Problem:
Score : 500 points

Problem Statement

You are given a string S of length N consisting of lowercase English letters.

Process Q queries of the following two types:

Type 1: change the i_q-th character of S to c_q. (Do nothing if the i_q-th character is already c_q.)

Type 2: answer the number of different characters occurring in the substring of S between the l_q-th and r_q-th characters (inclusive).

Constraints

N, Q, i_q, l_q, and r_q are integers.

S is a string consisting of lowercase English letters.

c_q is a lowercase English letter.

1 \leq N \leq 500000

1 \leq Q \leq 20000

|S| = N

1 \leq i_q \leq N

1 \leq l_q \leq r_q \leq N

There is at least one query of type 2 in each testcase.

Input

Input is given from Standard Input in the following format:

N
S
Q
Query_1
\vdots
Query_Q

Here, Query_i in the 4-th through (Q+3)-th lines is one of the following:

1 i_q c_q

2 l_q r_q

Output

For each query of type 2, print a line containing the answer.

Sample Input 1

7
abcdbbd
6
2 3 6
1 5 z
2 1 1
1 4 a
1 7 d
2 1 7

Sample Output 1

3
1
5

In the first query, cdbb contains three kinds of letters: b , c , and d, so we print 3.

In the second query, S is modified to abcdzbd.

In the third query, a contains one kind of letter: a, so we print 1.

In the fourth query, S is modified to abcazbd.

In the fifth query, S does not change and is still abcazbd.

In the sixth query, abcazbd contains five kinds of letters: a, b, c, d, and z, so we print 5.

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)



import java.util.*;

class Order{
    int a,b,c;
}

class Main {

    static  int MM=100;
    public static void main(String args[]) {
        Scanner s = new Scanner(System.in);

        int n=s.nextInt();
        char[] ss=s.next().toCharArray();
        int q=s.nextInt();

        List<Order> querys=new ArrayList<>();

        HashMap<Integer,Set<Integer>> memory=new HashMap<>();

        for (int i = 0; i <q ; i++) {
            Order o=new Order();
            o.a=s.nextInt();
            o.b=s.nextInt();
            if(o.a==1){
                o.c=s.next().charAt(0);
            }else{
                o.c=s.nextInt();
            }
            querys.add(o);
        }
        // init moem

        for (int i = 0; i <ss.length ; i++) {
            int index=i/MM;
            Set ssss= memory.getOrDefault(index,new HashSet<>());
            ssss.add(ss[i]);
            memory.put(index,ssss);
        }


        for (int i = 0; i <q ; i++) {
            Order o=querys.get(i);
            if(o.a==1){
                int old=ss[o.b-1];
                ss[o.b-1]= (char) o.c;
                int index=o.b/MM;
                //System.out.println(ss);
                Set ssss= memory.getOrDefault(index,new HashSet<>());
                ssss.remove(old);
                ssss.add(o.c);
                memory.put(index,ssss);
            }else{
                int ret=0;
                boolean isMem=(o.b/MM)!=(o.c/MM);
                int start1=o.b;
                int end1=(o.b/MM+1)*MM-1;
                int start2=o.c/MM*MM;
                int end2=o.c;
                Set<Integer>  to =new HashSet<>();


                if(!isMem){
                    end1=end2;
                }

                count(start1,end1,ss,to);
                if (isMem){
                    count(start2,end2,ss,to);
                    for (int j = start1/MM+1; j <end2/MM ; j++) {
                        ret+=memory.get(j).size();
                    }
                }
                System.out.println(ret+to.size());
            }
        }
    }
    static void count(int s, int e, char[] ss, Set to){
        for (int j = s; j <=e ; j++) {
            to.add((int) ss[j-1]);
        }
    }
}