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 Ruby (2.3.3) (judge verdict: Accepted)

#exec({'RUBY_THREAD_VM_STACK_SIZE'=>'100000000'},'/usr/bin/ruby', $0) if !ENV['RUBY_THREAD_VM_STACK_SIZE']

require 'prime'

def inpf() a=gets.chomp.split(" ").map(&:to_f)end

def inps() a=gets.chomp.split(" ")end

def copy(a) Marshal.load(Marshal.dump(a)) end

def kaijo(n,r = 10**9+7)(n < 2)? 1 : (2..n).inject{|memo,u|memo=(memo*u)%r} end

def na(n,d=0) Array.new(n,d)end

def na2(n,m,d=0) Array.new(n){Array.new(m,d)}end

def na3(n,m,l,d=0) Array.new(n){Array.new(m){Array.new(l,d)}}end

def inp() a=gets.chomp.split(" ").map(&:to_i)end

def r_up(a, b) (a+b-1)/b end

def sum(a) a.inject(:+) end

def big(a,b) return (a>b)? a:b end

def small(a,b) return (a<b)? a:b end

n = inp[0]
s = gets.chomp.split("").map &:to_i
at = na3(10,10,10,false)
dp = na(10,false)
dp2 = na(100,false)

s.reverse.each do |d|

  (0..99).each do |dd|
    if(dp2[dd] == true)
      at[d*100+dd] = true
    end
  end
  (0..9).each do |dd|
    if(dp[dd] == true)
      dp2[d*10 + dd] = true
    end
  end
  dp[d] = true
end

p at.flatten.count(true)