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.



Sample input:
7
abcdbbd
6
2 3 6
1 5 z
2 1 1
1 4 a
1 7 d
2 1 7


Sample output:
3
1
5

Submission status: Wrong Answer

Solution code:
def main(sample_file = ''):

    """ convenient functions
    # for i, a in enumerate(iterable)
    # q, mod = divmod(a, b)
    # divmod(x, y) returns the tuple (x//y, x%y)
    # Higher-order function: reduce(operator.mul, xyz_count, 1)
    # manage median(s) using two heapq https://atcoder.jp/contests/abc127/tasks/abc127_f
    """

    """convenient decorator
    # @functools.lru_cache():
    # to facilitate use of recursive function
        # ex:
        # from functools import lru_cache
        # import sys
        # sys.setrecursionlimit(10**9)
        # @lru_cache(maxsize=None)
        # def fib(n):
        #     if n < 2:
        #         return n
        #     return fib(n-1) + fib(n-2)
        # print(fib(1000))
    """
    
    # import numpy as np
    import sys
    sys.setrecursionlimit(10**7)
    from itertools import accumulate, combinations, permutations, product # https://docs.python.org/ja/3/library/itertools.html
    # accumulate() returns iterator! to get list: list(accumulate())
    from math import factorial, ceil, floor, sqrt
    def factorize(n):
        """return the factors of the Arg and count of each factor
        
        Args:
            n (long): number to be resolved into factors
        
        Returns:
            list of tuples: factorize(220) returns [(2, 2), (5, 1), (11, 1)]
        """
        fct = []  # prime factor
        b, e = 2, 0  # base, exponent
        while b * b <= n:
            while n % b == 0:
                n = n // b
                e = e + 1
            if e > 0:
                fct.append((b, e))
            b, e = b + 1, 0
        if n > 1:
            fct.append((n, 1))
        return fct
    def combinations_count(n, r):   
        """Return the number of selecting r pieces of items from n kinds of items.
        
        Args:
            n (long): number
            r (long): number
        
        Raises:
            Exception: not defined when n or r is negative
        
        Returns:
            long: number
        """
        # TODO: How should I do when n - r is negative?
        if n < 0 or r < 0:
            raise Exception('combinations_count(n, r) not defined when n or r is negative')
        if n - r < r: r = n - r
        if r < 0: return 0
        if r == 0: return 1
        if r == 1: return n
        numerator = [n - r + k + 1 for k in range(r)]
        denominator = [k + 1 for k in range(r)]
        for p in range(2,r+1):
            pivot = denominator[p - 1]
            if pivot > 1:
                offset = (n - r) % p
                for k in range(p-1,r,p):
                    numerator[k - offset] /= pivot
                    denominator[k] /= pivot
        result = 1
        for k in range(r):
            if numerator[k] > 1:
                result *= int(numerator[k])
        return result
    def combinations_with_replacement_count(n, r):
        """Return the number of selecting r pieces of items from n kinds of items allowing individual elements to be repeated more than once.
        
        Args:
            n (long): number
            r (long): number
        
        Raises:
            Exception: not defined when n or r is negative
        
        Returns:
            long: number
        """
        if n < 0 or r < 0:
            raise Exception('combinations_with_replacement_count(n, r) not defined when n or r is negative')
        elif n == 0:
            return 1
        else:
            return combinations_count(n + r - 1, r)
    from bisect import bisect_left, bisect_right
    from collections import deque, Counter, defaultdict # https://docs.python.org/ja/3/library/collections.html#collections.deque
    from heapq import heapify, heappop, heappush, heappushpop, heapreplace,nlargest,nsmallest # https://docs.python.org/ja/3/library/heapq.html
    from copy import deepcopy, copy # https://docs.python.org/ja/3/library/copy.html
    import operator
    from operator import itemgetter #sort 
    # ex1: List.sort(key=itemgetter(1))
    # ex2: sorted(tuples, key=itemgetter(1,2))
    from functools import reduce, lru_cache
    def chmin(x, y):
        """change minimum
        if x > y, x = y and return (x, True).
        convenient when solving problems of dp[i]
        
