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.Arrays;
import java.util.List;
import java.util.Scanner;
import java.util.function.BiFunction;
import java.util.stream.Collectors;
import java.util.stream.IntStream;

public class Main {
    public static void main(final String[] args) {
        final Scanner scanner = new Scanner(System.in);
        final int n = scanner.nextInt();
        final String s = scanner.next();
        final List<Long> list = IntStream.range(0, n)
                .mapToObj(i -> 1L << (s.charAt(i) - 'a'))
                .collect(Collectors.toList());

        final SegmentTree tree = new SegmentTree(list, 0, (a, b) -> a | b);
        final int q = scanner.nextInt();
        for (int _ = 0; _ < q; _++) {
            if (scanner.nextInt() == 1) {
                final int i = scanner.nextInt();
                final int c = scanner.next().toCharArray()[0] - 'a';
                tree.update(i - 1, 1 << c);
            } else {
                final int l = scanner.nextInt();
                final int r = scanner.nextInt();
                final long result = tree.query(l - 1, r - 1);
                final String tmp = Long.toBinaryString(result);
                final long answer = IntStream.range(0, tmp.length())
                        .map(tmp::charAt)
                        .filter(c -> c == '1')
                        .count();

                System.out.println(answer);
            }
        }
    }

    private static class SegmentTree {
        private final long[] internalTree;
        private final int exponent;
        private BiFunction<Long, Long, Long> updater;

        SegmentTree(final List<Long> list, final long initialValue, final BiFunction<Long, Long, Long> updater) {
            this.exponent = calcExponent(list.size());
            this.updater = updater;
            internalTree = initTree(list, initialValue);
        }

        private long[] initTree(final List<Long> list, final long initialValue) {
            final long[] array = new long[exponent * 2];
            Arrays.fill(array, initialValue);
            for (int i = 0; i < list.size(); i++) {
                array[i + exponent] = list.get(i);
            }

            for (int i = exponent - 1; i > 0; i--) {
                array[i] = updater.apply(array[i * 2], array[i * 2 + 1]);
            }

            return array;
        }

        void update(final int index, final long value) {
            internalTree[index + exponent] = value;
            int current = index / 2;
            while (current > 0) {
                internalTree[current] = updater.apply(internalTree[current * 2], internalTree[current * 2 + 1]);
                current /= 2;
            }
        }

        long query(final int left, final int right) {
            return query(left, right, 0, exponent - 1);
        }

        long query(final int left, final int right, final int begin, final int end) {
            if (left == begin && right == end) {
                return internalTree[(exponent + left) / (right - left + 1)];
            }

            final int mid = (begin + end) / 2;
            if (left <= mid && right <= mid) {
                return query(left, right, begin, mid);
            } else if (left <= mid) {
                return updater.apply(query(left, mid, begin, mid), query(mid + 1, right, mid + 1, end));
            } else {
                return query(left, right, mid + 1, end);
            }
        }

        private int calcExponent(final int n) {
            int exp = 1;
            while (exp < n) {
                exp <<= 1;
            }
            return exp;
        }
    }
}