JOIOJI

  1/**
  2 *    author: Akvicor
  3 *    created: 2019-07-17 10-09-13
  4**/
  5 
  6#include <bits/stdc++.h>
  7 
  8using namespace std;
  9 
 10#ifdef DEBUG
 11string to_string(string s) {
 12    return '"' + s + '"';
 13}
 14 
 15string to_string(const char* s) {
 16    return to_string((string) s);
 17}
 18 
 19string to_string(bool b) {
 20return (b ? "true" : "false");
 21}
 22 
 23template <typename A, typename B>
 24string to_string(pair<A, B> p) {
 25    return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
 26}
 27 
 28template <typename A>
 29string to_string(A v) {
 30    bool first = true;
 31    string res = "{";
 32    for (const auto &x : v) {
 33        if (!first) {
 34            res += ", ";
 35        }
 36        first = false;
 37        res += to_string(x);
 38    }
 39    res += "}";
 40    return res;
 41}
 42 
 43void debug_out() { cerr << endl; }
 44 
 45template <typename Head, typename... Tail>
 46void debug_out(Head H, Tail... T) {
 47    cerr << " " << to_string(H);
 48    debug_out(T...);
 49}
 50#endif
 51 
 52#ifdef DEBUG
 53#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
 54#else
 55#define debug(...) 17
 56#endif
 57 
 58#ifdef DEBUG
 59#define FAST_IO 17
 60#else
 61#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
 62#define endl '\n'
 63#endif
 64 
 65#define LL long long
 66#define ULL unsigned long long
 67#define rep(i, n) for(int i = 0; i < (n); ++i)
 68#define reep(i, n) for(int i = 0; i <= (n); ++i)
 69#define lop(i, a, n) for(int i = a; i < (n); ++i)
 70#define loop(i, a, n) for(int i = a; i <= (n); ++i)
 71#define ALL(v) (v).begin(), (v).end()
 72#define PB push_back
 73#define VI vector<int>
 74#define PII pair<int,int>
 75#define FI first
 76#define SE second
 77#define SZ(x) ((int)(x).size())
 78 
 79const double EPS = 1e-6;
 80const double PI = acos(-1.0);
 81const int INF = 0x3f3f3f3f;
 82const LL LINF = 0x7f7f7f7f7f7f7f7f;
 83const int MAXN = (int)1e6 + 10;
 84const int MOD = (int)1e9 + 7;
 85 
 86string s;
 87map<pair<int, int>, int> m;
 88int ans, J, O, I, n;
 89 
 90int main(){
 91    FAST_IO;
 92 
 93    ans = J = O = I = n = 0;
 94 
 95    cin >> n >> s;
 96 
 97    m[make_pair(0, 0)] = -1;
 98 
 99    rep(i, n){
100        J += s[i]=='J';
101        O += s[i]=='O';
102        I += s[i]=='I';
103        // 如果出现了两次相同的 (J-O, I-O) 
104        // 那么就说明这两次之间的 J、O、I 的数量一定相等，符合题目要求，只需用 当前i-上次的i 即可
105        if(m.find(make_pair(J-O, I-O))==m.end()) m[make_pair(J-O, I-O)] = i;
106        else ans = max(ans, i-m[make_pair(J-O, I-O)]);
107    }
108 
109    cout << ans << endl;
110 
111    return 0;
112}