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 :
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))
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