This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include <bits/stdc++.h>
#ifdef DEBUG
#include "debug.hpp"
#else
#pragma GCC optimize("Ofast")
#define trace(...)
#include <bits/stdc++.h>
#define endl '\n'
#endif
using namespace std;
const string COL = "RGY";
int count_inversions(vector<int>& v, int l, int r) {
if (r - l == 1) return 0;
int mid = (l + r) / 2;
int ans = 0;
ans += count_inversions(v, l, mid);
ans += count_inversions(v, mid, r);
vector<int> m;
m.reserve(r - l);
int ptr1 = l, ptr2 = mid;
while (ptr1 < mid || ptr2 < r) {
if (ptr1 == mid) {
m.push_back(v[ptr2++]);
} else if (ptr2 == r) {
m.push_back(v[ptr1++]);
ans += ptr2 - mid;
} else if (v[ptr1] < v[ptr2]) {
m.push_back(v[ptr1++]);
ans += ptr2 - mid;
} else {
m.push_back(v[ptr2++]);
}
}
copy(m.begin(), m.end(), v.begin() + l);
return ans;
}
int count_swaps(string s, string const& t) {
int n = s.length();
vector<stack<int>> st(3);
for (int i = n - 1; i >= 0; --i) {
st[COL.find(t[i])].push(i);
}
vector<int> v(n);
for (int i = 0; i < n; ++i) {
int c = COL.find(s[i]);
if (st[c].empty()) return INT_MAX;
v[i] = st[c].top();
st[c].pop();
}
return count_inversions(v, 0, n);
}
int subtask_1(int n, string s) {
int ans = INT_MAX;
for (int i = 0; i < 3; ++i) {
for (int j = 0; j < 1 << (n - 1); ++j) {
string t;
int prev_c = i;
t.push_back(COL[prev_c]);
for (int k = 0; k < n - 1; ++k) {
int bit = (j >> k) & 1;
prev_c = (prev_c + bit + 1) % 3;
t.push_back(COL[prev_c]);
}
ans = min(count_swaps(s, t), ans);
}
}
if (ans == INT_MAX) ans = -1;
return ans;
}
int subtask_3(int n, string s) {
string t1, t2;
for (int i = 0; i < n; ++i) {
t1.push_back("RG"[i % 2]);
t2.push_back("GR"[i % 2]);
}
int ans = min(count_swaps(s, t1), count_swaps(s, t2));
if (ans == INT_MAX) ans = -1;
return ans;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(NULL);
int n;
string s;
cin >> n >> s;
int ans;
if (n <= 15) {
ans = subtask_1(n, s);
} else {
ans = subtask_3(n, s);
}
cout << ans << endl;
return 0;
}
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