Problem:
Score: 400 points

Problem Statement

AtCoder Inc. has decided to lock the door of its office with a 3-digit PIN code.

The company has an N-digit lucky number, S. Takahashi, the president, will erase N-3 digits from S and concatenate the remaining 3 digits without changing the order to set the PIN code.

How many different PIN codes can he set this way?

Both the lucky number and the PIN code may begin with a 0.

Constraints

4 \leq N \leq 30000

S is a string of length N consisting of digits.

Input

Input is given from Standard Input in the following format:

N
S

Output

Print the number of different PIN codes Takahashi can set.

Sample Input 1

4
0224

Sample Output 1

3

Takahashi has the following options:

Erase the first digit of S and set 224.

Erase the second digit of S and set 024.

Erase the third digit of S and set 024.

Erase the fourth digit of S and set 022.

Thus, he can set three different PIN codes: 022, 024, and 224.

Sample Input 2

6
123123

Sample Output 2

17

Sample Input 3

19
3141592653589793238

Sample Output 3

329

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <vector>
#include <iostream>
#include <cmath>
#include <map>
#include <unordered_map>
#include <algorithm>
#include <fstream>
#include <unistd.h>
#include <string.h>
#include <string>
#include <numeric>
#include <queue>
#include <deque>
#include <sstream>
#include <iomanip>
#include <set>
#include <stack>
#include <cassert>
#include <functional>
#include <random>
using namespace std;
typedef int64_t ll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<vvl> vvvl;
typedef vector<vvvl> vvvvl;
typedef vector<vi> vvi;
typedef vector<vvi> vvvi;
typedef vector<bool> vb;
typedef vector<vector<bool>> vvb;
typedef vector<vvb> vvvb;
typedef vector<vvvb> vvvvb;
typedef vector<string> vs;
typedef vector<vs> vvs;
typedef vector<vvs> vvvs;
typedef vector<char> vc;
typedef vector<vector<char>> vvc;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef vector<vvd> vvvd;
typedef pair<ll, ll> PL;
typedef vector<PL> vpl;
typedef vector<vector<PL>> vvpl;
typedef vector<vvpl> vvvpl;
typedef pair<ll, PL> TL;
typedef vector<TL> vtl;
typedef vector<vtl> vvtl;
typedef long double ld;
typedef vector<ld> vld;
const int INF = 1001001001;
const ll LINF = 1e18;
const double pi = 3.1415926535897932;
const string endstr = "\n";
#define FOR(i, a, b) for(ll i = (a); i < b; i++)
#define RFOR(i, a, b) for(ll i = (a); i > b; i--)
#define rep(i, n) for(ll i = 0; i < n; i++)
#define rrep(i, n) for(ll i = n-1; i > -1; i--)
#define FORMAP(it, m) for(auto it = m.begin(); it != m.end(); it++)
#define ff first
#define ss second
#define pb push_back
#define epb emplace_back
#define ALL(X) (X).begin(),(X).end()

template <typename T>
T gcd(T a, T b) {
    return (a == 0) ? b : gcd(b%a, a);
}
template <typename T>
T lcm(T a, T b) {
    return a / gcd(a, b) * b;
}

bool p_comp_fs(const PL p1, const PL p2){ return p1.first < p2.first;};
bool p_comp_fg(const PL p1, const PL p2){ return p1.first > p2.first;};
bool p_comp_ss(const PL p1, const PL p2){ return p1.second < p2.second;};
bool p_comp_sg(const PL p1, const PL p2){ return p1.second > p2.second;};
template <typename T>
vector<T> uniquen(vector<T> vec /* copy */){
    sort(ALL(vec));
    vec.erase(unique(vec.begin(), vec.end()), vec.end());
    return vec;
}

inline ll popcnt(ll x){return __builtin_popcountll((unsigned long long)x);};
template<class T> bool chmax(T &a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> bool chmin(T &a, T b) { if (a > b) { a = b; return true; } return false; }
//friend bool operator<(const tpl& lhs, const tpl& rhs){
//    return std::tie(lhs.l, lhs.r) < std::tie(rhs.l, rhs.r);
//}



int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    ll N; cin >> N; string S; cin >> S;
    vvl idx(10);
    rep(i, N){
        ll d = S[i]-'0';
        idx[d].pb(i);
    }
    
    ll ans = 0;
    rep(l, 10) rep(m, 10) rep(r, 10){
        if(idx[l].size() == 0 || idx[m].size() == 0 || idx[r].size() == 0){
            continue;
        }
        ll sti = idx[l].front();
        ll tei = idx[r].back();
        bool ok = false;
        for(auto idxm : idx[m]){
            if(sti < idxm && idxm < tei) ok = true;
        }
        
        if(ok) ans++;
    }
    cout << ans << endstr;
    return 0;
}