# New Year Chaos

### 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 from 1 to n. Any person can bribe the person directly in front of them to swap positions, but they still wear their original sticker. One person can bribe at most two others.

Determine the minimum number of bribes that took place to get to a given queue order. Print the number of bribes, or, if anyone has bribed more than two people, print Too chaotic.

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 2  people.

Input Format

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

Each of the next t pairs of lines are as follows:
- The first line contains an integer t, the number of people in the queue
- The second line has n space-separated integers describing the final state of the queue.

Constraints

1  <=  t  <=  10
1  <=  n  <=  10^5

Sample Input

STDIN                 Function
-----                       --------
2                         t = 2
5                         n = 5
2 1 5 3 4            q = [2, 1, 5, 3, 4]
5                         n = 5
2 5 1 3 4            q = [2, 5, 1, 3, 4]

Sample Output

3
Too chaotic```

### Solution :

```                            ```Solution in C :

In   C  :

#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>

int main(){
int f,T,tmp,i,j,k,ans,c;
scanf("%d",&T);
for(int a0 = 0; a0 < T; a0++){
int n;
ans=c=f=0;
scanf("%d",&n);
int *q = malloc(sizeof(int) * n);
for(int q_i = 0; q_i < n; q_i++){
scanf("%d",&q[q_i]);
}
for(i=0;i<n-1;i++)
{

for(j=i+1;q[i]>q[j]&&j<n;j++)
{
ans++;
c++;
// printf("i=%d,ans=%d,c=%d\n",i,ans,c);

if(c>2)
{
f=1;
break;
}
}
c=0;
if(f==1)
break;
}
if(f==1)
printf("Too chaotic\n");
else
{
ans=0;
for(i=1;i<n;i++)
{
k=q[i];
for(j=i-1;j>=0&&k<q[j];j--)
{q[j+1]=q[j];
ans++;}
q[j+1]=k;
}
printf("%d\n",ans);
}

}
return 0;
}```
```

```                        ```Solution in C++ :

In   C++  :

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <string>
#include <bitset>
#include <cstdio>
#include <limits>
#include <vector>
#include <climits>
#include <cstring>
#include <cstdlib>
#include <fstream>
#include <numeric>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <unordered_map>

using namespace std;

int main(){
int T;
cin >> T;
for(int a0 = 0; a0 < T; a0++){
int n;
cin >> n;
vector<int> q(n);
for(int q_i = 0;q_i < n;q_i++){
cin >> q[q_i];
}
int ans = 0;
for (int i = n - 1; i >= 0; i--){
if (ans == -1)
break;
int k = i;
while (q[k] != i + 1)
k--;
if (i - k > 2){
ans = -1;
break;
} else {
while (k != i){
swap(q[k], q[k + 1]);
k++;
ans++;
}
}
}
if (ans == -1)
puts("Too chaotic");
else
cout << ans << "\n";
}
return 0;
}```
```

```                        ```Solution in Java :

In   Java  :

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int T = in.nextInt();
for(int a0 = 0; a0 < T; a0++){
int n = in.nextInt();
int q[] = new int[n];
for(int q_i=0; q_i < n; q_i++){
q[q_i] = in.nextInt();
}

int bribes = 0;
for (int i = 1; i <= n; ++i) {
if (q[i - 1] - i > 2) {
System.out.println("Too chaotic");
bribes = -1;
break;
}
}

if (bribes == 0)
{
for (int i = 0; i < n; ++i) {
for (int j = i + 1; j < i + 100 && j < n; ++j) {
if (q[j] < q[i]) {
bribes++;
}
}
}
System.out.println(bribes);
}
}
}
}```
```

```                        ```Solution in Python :

In   Python3  :

#!/bin/python3

import math
import os
import random
import re
import sys

# Complete the minimumBribes function below.
def minimumBribes(q):
m = 0
Q = [i-1 for i in q]
for i,j in enumerate(Q):
if j-i > 2:
print('Too chaotic')
return
for k in range(max(j-1, 0), i):
if Q[k] > j:
m += 1
print(m)

if __name__ == '__main__':
t = int(input())

for t_itr in range(t):
n = int(input())

q = list(map(int, input().rstrip().split()))

minimumBribes(q)```
```

## Delete a Node

Delete the node at a given position in a linked list and return a reference to the head node. The head is at position 0. The list may be empty after you delete the node. In that case, return a null value. Example: list=0->1->2->3 position=2 After removing the node at position 2, list'= 0->1->-3. Function Description: Complete the deleteNode function in the editor below. deleteNo

## Print in Reverse

Given a pointer to the head of a singly-linked list, print each data value from the reversed list. If the given list is empty, do not print anything. Example head* refers to the linked list with data values 1->2->3->Null Print the following: 3 2 1 Function Description: Complete the reversePrint function in the editor below. reversePrint has the following parameters: Sing

Given the pointer to the head node of a linked list, change the next pointers of the nodes so that their order is reversed. The head pointer given may be null meaning that the initial list is empty. Example: head references the list 1->2->3->Null. Manipulate the next pointers of each node in place and return head, now referencing the head of the list 3->2->1->Null. Function Descriptio