        Args:
            x (long): current minimum value
            y (long): potential minimum value
        
        Returns:
            (x, bool): (x, True) when updated, else (x, False)
        """
        if x > y:
            x = y
            return (x, True)
        else:
            return (x, False)
    def chmax(x, y):
        """change maximum
        if x < y, x = y and return (x, True).
        convenient when solving problems of dp[i]
        
        Args:
            x (long): current maximum value
            y (long): potential maximum value
        
        Returns:
            (x, bool): (x, True) when updated, else (x, False)
        """
        if x < y:
            x = y
            return (x, True)
        else:
            return (x, False)

    from fractions import gcd # Deprecated since version 3.5: Use math.gcd() instead.
    def gcds(numbers):
        return reduce(gcd, numbers)
    def lcm(x, y):
        return (x * y) // gcd(x, y)
    def lcms(numbers):
        return reduce(lcm, numbers, 1)
    def make_divisors(n, reversed=False):
        """create list of divisors
        
        Args:
            number (int): number from which list of divisors is created
            reversed (bool, optional): ascending order if False. Defaults to False.
        
        Returns:
            list: list of divisors
        """
        divisors = set()
        for i in range(1, int(n**0.5)+1):
            if n % i == 0:
                divisors.add(i)
                divisors.add(n//i)
        return sorted(list(divisors),reverse=reversed)
    
    # first create factorial_list
    # fac_list = mod_factorial_list(n)
    INF = 10 ** 18
    MOD = 10 ** 9 + 7
    AtoZ = list('abcdefghijklmnopqrsruvwxyz') # 26
    modpow = lambda a, n, p = MOD: pow(a, n, p) # Recursive function in python is slow!
    def modinv(a, p = MOD):
        # evaluate reciprocal using Fermat's little theorem:
        # a**(p-1) is identical to 1 (mod p) when a and p is coprime
        return modpow(a, p-2, p)
    def modinv_list(n, p = MOD):
        if n <= 1:
            return [0,1][:n+1]
        else:
            inv_t = [0,1]
            for i in range(2, n+1):
                inv_t += [inv_t[p % i] * (p - int(p / i)) % p]
            return inv_t
    def modfactorial_list(n, p = MOD):
        if n == 0:
            return [1]
        else:
            l = [0] * (n+1)
            tmp = 1
            for i in range(1, n+1):
                tmp = tmp * i % p
                l[i] = tmp
            return l
    def modcomb(n, k, fac_list = [], p = MOD):
        # fac_list = modfactorial_list(100)
        # print(modcomb(100, 5, modfactorial_list(100)))
        from math import factorial
        if n < 0 or k < 0 or n < k: return 0
        if n == 0 or k == 0: return 1
        if len(fac_list) <= n:
            a = factorial(n) % p
            b = factorial(k) % p
            c = factorial(n-k) % p
        else:
            a = fac_list[n]
            b = fac_list[k]
            c = fac_list[n-k]
        return (a * modpow(b, p-2, p) * modpow(c, p-2, p)) % p
    def modadd(a, b, p = MOD):
        return (a + b) % MOD
    def modsub(a, b, p = MOD):
        return (a - b) % p
    def modmul(a, b, p = MOD):
        return ((a % p) * (b % p)) % p
    def moddiv(a, b, p = MOD):
        return modmul(a, modpow(b, p-2, p))

    class UnionFindTree:
        """union find tree class
        TODO: fix this description...
        how to use (example):
        >>  uf = UnionFindTree(N) 
        >>  if uf.find_root(a) == uf.find_root(b):
        >>      do something
        >>  else:
        >>      do something
        >>      uf.unite(a, b)
        """
        def __init__(self, N):
            self.root = [-1] * (N+1)
            self.rank   = [0]  * (N+1)
            self.connected_num = [1] * (N+1)
    
        def find_root(self,x):
            root = self.root
            while root[x] != -1:
                x = root[x]
            return x
    
        def unite(self,x,y):
            root = self.root
            rank = self.rank
            connected_num = self.connected_num
            find_root = self.find_root
    
            rx = find_root(x)
            ry = find_root(y)
            if rx != ry:
                if rank[rx] < rank[ry]:
                    root[rx] = ry
                    rx,ry = ry,rx
                else:
                    if rank[rx] == rank[ry]:
                        rank[rx] += 1
                    root[ry] = rx       
                connected_num[rx] += connected_num[ry]

