Implementation
Problem_2_COW_Splits.cpp
C++
#include <bits/stdc++.h>
using namespace std;
void solve() {
int n; string s; cin >> n >> s;
//notice that if it is possilbe we can always do it in 1 or 2 steps
if (n & 1) {
cout << -1 << '\n';
return;
}
// partition in half
if (s.substr(0, 3 * n / 2) == s.substr(3 * n / 2)) {
cout << 1 << '\n';
for (int i = 0; i < 3 * n; i++) {
cout << 1 << " \n"[i == 3 * n - 1];
}
return;
}
vector<int> ans(3 * n, 2);
for (int i = 0; i < n / 2; i++) {
int l = 3 * i, r = 3 * n / 2 + 3 * i;
string a = s.substr(l, 3), b = s.substr(r, 3);
if (a == b) {
for (int j = 0; j < 3; j++) {
ans[l + j] = 1;
ans[r + j] = 1;
}
} else {
bool done = false;
for (int j = 0; j < 3 && !done; j++) {
for (int k = j + 1; k < 3 && !done; k++) {
string t = "";
t += a[j]; t += a[k];
int p = 0;
for (char c : b) {
if (p < 2 && c == t[p]) p++;
}
if (p == 2) {
ans[l + j] = ans[l + k] = 1;
for (int x = 0; x < 3; x++) {
if (b[x] == t[0] || b[x] == t[1]) {
ans[r + x] = 1;
}
}
done = 1;
}
}
}
}
}
cout << 2 << '\n';
for (int i = 0; i < ans.size(); i++) {
cout << ans[i] << " \n"[i == ans.size() - 1];
}
}
int main() {
ios::sync_with_stdio(0); cin.tie(0);
int t, k; cin >> t >> k;
while (t--) {
solve();
}
}