# New Year Chaos, Hackerrank

We will discuss Find the **New Year Chaos** problem from Hackerrank here.

## Problem Statement

It is New Year’s Day and people are in line for the Wonderland rollercoaster ride. Each person wears a sticker indicating their *initial* position in the queue. Initial positions increment by from at the front of the line to at the back.

### Rules –

- Any person in the queue can bribe the person
*directly in front*of them to swap positions. - If two people swap positions, they still wear the same sticker denoting their original places in line.
- One person can bribe
*at most two others*.

Fascinated by this chaotic queue, you decide you must know the minimum number of bribes that took place to get the queue into its current state. If anyone has bribed more than two people, the line is too chaotic to compute the answer.

**Function Description**

Complete the function *minimumBribes* in the editor below.

minimumBribes has the following parameter(s):

*int q[n]*: the positions of the people after all bribes

**Returns**

- No value is returned. Print the minimum number of bribes necessary or
`Too chaotic`

if someone has bribed more than people.

**Input Format**

The first line contains an integer , the number of test cases.

Each of the next pairs of lines are as follows:

– The first line contains an integer , the number of people in the queue

– The second line has space-separated integers describing the final state of the queue.

**Constraints**

- 1 <= t <= 10
- 1 <= n <= 10
^{5}

Input

5

2 1 5 3 4

Output

3

Input

5

2 5 1 3 4

Output

Too chaotic

In this problem, we need to check if the given array can be rearranged into a sorted array by following the **rules**.

## Solution: New year chaos

As the question only asks about the number of bribes all you need to do is to count the number of people who overtake a person.

Two points to keep in mind while solving this problem:

- A person can bribe the one who is sitting just in front of him. The opposite is not possible.
- A person can bribe at most 2 different persons.

## Approach

We are using two loops here.

Outer loop runs backwards and if the difference between current value and current index is greater than 2, then print “Too chaotic” and return. Why we are checking specifically 2, because as per the rule we can bribe as most 2 different persons.

Inner loop runs from, and if value and j^{th} position is greater than value at i^{th} position than update the sum.