    # ************ Graph ************
    # Graph: https://en.wikipedia.org/wiki/Directed_graph
    # 最短経路問題: https://ja.wikipedia.org/wiki/%E6%9C%80%E7%9F%AD%E7%B5%8C%E8%B7%AF%E5%95%8F%E9%A1%8C
    
    # Bellman-Ford: O(|V||E|). Use this if there exists an edge with negative length in the graph
    # After N steps, the shortest path has converded if there doesn't exist an cycle of edges with negative
    # Watch out: d[N] == d[2*N] doesn't necessarily mean the graph doesn't have negative cycle
    # ref: https://www.youtube.com/watch?v=1Z6ofKN03_Y

    def BellmanFord(N, M, ABC, vertex_start, vertex_end, value_if_inf = -1, find_shortest = False):
        """to calculate furthest or shortest length between vertex_start and vertex_end using BellmanFord algorithm
        
        Args:
            N (int): number of vertices
            M (int): number of edges
            ABC (list): [(ai, bi, ci) for _ in range(N)] where i-th edge is directed from vertex ai to vertex bi and the length is ci 
            vertex_start (int): start vertex. usually use 0.
            vertex_end (int): end vertex. usually use N-1.
            value_if_inf (int or string as you like, optional): value you want when the furthest (or shortest) distance is infinite (or -infinite). Defaults to -1.
            find_shortest (bool, optional): choose False to find furthest path. Defaults to False.
        
        Returns:
            int or string: normally int (but can be str if you set value_if_inf to str)
            
        Example:
                N, M, P = R()
                ABC = [R() for _ in range(M)]
                ABC = [(a-1, b-1, c-P) for a, b, c in ABC]
                print(BellmanFord(N, M, ABC, 0, N-1, value_if_inf = 'inf'))

        """

        def make_reachable_list(N, M, ABC, vertex_start, vertex_end):
            reachable_to_direct = defaultdict(list)
            reachable_from_direct = defaultdict(list)
            reachable_from_start = [False] * N
            reachable_to_end = [False] * N
            reachable_from_start[vertex_start] = True
            reachable_to_end[vertex_end] = True
            reachable_from_both_sides = [False] * N
            dfs_from_start = []
            dfs_to_end = []
            for a, b, c in ABC:
                reachable_to_direct[a].append(b)
                reachable_from_direct[b].append(a)
                if a == vertex_start:
                    dfs_from_start.append(b)
                    reachable_from_start[b] = True
                if b == vertex_end:
                    dfs_to_end.append(a)
                    reachable_to_end[a] = True
            while dfs_from_start:
                v = dfs_from_start.pop()
                for i in reachable_to_direct[v]:
                    if not reachable_from_start[i]:
                        reachable_from_start[i] = True
                        dfs_from_start.append(i)
            while dfs_to_end:
                v = dfs_to_end.pop()
                for i in reachable_from_direct[v]:
                    if not reachable_to_end[i]:
                        reachable_to_end[i] = True
                        dfs_to_end.append(i)
            for i in range(N):
                reachable_from_both_sides[i] = reachable_from_start[i] and reachable_to_end[i]
            return reachable_from_both_sides
        
        reachable_from_both_sides = make_reachable_list(N, M, ABC, vertex_start, vertex_end)
        
        if find_shortest:
            dist = [INF for i in range(N)]
        else:
            dist = [-INF for i in range(N)]
        
        dist[vertex_start] = 0
        for i in range(N):
            updated = False
            for a, b, c in ABC:
                if not reachable_from_both_sides[a]:
                    continue
                elif find_shortest:
                    update_condition = dist[a] + c < dist[b]
                else:
                    update_condition = dist[a] + c > dist[b]
                if dist[a] != INF and update_condition:
                    dist[b] = dist[a] + c
                    updated = True
                    if i == N-1:
                        return value_if_inf
            if not updated:
                break
        return dist[vertex_end]

