Problem:
Score : 300 points

Problem Statement

There are N people standing in a row from west to east.
Each person is facing east or west.
The directions of the people is given as a string S of length N.
The i-th person from the west is facing east if S_i = E, and west if S_i = W.

You will appoint one of the N people as the leader, then command the rest of them to face in the direction of the leader.
Here, we do not care which direction the leader is facing.

The people in the row hate to change their directions, so you would like to select the leader so that the number of people who have to change their directions is minimized.
Find the minimum number of people who have to change their directions.

Constraints

2 \leq N \leq 3 \times 10^5

|S| = N

S_i is E or W.

Input

Input is given from Standard Input in the following format:

N
S

Output

Print the minimum number of people who have to change their directions.

Sample Input 1

5
WEEWW

Sample Output 1

1

Assume that we appoint the third person from the west as the leader.
Then, the first person from the west needs to face east and has to turn around.
The other people do not need to change their directions, so the number of people who have to change their directions is 1 in this case.
It is not possible to have 0 people who have to change their directions, so the answer is 1.

Sample Input 2

12
WEWEWEEEWWWE

Sample Output 2

4

Sample Input 3

8
WWWWWEEE

Sample Output 3

3

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

#include <bits/stdc++.h>
#include <cstdlib>  // abs() for integer 絶対値求めやつ
#include <cmath>    // abs() for float, and fabs()
#include <algorithm>
#include <vector>
#include <string>
#include<iomanip>//浮動小数点数を出力する制度の指定
//使い方　数値をnumとして　cout<<setprecision(n)<<num ;nは桁数
#define rep(i,n) for(int i = 0; i < (n);i++)
#define SORT(a) sort((a).begin(),(a).end())
#define REV(a) reverse((a).begin(),(a).end())
using namespace std;
using ll =long long;
#define INF 2000000000
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
#define che(a,string) cout<<"//"<<string <<"=="<<(a)<<"//" <<endl;
bool IsInt(double a){int b = a/1;if(a==b){return true;}else {return false;}}
/*覚えてないことメa
数値型から文字列に変換　to_string(number);これは簡単
文字列から数値型に変換　stoi(number)
文字列からllに変換　stoll(number)*/

//ここから書き始める
int main(){
	int n;cin>>n;
	string s; cin >> s;
	vector<int> A(n+1);
	A[0]=0;
	ll count = 0;
	for(int i = 1 ; i < n+1  ; i++){
		if(s.at(i-1) == 'W' ){
			count++;
		}
		A[i] = count;
	}
	ll ans = n;
	//i番目まででWを向いてる人の数とi+1番目からEを向いてる人の数を足し合わせたものを求める
	int bef,aft;
	for(int i = 0 ; i < n ; i ++ ){
		bef = A[i];
		aft = ((n-1)-i)-(A[n]-A[i+1]);
		ll k = bef + aft;
		chmin(ans,k);
	}
	cout<<ans<<endl;
}