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.

Example

q = [1,2,3,5,4,6,7,8]
If person 5 bribes person 4, the queue will look like this: 1,2,3,5,4,6,7,8. Only 1 bribe is required. Print 1.
q = [4,1,2,3]

Person 4 had to bribe 3 people to get to the current position. 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


Solution :



title-img


                            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;
}









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();
        int a,c;
        for(int a0 = 0; a0 < T; a0++){
            int n = in.nextInt();
            int q[] = new int[n], q2[] = new int[n];
            for(int q_i=0; q_i < n; q_i++){
                q[q_i] = in.nextInt()-1;
                q2[q_i] = q_i;
            }
            c=0;
            for(int i=0;i<n;i++) {
                if (q[i]==q2[i]) {
                    
                } else if (q[i]==q2[i+1]) {
                    a=q2[i];
                    q2[i]=q2[i+1];
                    q2[i+1]=a;
                    c++;
                } else if (q[i]==q2[i+2]) {
                    a=q2[i+1];
                    q2[i+1]=q2[i+2];
                    q2[i+2]=a;
                    a=q2[i];
                    q2[i]=q2[i+1];
                    q2[i+1]=a;
                    c+=2;
                } else {
                    System.out.println("Too chaotic");
                    i=n;
                }
                if (i==n-1) {
                    System.out.println(c);
                }
            }
        }
    }
}









In C :





#include <stdio.h>
int N;
int ar[100000];

int scan_swap ( int j ) {
	int temp;
	if ( j == ar[j - 1] ){
		return 0;
	}
	else if ( j == ar [j - 2] ){
		temp = ar[j- 2];
		ar[j - 2] = ar[j - 1];
		ar[j - 1] = temp;
		return 1;
	}
	else if (j == ar [j - 3] ){
		temp = ar[j- 3];
		ar[j - 3] = ar[j - 2];
		ar[j - 2] = ar[j - 1];
		ar[j - 1] = temp;
		return 2;
	}
	else return -1;
}



int main (void){
	int T, i, j, k, point;
	scanf ("%d", &T);
	for (i=0;i<T;++i){
		scanf ("%d", &N);
		for (j=0; j<N; ++j) 
			scanf("%d", &ar[j] );
			int total = 0;
		for (j=N; j>0; --j){
			point = scan_swap ( j );
			/*for (k=0; k<N; ++k) 
				printf("%d ", ar[k] );puts("");*/
			if ( point != -1 ){
				total += point;
			}
			else {
				total = -1;
				break;
			}
		}
		if ( -1 == total ){
			puts ("Too chaotic");
		}
		else printf ("%d\n", total);
	}
}









In Python3 :






def count_inversion(lst):
    return merge_count_inversion(lst)[1]

def merge_count_inversion(lst):
    if len(lst) <= 1:
        return lst, 0
    middle = int( len(lst) / 2 )
    left, a = merge_count_inversion(lst[:middle])
    right, b = merge_count_inversion(lst[middle:])
    result, c = merge_count_split_inversion(left, right)
    return result, (a + b + c)

def merge_count_split_inversion(left, right):
    result = []
    count = 0
    i, j = 0, 0
    left_len = len(left)
    while i < left_len and j < len(right):
        if left[i] <= right[j]:
            result.append(left[i])
            i += 1
        else:
            result.append(right[j])
            count += left_len - i
            j += 1
    result += left[i:]
    result += right[j:]
    return result, count
for _ in range(int(input())):
    N=int(input())
    a=list(map(int,input().split()))
    if any(a[i]-(i+1)>2 for i in range(N)):
        print("Too chaotic")
    else:
        print(count_inversion(a))
                        




View More Similar Problems

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

View Solution →

Reverse a linked list

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

View Solution →

Compare two linked lists

You’re given the pointer to the head nodes of two linked lists. Compare the data in the nodes of the linked lists to check if they are equal. If all data attributes are equal and the lists are the same length, return 1. Otherwise, return 0. Example: list1=1->2->3->Null list2=1->2->3->4->Null The two lists have equal data attributes for the first 3 nodes. list2 is longer, though, so the lis

View Solution →

Merge two sorted linked lists

This challenge is part of a tutorial track by MyCodeSchool Given pointers to the heads of two sorted linked lists, merge them into a single, sorted linked list. Either head pointer may be null meaning that the corresponding list is empty. Example headA refers to 1 -> 3 -> 7 -> NULL headB refers to 1 -> 2 -> NULL The new list is 1 -> 1 -> 2 -> 3 -> 7 -> NULL. Function Description C

View Solution →

Get Node Value

This challenge is part of a tutorial track by MyCodeSchool Given a pointer to the head of a linked list and a specific position, determine the data value at that position. Count backwards from the tail node. The tail is at postion 0, its parent is at 1 and so on. Example head refers to 3 -> 2 -> 1 -> 0 -> NULL positionFromTail = 2 Each of the data values matches its distance from the t

View Solution →

Delete duplicate-value nodes from a sorted linked list

This challenge is part of a tutorial track by MyCodeSchool You are given the pointer to the head node of a sorted linked list, where the data in the nodes is in ascending order. Delete nodes and return a sorted list with each distinct value in the original list. The given head pointer may be null indicating that the list is empty. Example head refers to the first node in the list 1 -> 2 -

View Solution →