    # Warshall Floyd O(V**3)
    def warshall_floyd(number_vertex, XYD, shift_one_for_vertex=False):
        # ref: https://juppy.hatenablog.com/entry/2018/11/01/%E8%9F%BB%E6%9C%AC_python_%E5%85%A8%E7%82%B9%E5%AF%BE%E6%9C%80%E7%9F%AD%E7%B5%8C%E8%B7%AF%E6%B3%95%EF%BC%88%E3%83%AF%E3%83%BC%E3%82%B7%E3%83%A3%E3%83%AB%E3%83%95%E3%83%AD%E3%82%A4%E3%83%89%E6%B3%95
        # number_vertex: len(vertex)
        # XYD: list of distance from x to y
        # shift_one_for_vertex: XYDの値が1~Nの時(0~N-1でない時)に調整するフラグ。
        # d[i][j]: iからjへの最短距離
        d = [[float("inf")]*number_vertex for i in range(number_vertex)]
        for x, y, dist in XYD:
            if shift_one_for_vertex:
                x = x - 1
                y = y - 1
            d[x][y] = dist
            d[y][x] = dist
        for i in range(number_vertex):
            d[i][i] = 0 # distance of same vertex is 0
        for k in range(number_vertex):
            for i in range(number_vertex):
                for j in range(number_vertex):
                    d[i][j] = min(d[i][j],d[i][k] + d[k][j])
        return d


    """ initialize variables and set inputs
    # initialize variables
        # to initialize list, use [0] * n
        # to initialize two dimentional array:
            # ex) [[0] * N for _ in range(N)]
            # ex2) dp = [[0] * (N+1) for _ in range(W*2)]
    # set inputs
        # put inputs between specific values (almost as quickly)
        # ex) S = [-INF] + [int(r()) for _ in range(A)] + [INF]
        # open(0).read() is sometimes useful:
        # ex) n, m, *x = map(int, open(0).read().split())
        #     min(x[::2]) - max(x[1::2])
        # ex2) *x, = map(int, open(0).read().split())
        #     don't forget to add comma after *x if only one variable is used
    # preprocessing
        # transpose = [x for x in zip(*data)]
        # ex) [[1, 2, 3], [4, 5, 6], [7, 8, 9]] => [(1, 4, 7), (2, 5, 8), (3, 6, 9)]
        # flat = [flatten for inner in data for flatten in inner]
        # ex) [[1, 2, 3], [4, 5, 6], [7, 8, 9]] => [1, 2, 3, 4, 5, 6, 7, 8, 9]
    # calculate and output
        # output pattern
        # ex1) print(*l) => when l = [2, 5, 6], printed 2 5 6
    """

    # functions to read input
    r = lambda: sys.stdin.readline().strip()
    r_int = lambda: int(r())
    r_float = lambda: float(r())
    R = lambda: list(map(int, r().split()))
    R_str = lambda: r().split()
    R_map = lambda: map(int, r().split())
    R_float = lambda: list(map(float, r().split()))
    R_tuple = lambda: tuple(map(int, r().split()))

    """ how to treat input
    # single int: int(r())
    # single string: r()
    # single float: float(r())
    # line int: R()
    # line string: r().split()
    # line (str, int, int): [j if i == 0 else int(j) for i, j in enumerate(r().split())]
    # lines int: [R() for _ in range(n)]
    """

    # for test
    if sample_file:
        sys.stdin = open(sample_file)
        
    # ----------------------------------
    # main
    #%%
    N = r_int()
    S = r()
    Q = r_int()
    Que = [R_str() for _ in range(Q)]
    
