Facebook Interview Question for Software Engineers

Country: United States
Interview Type: In-Person

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of 2 vote

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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;


- aonecoding July 15, 2017 | Flag Reply
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of 0 vote

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 {
	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))
	//O(log(n)) Binary Search
	public static boolean findInSmallArray(int[] arr , int value){ // serachArray, valueTobeSearched 
		int low=0;
		int high=arr.length-1;
			int mid = (low+high)/2;
			if(arr[mid] == value){
				return true;
			}else if(arr[mid] < value){
		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

- Mayank Jain August 06, 2017 | Flag Reply

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