Question
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
Explanation
Hii Guys Solution of the problem and its explanation is here !!
So Lets begin
Take Input "n and t" where n is number of school childrens and t is time in seconds.
Take a string s or "char arr[n]" of size
Now you have to take the for loop of t times because you are going to swap the Boys and Girls if Boys is standing in front of Girl.
and take another for loop inside the previous for loop for size n-1 and start it from 0 and lets consider variale " j "
and put condition inside this for loop only that
if(arr[j]=='B' and arr[j+1]=='G'){
swap(arr[j+1],arr[j]);
++j;
}
after which the loop will look like this
for(int i=0;i<t;i++){
for(int j=0;j<n-1;j++){
if(arr[j]=='B' and arr[j+1]=='G'){
swap(arr[j+1],arr[j]); ++j;
}
}
}
Then after this it will swap and give you desired output of the given array and simple print the array you will get the desired and required output
Code
#include<bits/stdc++.h>
using namespace std;
#define test long long T;cin>>T;while(T--)
void solve(){
int n,t; cin>>n>>t;
char arr[n];
for(int i=0;i<n;i++) cin>>arr[i];
for(int i=0;i<t;i++){
for(int j=0;j<n-1;j++){
if(arr[j]=='B' and arr[j+1]=='G'){
swap(arr[j+1],arr[j]); ++j;
}
}
}
for(int i=0;i<n;i++) cout<<arr[i]; cout<<endl;
}
signed main() {
//test (if you want to take the more test cases you may uncomment it out)
solve();
}
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