Problem:
Score : 200 points

Problem Statement

There are N people numbered 1 to N. Each person wears a red hat or a blue hat.

You are given a string s representing the colors of the people. Person i wears a red hat if s_i is R, and a blue hat if s_i is B.

Determine if there are more people wearing a red hat than people wearing a blue hat.

Constraints

1 \leq N \leq 100

|s| = N

s_i is R or B.

Input

Input is given from Standard Input in the following format:

N
s

Output

If there are more people wearing a red hat than there are people wearing a blue hat, print Yes; otherwise, print No.

Sample Input 1

4
RRBR

Sample Output 1

Yes

There are three people wearing a red hat, and one person wearing a blue hat.

Since there are more people wearing a red hat than people wearing a blue hat, the answer is Yes.

Sample Input 2

4
BRBR

Sample Output 2

No

There are two people wearing a red hat, and two people wearing a blue hat.

Since there are as many people wearing a red hat as people wearing a blue hat, the answer is No.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <string>
#include <cstdlib>
#include <vector>
#include <array>
#include <map>
#include <algorithm>
#include <queue>
#include <stack>
#include <cmath>

using namespace std;
typedef long long int ll;

int main(int argc, char const *argv[])
{
	char s;
	int n;
	cin>>n;
	int r=0,b=0;
	for(int i = 0; i<n;i++)
	{
		cin>>s;
		if(s=='R')r++;
		else b++;
	}
	
	if(r <= b)
	{
		cout<< "No" <<endl;
	}
	else
	{
		cout << "Yes" << endl;
	}
}