    # 小文字をカウントしたリストを持つセグ木で行けそう。
    class SegTree:
        """
        ref: https://qiita.com/takayg1/items/c811bd07c21923d7ec69
        init(init_val, ide_ele): 配列init_valで初期化 O(N)
        update(k, x): k番目の値をxに更新 O(logN)
        query(l, r): 区間[l, r)をsegfuncしたものを返す O(logN)
        """
        def __init__(self, init_val, segfunc, ide_ele):
            """
            init_val: 配列の初期値
            segfunc: 区間にしたい操作
            ide_ele: 単位元
            n: 要素数
            num: n以上の最小の2のべき乗
            tree: セグメント木(1-index)
            """
            n = len(init_val)
            self.segfunc = segfunc
            self.ide_ele = ide_ele
            self.num = 1 << (n - 1).bit_length()
            self.tree = [ide_ele] * 2 * self.num
            # 配列の値を葉にセット
            for i in range(n):
                self.tree[self.num + i] = init_val[i]
            # 構築していく
            for i in range(self.num - 1, 0, -1):
                self.tree[i] = self.segfunc(self.tree[2 * i], self.tree[2 * i + 1])

        def update(self, k, x):
            """
            k番目の値をxに更新
            k: index(0-index)
            x: update value
            """
            k += self.num
            self.tree[k] = x
            while k > 1:
                self.tree[k >> 1] = self.segfunc(self.tree[k], self.tree[k ^ 1])
                k >>= 1

        def query(self, l, r):
            """
            [l, r)のsegfuncしたものを得る
            l: index(0-index)
            r: index(0-index)
            """
            res = self.ide_ele

            l += self.num
            r += self.num
            while l < r:
                if l & 1:
                    res = self.segfunc(res, self.tree[l])
                    l += 1
                if r & 1:
                    res = self.segfunc(res, self.tree[r - 1])
                l >>= 1
                r >>= 1
            return res
    
    # a = [14, 5, 9, 13, 7, 12, 11, 1, 7, 8]
    
    def segfunc(x, y):
        return x | y
    
    def count_ones_by_bitmask(num):
        count = 0
        while num:
            count += num & 1
            num >>= 1
        return count
    
    S_int = [2**(ord(i)-97) for i in S]
    ide_ele = 0
    seg = SegTree(S_int, segfunc, ide_ele)
    for i, a, b in Que:
        if i == '1':
            seg.update(int(a), 2**(ord(b)-97))
        else:
            print(count_ones_by_bitmask(seg.query(int(a)-1, int(b))))

    #%%
    # end of main
    # ----------------------------------

    """memo: how to use defaultdict of list
    # initialize
    Dic = defaultdict(list)
    # append / extend
    Dic[x].append(y)
    # three methods for loop: keys(), values(), items()
    for k, v in Dic.items():
    """
    
    """memo: how to solve binary problems
    # to make binary digits text
    >>> a = 5
    >>> bin_str_a = format(a, '#06b')
    >>> print(bin_str_a)
    0b0101 # first 2 strings (='0b') indicates it is binary
    """
    
    """memo: how to solve the problem
    creating simple test/answer
    greed
    simple dp
    graph
    """

if __name__ == '__main__':
    main()

Submission status: Wrong Answer

Solution code:
#query=10**4*2
#alph=26
#insert=500000
#O(|insert*query*alph|)?
def main():
	import bisect,sys
	input=sys.stdin.readline
	n=int(input())
	s=list(input().rstrip())
	q=int(input())
	word=[[]for _ in range(26)]
	for i,j in enumerate(s):
		ch=ord(j)-ord("a")
		word[ch].append(i)
	for _ in range(q):
		query=input().split()
		if query[0]=="1":
			i,x=int(query[1])-1,ord(query[2])-ord("a")
			old=ord(s[i])-ord("a")
			if s==x:
				continue
			word[old].remove(i)
			new=bisect.bisect_left(word[x],i)
			word[x].insert(new,i)
			s[i]=chr(x+ord("a"))
		else:
			l,r=int(query[1])-1,int(query[2])-1
			kind=0
			for i in range(26):
				l=bisect.bisect_left(word[i],l)
				if l==len(word[i]):
					continue
				if word[i][l]<=r:
					kind+=1
			print(kind)
if __name__ == '__main__':
	main()

