## tihor

BAN USER- 2of 2 votes

AnswersGiven an array arr[] of size n. Three elements arr[i], arr[j] and arr[k] form an inversion of size 3 if a[i] > a[j] >a[k] and i < j < k. Find total number of inversions of size 3

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e.g.

Input: 23, 10, 24, 7, 12, 19

Ans: 23, 10, 7| Report Duplicate | Flag | PURGE

Deshaw Inc Applications Developer - 1of 1 vote

AnswersYou have a binary search tree and you have to return the two nodes such that there sum i equal to ‘K’. Pseudo code is to be given.

- tihor in India

O(n)time & O(n) sppace is easy but challenge O(n) time & O(1) space.| Report Duplicate | Flag | PURGE

Amazon Java Developer Algorithm - 1of 3 votes

AnswersYou are given a 2D Array that contains only 0s and 1s in sorted order. i.e. First Os and then 1s.

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Array:

0 0 0 1

1 1 1 1

0 0 1 1

0 1 1 1

You have to figure out the row that contains maximum number of 1s.

e.g. in above case we have row 2 as the answer.| Report Duplicate | Flag | PURGE

Amazon Software Engineer / Developer Algorithm - 1of 1 vote

AnswersIn Amazon's interview, Round 2 they asked question:

Write a program to print inorder traversal of two trees.

Both values must be printed alternatively.

e.g. if inorder traversal of tree 1 is 10, 15, 20 and tree 2 is 100, 150, 200 then it should print

10, 100, 15, 150, 20, 200.

I tried printing it recursively by calling 2 functions recursively (f1 calls f2, then f2 calls f1 and so on).

I don't know where my approach is going wrong? I know there are other ways to do it as well but in this approach what is the mistake?

- tihor in India`public class HelloWorld{ //this is the Node used in the tree static class Node{ private int data; private Node left; private Node right; public Node(int data){ this.data = data; left = null; right = null; } public void setLeft(Node left){ this.left = left; } public void setRight(Node right){ this.right = right; } public Node getLeft(){ return this.left; } public Node getRight(){ return this.right; } public int getData(){ return this.data; } public boolean equals(Node n){ if(this.data ==(int) n.getData()) return true; else return false; } } public static void main(String[] args){ HelloWorld bts = new HelloWorld(); bts.run(); } //execute the test case public void run(){ Node root = new Node(10); insert(root,new Node(20)); insert(root,new Node(50)); insert(root,new Node(40)); insert(root,new Node(15)); Node node2 = new Node(100); insert(node2,new Node(200)); insert(node2,new Node(500)); insert(node2,new Node(400)); insert(node2,new Node(150)); //inOrderTraverse(node2); inOrderTraverse1(root, node2); } // insert a node to the binary search tree public void insert(Node root, Node n){ if(root == null|| n == null) return; if(root.getData() > n.getData()){ if(root.getLeft() == null){ root.setLeft(n); }else{ insert(root.getLeft(),n); } }else if(root.getData() < n.getData()){ if(root.getRight() == null){ root.setRight(n); }else{ insert(root.getRight(),n); } } } public void inOrderTraverse(Node node){ if(node != null){ if(node.getLeft() != null) inOrderTraverse(node.getLeft()); System.out.print(" "+node.getData()); if(node.getRight() != null) inOrderTraverse(node.getRight()); } } //in-order Traversal public void inOrderTraverse1(Node node1, Node node2){ if(node2 != null){ inOrderTraverse2(node1, node2.getLeft()); System.out.println(" "+node2.getData()); inOrderTraverse2(node1, node2.getRight()); } } public void inOrderTraverse2(Node node1, Node node2){ if(node1 != null ){ inOrderTraverse1(node1.getLeft(), node2); System.out.println(" "+node1.getData()); inOrderTraverse1(node1.getRight(), node2); } }`

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- tihor December 31, 2015