Codeforces 266B Queue at the School Solution & Explanation

Difficulty : 800

Problem Description

During the break the schoolchildren, boys and girls, formed a queue of n people in the canteen. Initially the children stood in the order they entered the canteen. However, after a while the boys started feeling awkward for standing in front of the girls in the queue and they started letting the girls move forward each second.

Let’s describe the process more precisely. Let’s say that the positions in the queue are sequentially numbered by integers from 1 to n, at that the person in the position number 1 is served first. Then, if at time x a boy stands on the i-th position and a girl stands on the (i + 1)-th position, then at time x + 1 the i-th position will have a girl and the (i + 1)-th position will have a boy. The time is given in seconds.

You’ve got the initial position of the children, at the initial moment of time. Determine the way the queue is going to look after t seconds.

Input

The first line contains two integers n and t (1 ≤ n, t ≤ 50), which represent the number of children in the queue and the time after which the queue will transform into the arrangement you need to find.

The next line contains string s, which represents the schoolchildren’s initial arrangement. If the i-th position in the queue contains a boy, then the i-th character of string s equals “B”, otherwise the i-th character equals “G”.

Output

Print string a, which describes the arrangement after t seconds. If the i-th position has a boy after the needed time, then the i-th character a must equal “B”, otherwise it must equal “G”.

Examples

Solution

We can use a for loop to simulate the passage of time and a while loop to check each student in the queue.

C# Solution

Solution1

string[] arr = Console.ReadLine().Split(" ");
int n = int.Parse(arr[0]);
int t = int.Parse(arr[1]);

char[] s = Console.ReadLine().ToCharArray();

for (int time = 0; time < t; time++)
{
    int i = 0;
    while (i < n - 1)
    {
        if (s[i] == 'B' && s[i + 1] == 'G')
        {
            char tmp = s[i];
            s[i] = s[i + 1];
            s[i + 1] = tmp;

            i += 2;
        }
        else
        {
            i += 1;
        }
    }
}

Console.WriteLine(new string(s));

If there is a boy positioned before a girl, they will exchange their positions. To handle this, we use ‘i += 2‘ to skip to the next pair.

Conclusion

To be honest, I believe the difficulty of this problem is definitely more than 800…

This problem came from codeforces contest – Dashboard – Codeforces Round 163 (Div. 2) – Codeforces

Here to see other problem of this contest:

• Codeforces 266A Stones on the Table

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The problem link : Problem – 266B – Codeforces

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