Submission status: Wrong Answer

Solution code:
#query=10**4*2
#alph=26
#insert=500000
#O(|insert*query*alph|)?
def main():
	import bisect,sys
	input=sys.stdin.readline
	n=int(input())
	s=list(input().rstrip())
	q=int(input())
	word=[[]for _ in range(26)]
	for i,j in enumerate(s):
		ch=ord(j)-ord("a")
		word[ch].append(i)
	for _ in range(q):
		query=input().split()
		if query[0]=="1":
			i,x=int(query[1])-1,ord(query[2])-ord("a")
			old=ord(s[i])-ord("a")
			word[old].remove(i)
			new=bisect.bisect_left(word[x],i)
			word[x].insert(new,i)
			s[i]=chr(x+ord("a"))
		else:
			l,r=int(query[1])-1,int(query[2])-1
			kind=0
			for i in range(26):
				l=bisect.bisect_left(word[i],l)
				if l==len(word[i]):
					continue
				if word[i][l]<=r:
					kind+=1
			print(kind)
if __name__ == '__main__':
	main()

Submission status: Wrong Answer

Solution code:
#query=10**4*2
#alph=26
#insert=500000
#O(|insert*query*alph|)?
def main():
	import bisect,sys
	input=sys.stdin.readline
	n=int(input())
	s=list(input().rstrip())
	q=int(input())
	word=[[]for _ in range(26)]
	for i,j in enumerate(s):
		ch=ord(j)-ord("a")
		word[ch].append(i)
	for _ in range(q):
		query=input().split()
		if query[0]=="1":
			# i,x=int(query[1])-1,ord(query[2])-ord("a")
			# old=ord(s[i])-ord("a")
			# word[old].remove(i)
			# new=bisect.bisect_left(word[x],i)
			# word[x].insert(new,i)
			# s[i]=x
			pass
		else:
			l,r=int(query[1])-1,int(query[2])-1
			kind=0
			for i in range(26):
				l=bisect.bisect_left(word[i],l)
				if l==len(word[i]):
					continue
				if word[i][l]<=r:
					kind+=1
			print(kind)
if __name__ == '__main__':
	main()

Submission status: Wrong Answer

Solution code:
n=int(input())
l=list(input())
q=int(input())
bit=[[0]*(n+1) for i in range(26)]
def bit_sum(o,i):
  s=0
  while i:
    s+=bit[o][i]
    i-=i&-i
  return s
def bit_add(o,i,x):
  while i<=n:
    bit[o][i] += x
    i+=i&-i
for i in range(n):
  bit_add(ord(l[i])-97,i+1,1)
for i in range(q):
  a,b,c=input().split()
  b=int(b)-1
  if a=='1':
    bit_add(ord(l[b])-97,i+1,-1)
    l[b]=c
    bit_add(ord(c)-97,i+1,1)
  else:
    c=int(c)
    print(sum(1 for o in range(26) if bit_sum(o,c)-bit_sum(o,b)))

Submission status: Wrong Answer

Solution code:
# coding: utf-8
import sys
from bisect import bisect_left, bisect_right, insort_left
from collections import defaultdict

sr = lambda: sys.stdin.readline().rstrip()
ir = lambda: int(sr())
lr = lambda: list(map(int, sr().split()))

# 各アルファベットのindexのリストを作り,bisect_left()
N = ir()
S = sr()
dic = defaultdict(list)
for i, s in enumerate(S):
    dic[s].append(i)

Q = ir()
for _ in range(Q):
    a, b, c = sr().split()
    if a == '1':
        b = int(b) - 1
        if S[b] == c:
            continue
        if b in dic[S[b]]:
            dic[S[b]].remove(b)        
        insort_left(dic[c], b)
    else:
        b = int(b) - 1; c = int(c) - 1
        kind = 0
        for x in range(26):
            word = chr(ord('a')+x)
            if len(dic[word]) == 0:
                continue
            i = bisect_left(dic[word], b)
            j = bisect_right(dic[word], c)
            if i != j:
                kind += 1
        print(kind)

# 07

Submission status: Wrong Answer

Solution code:
TO_ASCII = 97

N = int(input())
S = input()
Q = int(input())

