1191B - Tokitsukaze and Mahjong - CodeForces Solution


brute force implementation *1200

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Python Code:

lst = input().split()
dicts = {}
max = 0
for i in lst:
    if i in dicts:
        dicts[i] += 1
    else:
        dicts[i] = 1
    if dicts[i] > max:
        max = dicts[i]
if max == 3:
    print(0)
elif max == 2:
    print(1)
else:
    max = 0
    for i in range(2):
        for j in range(i + 1, 3):
            if lst[i][1] == lst[j][1] and abs(ord(lst[i][0]) - ord(lst[j][0])) < 3:
                max += 1
    if max == 0:
        print(2)
    elif max > 0 and max < 3:
        print(1)
    else:
        print(0)

C++ Code:

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int ui;
typedef long double ld;
typedef pair<int, int> ii;
typedef pair<int, ii> iii;
typedef pair<ll, ll> pll;
typedef pair<ll, ii> pli;
typedef vector<int> vi;
typedef vector<ll> vll;
#define fi first
#define se second

const int INF = 1e9;
const ll LINF = 1e18;
const ll MOD = 1e9 + 7;
const double EPS = 1e-9;
const int N = 1e4;

int f[127][10];

void solve() {
    string a, b, c;
    
    cin >> a >> b >> c;
    f[a[1]][a[0]-'0']++;
    f[b[1]][b[0]-'0']++;
    f[c[1]][c[0]-'0']++;
    int mx = 0;
    int p[3] = {'m', 'p', 's'};
    for(int i = 0; i < 3; i++) {
        for(int j = 1; j <= 9; j++) {
            mx = max(mx, f[p[i]][j]);
        }
    }
    for(int i = 0; i < 3; i++) {
        for(int j = 3; j <= 9; j++) {
            mx = max(mx, (f[p[i]][j]!= 0)+(f[p[i]][j-1] != 0)+(f[p[i]][j-2] != 0));
        }
    }
    cout << 3 - mx << "\n";
}

int main() {
    ios::sync_with_stdio(false); cin.tie(nullptr);
    int T = 1;
    
    //cin >> T;
    for(int i = 1; i <= T; i++) {
        //cout << "Case #" << i << ":\n";
        solve();
    }
    return 0;
}


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