, root};
    

        if(prev != null) {

            prev[1].right = root;

            root.left = prev[1];

            res[0] = prev[0];

        }
        
if(next != null) {

            next[0].left = root;

            root.right = next[0];

            res[1] = next[1];

        }

        return res;

    }

Let say
Small array size = m
Large array size = n

Steps: Loop through each element of small array(O(m)) and do binary search on large array O(log(n)) = O(mlog(n))

public class InterSectionOfSortedArrays {
	
	//O(mlog(n))
	public static void getIntersection(int[] arr1 , int[] arr2){
		
		for(int i=0;i<=arr2.length-1;i++){ //O(m)
			if(findInSmallArray(arr1,arr2[i])){ // O(log(n))
				System.out.println(arr2[i]);
			}
		}
	}
	
	//O(log(n)) Binary Search
	public static boolean findInSmallArray(int[] arr , int value){ // serachArray, valueTobeSearched 
		int low=0;
		int high=arr.length-1;
		
		while(low<=high){
			int mid = (low+high)/2;
			if(arr[mid] == value){
				return true;
			}else if(arr[mid] < value){
				low=mid+1;
			}else{
				high=mid-1;
			}
		}
		return false;
	}
	
	
	public static void main(String[] args) {
	
		int [] arr1 = {1, 3, 4, 5, 7,8,9,10,11}; // very large
		int [] arr2 = {2, 3, 5, 6, 9, 10}; 	// small
		getIntersection(arr1,arr2);
	}
}

4.1 is a good one.

data Tree a = Leaf a | Bin (Tree a) (Tree a)

data Circular a = Node a (Circular a) (Circular a) 

-- pre-order traversal
--
--   .
--  / \
-- 0   .
--    / \
--   .   .
--  / \  |\
-- 1   5 7 3
--
-- 0 <-> 1 <-> 5 <-> 7 <-> 3 <-> 0 <-> ...
--
toCircular :: Tree a -> Circular a
toCircular t =
  let
    go :: Tree a -> Circular a -> Circular a
    go (Leaf a) n =
      let
        x =
          Node a x n
      in
        x
    go (Bin t1 t2) n =
      let
        x1 =
          go t1 x2
        x2 =
          go t2 n
      in
        x1
  in
    go t (toCircular t)

-- view 8 . toCircular $ t
-- [0,1,5,7,3,0,1,5]
view :: Int -> Circular a -> [a]
view 0 _ =
  []
view n (Node a _ t2) =
  a : view (n-1) t2