Count = [[0 for i in range(0,26,1)]for j in range(0,N+1,1)]

for i in range(0,N,1):
    Count[i][ord(S[i])-TO_ASCII]=1

for i in range(0,N-1,1):
    for j in range(0,26,1):
        Count[i+1][j] += Count[i][j]

for i in range(0,Q,1):
    q = input()

Submission status: Wrong Answer

Solution code:
# coding: utf-8
import sys
#from operator import itemgetter
sysread = sys.stdin.readline
#from heapq import heappop, heappush
from collections import defaultdict
#from itertools import combinations
sys.setrecursionlimit(10**7)

def run():
    # import sys
    # A = list(map(int, sys.stdin.readline().split()))
    # all = sys.stdin.readlines()
    #N, *A = map(int, open(0))
    #N = int(input())
    ###
    def bin_search(A,low_val):
        high = len(A) -1
        low = 0
        current = high // 2
        if A[-1] < low_val:
            return None, None
        if A[0] > low_val:
            return 0, A[0]
        while high-low >1:
            if A[current] == low_val:return current, A[current]
            if A[current] > low_val:
                high = current
                current += low
                current //= 2
            else:
                low = current
                current += high
                current //= 2
        return high, A[high]

    N = int(sysread())
    S = list('0'+input())
    dic = defaultdict(lambda:[])
    for idx, s in enumerate(S[1:],1):
        dic[s].append(idx)

    Q = int(sysread())
    queries = []
    for _ in range(Q):
        queries.append(sysread().split())

    for q,i,c in queries:
        if q == '1':
            i = int(i)
            old = S[i]
            dic[old].remove(i)
            if dic[c] == []:
                dic[c].append(i)
            else:
                h, _ = bin_search(dic[c], i)
                if h is None:
                    dic[c].append(i)
                else:dic[c].insert(h,i)
            S[i] = c
        else:
            count = 0
            i,c = int(i), int(c)
            for key in dic.keys():
                tmp = dic[key]
                if tmp == []:continue
                _,v = bin_search(tmp, i)
                if v is None:continue
                if v <= c:
                    count += 1
            print(count)

if __name__ == "__main__":
    run()

Submission status: Wrong Answer

Solution code:
# coding: utf-8
import sys
#from operator import itemgetter
sysread = sys.stdin.readline
#from heapq import heappop, heappush
#from collections import defaultdict
#from itertools import combinations
sys.setrecursionlimit(10**7)
import numpy as np
def run():

    N = int(sysread())
    S = list('0'+input())
    Q = int(sysread())
    queries = []
    for _ in range(Q):
        queries.append(sysread().split())

    a2z = 'abcdefghijklmnopqrstuvwxyz'
    a2num = {val:idx for idx, val in enumerate(a2z)}
    counts = np.array([[0] * len(a2z) for _ in range(N+1)])
    counts[1][a2num[S[1]]] += 1
    for idx, s in enumerate(S[2:],2):
        counts[idx:, a2num[s]] += 1

    for t, i, c in queries:
        if t == '1':
            i = int(i)
            counts[i:, a2num[S[i]]] -= 1
            counts[i:, a2num[c]] += 1
            S[i] = c
        else:
            i,c = int(i), int(c)
            count = (counts[c, :] - counts[i-1, :] > 0).sum()
            print(count)
if __name__ == "__main__":
    run()

Submission status: Wrong Answer

Solution code:
from bisect import bisect_left, bisect_right
from heapq import heapify, heappush

n = int(input())
s = list(input())
q = int(input())

d = {}
flag = {}
for i in list('abcdefghijklmnopqrstuvwxyz'):
	d.setdefault(i, []).append(-1)
	flag.setdefault(i, []).append(-1)
for i in range(n):
	d[s[i]].append(i)
for i in list('abcdefghijklmnopqrstuvwxyz'):
	heapify(d[i])
	heapify(flag[i])

