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.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int[][] matrix = new int[n + 1][26];
		//a =97
		String s = sc.next();
		char[] cAry = s.toCharArray();
		int count = 1;
		for(char c : s.toCharArray()){
			for(int i = count; i <= n; i++){
				matrix[i][c - 97]++;
			}
			count++;
		}
		int q = sc.nextInt();
		for(int i = 0; i < q; i++){
			if(sc.nextInt() == 1){
				int p = sc.nextInt();
				char c = sc.next().toCharArray()[0];
				char cBefore = cAry[p - 1];
				if(c != cBefore){
					for(int k = p; k <= n; k++){
						matrix[k][c - 97]++;
						matrix[k][cBefore - 97]--;
					}
				}
			}else{
				int p1 = sc.nextInt();
				int p2 = sc.nextInt();
				int count2 = 0;
				for(int k = 0; k < 26; k++){
					if(matrix[p2][k] - matrix[p1 - 1][k] > 0){
						count2++;
					}
				}
				System.out.println(count2);
			}
		}
	}
}