Problem:


Score : 100 points

Problem Statement

You are given an array a_0, a_1, ..., a_{N-1} of length N. Process Q queries of the following types.

The type of i-th query is represented by T_i.

T_i=1: You are given two integers X_i,V_i. Replace the value of A_{X_i} with V_i.

T_i=2: You are given two integers L_i,R_i. Calculate the maximum value among A_{L_i},A_{L_i+1},\cdots,A_{R_i}.

T_i=3: You are given two integers X_i,V_i. Calculate the minimum j such that X_i \leq j \leq N, V_i \leq A_j. If there is no such j, answer j=N+1 instead.

Constraints

1 \leq N \leq 2 \times 10^5

0 \leq A_i \leq 10^9

1 \leq Q \leq 2 \times 10^5

1 \leq T_i \leq 3

1 \leq X_i \leq N, 0 \leq V_i \leq 10^9 (T_i=1,3)

1 \leq L_i \leq R_i \leq N (T_i=2)

All values in Input are integer.

Input

Input is given from Standard Input in the following format:

N Q
A_1 A_2 \cdots A_N
First query
Second query
\vdots
Q-th query

Each query is given in the following format:

If T_i=1,3,

T_i X_i V_i

If T_i=2,

T_i L_i R_i

Output

For each query with T_i=2, 3, print the answer.

Sample Input 1

5 5
1 2 3 2 1
2 1 5
3 2 3
1 3 1
2 2 4
3 1 3

Sample Output 1

3
3
2
6

First query: Print 3, which is the maximum of (A_1,A_2,A_3,A_4,A_5)=(1,2,3,2,1).

Second query: Since 3>A_2, j=2 does not satisfy the condition．Since 3 \leq A_3, print j=3.

Third query: Replace the value of A_3 with 1. It becomes A=(1,2,1,2,1).

Fourth query: Print 2, which is the maximum of (A_2,A_3,A_4)=(2,1,2).

Fifth query: Since there is no j that satisfies the condition, print j=N+1=6.



Sample input:
5 5
1 2 3 2 1
2 1 5
3 2 3
1 3 1
2 2 4
3 1 3


Sample output:
3
3
2
6

Submission status: Accepted

Solution code:
class LazySegmentTree():

    def __init__(self, n, f, g, h, ef, eh):
        """
        :param n: 配列の要素数
        :param f: 取得半群の元同士の積を定義
        :param g: 更新半群の元 xh が配列上の実際の値にどのように作用するかを定義
        :param h: 更新半群の元同士の積を定義　（更新半群の元を xh と表記）
        :param x: 配列の各要素の値。treeの葉以外は xf(x1,x2,...)
        """
        self.n = n
        self.f = f
        self.g = lambda xh, x: g(xh, x) if xh != eh else x
        self.h = h
        self.ef = ef
        self.eh = eh
        l = (self.n - 1).bit_length()
        self.size = 1 << l
        self.tree = [self.ef] * (self.size << 1)
        self.lazy = [self.eh] * ((self.size << 1) + 1)
        self.plt_cnt = 0

    def built(self, array):
        """
        arrayを初期値とするセグメント木を構築
        """
        for i in range(self.n):
            self.tree[self.size + i] = array[i]
        for i in range(self.size - 1, 0, -1):
            self.tree[i] = self.f(self.tree[i<<1], self.tree[(i<<1)|1])

    def update(self, i, x):
        """
        i 番目の要素を x に更新する
        """
        i += self.size
        self.propagate_lazy(i)
        self.tree[i] = x
        self.lazy[i] = self.eh
        self.propagate_tree(i)

    def get(self, i):
        """
        i 番目の値を取得（ 0-indexed ） ( O(logN) )
        """
        i += self.size
        self.propagate_lazy(i)
        return self.g(self.lazy[i], self.tree[i])

    def update_range(self, l, r, x):
        """
        半開区間 [l, r) の各々の要素 a に op(x, a)を作用させる （ 0-indexed ）　（ O(logN) ）
        """
        if l >= r:
            return
        l += self.size
        r += self.size
        l0 = l//(l&-l)
        r0 = r//(r&-r)
        self.propagate_lazy(l0)
        self.propagate_lazy(r0-1)
        while l < r:
            if r&1:
                r -= 1              # 半開区間なので先に引いてる
                self.lazy[r] = self.h(x, self.lazy[r])
            if l&1:
                self.lazy[l] = self.h(x, self.lazy[l])
                l += 1
            l >>= 1
            r >>= 1
        self.propagate_tree(l0)
        self.propagate_tree(r0-1)

