Codeforces.com is really a nice site, if want to try yourself vs. the top programming students in the world. Just to check your level in a way š

Thus, today I have decided to take a look at the problems of the second division of the competition, here.

The problem initially seems rather complicated, but after careful reading, you need to understand that it is not that difficult. Here is what it says:

*In Berland each high school student is characterized by academic performance ā integer value between 1 and 5.*

*In high school 0xFF there are two groups of pupils: the group A and the group B. Each group consists of exactly nstudents. An academic performance of each student is known ā integer value between 1 and 5.*

*The school director wants to redistribute students between groups so that each of the two groups has the same number of students whose academic performance is equal to 1, the same number of students whose academic performance is 2 and so on. In other words, the purpose of the school director is to change the composition of groups, so that for each value of academic performance the numbers of students in both groups are equal.*

*To achieve this, there is a plan to produce a series of exchanges of students between groups. During the single exchange the director selects one student from the class A and one student of class B. After that, they both change their groups.*

*Print the least number of exchanges, in order to achieve the desired equal numbers of students for each academic performance.*

*Input*

*The first line of the input contains integer number n (1āā¤ānāā¤ā100) ā number of students in both groups.*

*The second line contains sequence of integer numbers a _{1},āa_{2},ā…,āa_{n} (1āā¤āa_{i}āā¤ā5), where a_{i} is academic performance of the i-th student of the group A.*

*The third line contains sequence of integer numbers b _{1},āb_{2},ā…,āb_{n} (1āā¤āb_{i}āā¤ā5), where b_{i} is academic performance of the i-th student of the group B.*

*Output*

*Print the required minimum number of exchanges or -1, if the desired distribution of students can not be obtained.*

What we should do is the following:

- Read the input into
*nĀ*and two Lists –*a*and*b*. - Then make a dictionary with all possible grades – I use a cheap trick, adding a grade 0, because the lists start with iteration from 0.
- Then the master trick – iterate through the grades from the first classes and in the dictionary increase the position with 1 for each grade present in the first class and decrease with 1 for each grade present in the second class. Thus, you should not make any changes in these classes.
- Iterate through the grades and check for an odd number. If present, then the result is -1 directly.
- Use absolute values, of the counts of the grades in the dictionary. The absolute value is a must.
- The counts of the grades is what is left as a result. It should be divided by 4, because we have two classes and we switch two students there.
- That’s it! š

Enjoy the code here:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 |
using System; using System.Collections.Generic; using System.Linq; using System.Text; using System.Threading.Tasks; class ChangeOfStudents { static void Main() { int n = int.Parse(Console.ReadLine()); List<int> a = Console.ReadLine().Split().Select(int.Parse).ToList(); List<int> b = Console.ReadLine().Split().Select(int.Parse).ToList(); bool bNoChance = false; int sumTotal = 0; Dictionary<int, int> dTotal = new Dictionary<int, int> { {0,0}, {1,0}, {2,0}, {3,0}, {4,0}, {5,0}, {6,0}, }; for (int i = 0; i < n; i++) { dTotal[a[i]]++; dTotal[b[i]]--; } for(int i = 1; i <= 6; i++) { if (dTotal[i] %2 != 0) { bNoChance = true; } sumTotal += Math.Abs((dTotal[i])); } if (bNoChance) { Console.WriteLine(-1); } else { Console.WriteLine(sumTotal/4); } } } |

Have a nice day! š