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)
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