for i in range(q):
	q1,q2,q3 = map(str,input().split())
	if q1 == '1':
		q2 = int(q2) - 1
		if s[q2] != q3:
			heappush(flag[s[q2]],q2)
			heappush(d[q3],q2)
			s[q2] = q3
	else:
		ans = 0
		q2 = int(q2) - 1
		q3 = int(q3) - 1
		if q2 == q3:
			print(1)
			continue
		for string,l in d.items():
			res = 0
			if d[string] != [-1]:
				left = bisect_left(l,q2)
				right = bisect_right(l,q3)
			else:
				left = 0
				right = 0
			if flag[string] != [-1]:
				left2 = bisect_left(flag[string],q2)
				right2 = bisect_right(flag[string],q3)
			else:
				left2 = 0
				right2 = 0
			if left != right:
				if right - left > right2 - left2:
					res = 1
			ans += res
			#print(string,l,flag[string],res,[left,right,left2,right2])
		print(ans)
"""
3
aaa
4
1 1 z
1 2 z
1 3 z
2 1 3
"""

Submission status: Wrong Answer

Solution code:
from bisect import bisect_left
from heapq import heapify, heappush

n = int(input())
s = list(input())
q = int(input())

d = {}
flag = {}
for i in list('abcdefghijklmnopqrstuvwxyz'):
	d.setdefault(i, []).append(-1)
	flag.setdefault(i, []).append(-1)
for i in range(n):
	d[s[i]].append(i)
for i in list('abcdefghijklmnopqrstuvwxyz'):
	heapify(d[i])
	heapify(flag[i])

for i in range(q):
	q1,q2,q3 = map(str,input().split())
	if q1 == '1':
		q2 = int(q2) - 1
		if s[q2] != q3:
			heappush(flag[s[q2]],q2)
			heappush(d[q3],q2)
			s[q2] = q3
	else:
		ans = 0
		q2 = int(q2) - 1
		q3 = int(q3) - 1
		if q2 == q3:
			print(1)
			continue
		for string,l in d.items():
			res = 0
			if d[string] != [-1]:
				left = bisect_left(l,q2-1)
				right = bisect_left(l,q3+1)
			else:
				left = 0
				right = 0
			if flag[string] != [-1]:
				left2 = bisect_left(flag[string],q2-1)
				right2 = bisect_left(flag[string],q3+1)
			else:
				left2 = 0
				right2 = 0
			if left != right:
				if right - left > right2 - left2:
					res = 1
			ans += res
			#print(string,l,flag[string],res,[left,right,left2,right2])
		print(ans)
"""
3
aaa
4
1 1 z
1 2 z
1 3 z
2 1 3
"""

Submission status: Wrong Answer

Solution code:
from bisect import bisect_left
from heapq import heapify, heappush

n = int(input())
s = list(input())
q = int(input())

d = {}
flag = {}
for i in list('abcdefghijklmnopqrstuvwxyz'):
	d.setdefault(i, []).append(-1)
	flag.setdefault(i, []).append(-1)
for i in range(n):
	d[s[i]].append(i)
for i in list('abcdefghijklmnopqrstuvwxyz'):
	heapify(d[i])
	heapify(flag[i])

for i in range(q):
	q1,q2,q3 = map(str,input().split())
	if q1 == '1':
		q2 = int(q2) - 1
		if s[q2] != q3:
			heappush(flag[s[q2]],q2)
			heappush(d[q3],q2)
			s[q2] = q3
	else:
		ans = 0
		q2 = int(q2) - 1
		q3 = int(q3) - 1
		if q2 == q3:
			print(1)
			continue
		for string,l in d.items():
			res = 0
			if d[string] != [-1]:
				left = bisect_left(l,q2)
				right = bisect_left(l,q3+1)
			else:
				left = 0
				right = 0
			if flag[string] != [-1]:
				left2 = bisect_left(flag[string],q2)
				right2 = bisect_left(flag[string],q3+1)
			else:
				left2 = 0
				right2 = 0
			if left != right:
				if right - left > right2 - left2:
					res = 1
			ans += res
			#print(string,l,flag[string],res,[left,right,left2,right2])
		print(ans)
"""
3
aaa
4
1 1 z
1 2 z
1 3 z
2 1 3
"""