CareerCup

/*
## Setup

The flow of a dispute is as follows:
- A charge is created by an end customer.
- Stripe receives a dispute record from the bank.
- The business responds with evidence.
- If no second dispute is received within 30 days after evidence submission, the dispute is won. If a second dispute is received, the dispute is lost.

Charge
(Maybe) Dispute Record
(Maybe) Evidence submission
(Maybe) Second Dispute Record

The raw tables generated from the API look like:

```
Charges
+---------------+-----------+
| charge_id | varchar |
| created | timestamp |
| amount | int |
| seller_id | varchar |
| customer_id | varchar |
+---------------|-----------+

Dispute Records
+----------------+-----------+
| dispute_id | varchar |
| created | timestamp |
| charge_id | varchar |
+----------------|-----------+

Evidence Submission
+-------------------+-----------+
| evidence_id | varchar |
| created | timestamp |
| charge_id | varchar |
+-------------------|-----------+
```
*/

/*

1. Can you design a unified dispute table that would allow us to compute things like the win rate, dispute rate, evidence submission rate etc?

*/

A happy number is defined by the following process. Starting with any positive integer, replace the number by the sum of the squares of its digits, and repeat the process until the number equals 1 (where it will stay), or it loops endlessly in a cycle which does not include 1. Those numbers for which this process ends in 1 are happy numbers, while those that do not end in 1 are unhappy numbers.
Input:
Your program should read lines of text from standard input. Each line contains a single positive integer, N.
Output:
If the number is a happy number, print 1 to standard output. Otherwise, print 0.

Given two binary trees, explain how you would create a diff such that if you have that diff and either of the trees you should be able to generate the other binary tree. Implement a function which takes Node tree1, Node tree2 and returns that diff.

C program for the given two array as point of x and y
int x[] = { 2,3,2,4,2};
int y[] = {2, 2, 6,5,8}; count the pair
maximum area coverd in x y plane
Ouput:3
From the above input there are five coordinates possible (2,2)(3,2)(2,6()(4,5)and (2,8) in which (2,2)(2,6)(2,8)that is three coodinaters covers the maximum area in x y plane
Input: x[] ={1,2,3}
Y[] = {1,2,3}
Output: 1
there are three coordinates (1,1)(2,2)(3,3)in which only one coordinates covers maximum distancethat is (1,3)

Create an iterator class that stores a list of the built-in Iterators.
Implement the next() and hasNext() methods in a Round Robin pattern (pops next element in a circle).
Example:
Given a list [iterator1,iterator2, iterator3...]
when calling RoundIterator.next()
pops iterator1.next if iterator1.hasNext() is true
when calling RoundIterator.next()
pops iterator2.next() if iterator2.hasNext() is true
when calling RoundIterator.next()
pops iterator3.next if iterator3.hasNext() is true
...
when calling RoundIterator.next()
pops iterator1.next if iterator1.hasNext() is true
when calling RoundIterator.next()
pops iterator2.next if iterator2.hasNext() is true
when calling RoundIterator.next()
pops iterator3.next if iterator3.hasNext() is true
...
until there is no more element in any of the iterators