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>
using namespace std;

//#define int long long
typedef long long ll;

typedef unsigned long long ul;
typedef unsigned int ui;
const ll mod = 1000000007;
const ll INF = mod * mod;
const int INF_N = 1e+9;
typedef pair<int, int> P;
#define stop char nyaa;cin>>nyaa;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
#define all(v) (v).begin(),(v).end()
typedef pair<ll, ll> LP;
typedef long double ld;
typedef pair<ld, ld> LDP;
const ld eps = 1e-12;
const ld pi = acos(-1.0);
//typedef vector<vector<ll>> mat;
typedef vector<int> vec;

//繰り返し二乗法
ll mod_pow(ll a, ll n, ll m) {
	ll res = 1;
	while (n) {
		if (n & 1)res = res * a%m;
		a = a * a%m; n >>= 1;
	}
	return res;
}

struct modint {
	ll n;
	modint() :n(0) { ; }
	modint(ll m) :n(m) {
		if (n >= mod)n %= mod;
		else if (n < 0)n = (n%mod + mod) % mod;
	}
	operator int() { return n; }
};
bool operator==(modint a, modint b) { return a.n == b.n; }
modint operator+=(modint &a, modint b) { a.n += b.n; if (a.n >= mod)a.n -= mod; return a; }
modint operator-=(modint &a, modint b) { a.n -= b.n; if (a.n < 0)a.n += mod; return a; }
modint operator*=(modint &a, modint b) { a.n = ((ll)a.n*b.n) % mod; return a; }
modint operator+(modint a, modint b) { return a += b; }
modint operator-(modint a, modint b) { return a -= b; }
modint operator*(modint a, modint b) { return a *= b; }
modint operator^(modint a, int n) {
	if (n == 0)return modint(1);
	modint res = (a*a) ^ (n / 2);
	if (n % 2)res = res * a;
	return res;
}

//逆元(Eucledean algorithm)
ll inv(ll a, ll p) {
	return (a == 1 ? 1 : (1 - p * inv(p%a, a)) / a + p);
}
modint operator/(modint a, modint b) { return a * modint(inv(b, mod)); }

const int max_n = 1 << 18;
modint fact[max_n], factinv[max_n];
void init_f() {
	fact[0] = modint(1);
	for (int i = 0; i < max_n - 1; i++) {
		fact[i + 1] = fact[i] * modint(i + 1);
	}
	factinv[max_n - 1] = modint(1) / fact[max_n - 1];
	for (int i = max_n - 2; i >= 0; i--) {
		factinv[i] = factinv[i + 1] * modint(i + 1);
	}
}
modint comb(int a, int b) {
	if (a < 0 || b < 0 || a < b)return 0;
	return fact[a] * factinv[b] * factinv[a - b];
}
using mP = pair<modint, modint>;

int dx[4] = { 0,1,0,-1 };
int dy[4] = { 1,0,-1,0 };




void solve() {
    int N; cin >> N;
    string s; cin >> s;
    vec ec(N+1, 0), wc(N+1, 0);
    rep(i, N){
        if(s[i] == 'E'){
            ec[i+1] = ec[i] + 1;
            wc[i+1] = wc[i];
        }else{
            ec[i+1] = ec[i];
            wc[i+1] = wc[i] + 1;
        }
    }
    ll res = INF;
    rep(i, N){
        res = min(res, (ll)wc[i] + (ec[N]-ec[i+1]));
    }

    cout << res << endl;
}

signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  //cout << fixed << setprecision(10);
  //init_f();
  //init();
  //int t; cin >> t; rep(i, t)solve();
  solve();
//   stop
    return 0;
}