    def get_range(self, l, r):
        """
        [l, r)への作用の結果を返す　(0-indexed)
        """
        l += self.size
        r += self.size
        self.propagate_lazy(l//(l&-l))
        self.propagate_lazy((r//(r&-r))-1)
        res_l = self.ef
        res_r = self.ef
        while l < r:
            if l & 1:
                res_l = self.f(res_l, self.g(self.lazy[l], self.tree[l]))
                l += 1
            if r & 1:
                r -= 1
                res_r = self.f(self.g(self.lazy[r], self.tree[r]), res_r)
            l >>= 1
            r >>= 1
        return self.f(res_l, res_r)

    def max_right(self, l, z):
        """
        以下の条件を両方満たす r を(いずれか一つ)返す
            ・r = l or f(op(a[l], a[l + 1], ..., a[r - 1])) = true
            ・r = n or f(op(a[l], a[l + 1], ..., a[r])) = false
        """
        if l >= self.n: return self.n
        l += self.size
        s = self.ef
        while 1:
            while l % 2 == 0:
                l >>= 1
            if not z(self.f(s, self.g(self.lazy[l], self.tree[l]))):
                while l < self.size:
                    l *= 2
                    if z(self.f(s, self.g(self.lazy[l], self.tree[l]))):
                        s = self.f(s, self.g(self.lazy[l], self.tree[l]))
                        l += 1
                return l - self.size
            s = self.f(s, self.g(self.lazy[l], self.tree[l]))
            l += 1
            if l & -l == l: break
        return self.n

    def min_left(self, r, z):
        """
        以下の条件を両方満たす l を(いずれか一つ)返す
            ・l = r or f(op(a[l], a[l + 1], ..., a[r - 1])) = true
            ・l = 0 or f(op(a[l - 1], a[l], ..., a[r - 1])) = false
        """
        if r <= 0: return 0
        r += self.size
        s = self.ef
        while 1:
            r -= 1
            while r > 1 and r % 2:
                r >>= 1
            if not z(self.f(self.g(self.lazy[r], self.tree[r]), s)):
                while r < self.size:
                    r = r * 2 + 1
                    if z(self.f(self.g(self.lazy[r], self.tree[r]), s)):
                        s = self.f(self.g(self.lazy[r], self.tree[r]), s)
                        r -= 1
                return r + 1 - self.size
            s = self.f(self.g(self.lazy[r], self.tree[r]), s)
            if r & -r == r: break
        return 0

    def propagate_lazy(self, i):
        """
        lazy の値をトップダウンで更新する　（ O(logN) ）
        """
        for k in range(i.bit_length()-1,0,-1):
            x = i>>k
            if self.lazy[x] == self.eh:
                continue
            laz = self.lazy[x]
            self.lazy[(x<<1)|1] = self.h(laz, self.lazy[(x<<1)|1])
            self.lazy[x<<1] = self.h(laz, self.lazy[x<<1])
            self.tree[x] = self.g(laz, self.tree[x])   # get_range ではボトムアップの伝搬を行わないため、この処理をしないと tree が更新されない
            self.lazy[x] = self.eh


    def propagate_tree(self, i):
        """
        tree の値をボトムアップで更新する　（ O(logN) ）
        """
        while i>1:
            i>>=1
            self.tree[i] = self.f(self.g(self.lazy[i<<1], self.tree[i<<1]), self.g(self.lazy[(i<<1)|1], self.tree[(i<<1)|1]))

    def __getitem__(self, i):
        return self.get(i)

    def __iter__(self):
        for x in range(1, self.size):
            if self.lazy[x] == self.eh:
                continue
            self.lazy[(x<<1)|1] = self.h(self.lazy[x], self.lazy[(x<<1)|1])
            self.lazy[x<<1] = self.h(self.lazy[x], self.lazy[x<<1])
            self.tree[x] = self.g(self.lazy[x], self.tree[x])
            self.lazy[x] = self.eh
        for xh, x in zip(self.lazy[self.size:self.size+self.n], self.tree[self.size:self.size+self.n]):
            yield self.g(xh,x)

    def __str__(self):
        return str(list(self))


##################################################################################################################

N, Q = map(int, input().split())
A = list(map(int, input().split()))
ef = 0
eh = 0
f = lambda x, y : x if x > y else y
g = lambda x, y : x if x > y else y
h = lambda x, y : x if x > y else y
st = LazySegmentTree(N, f, g, h, ef, eh)
st.built(A)
res = []

for _ in range(Q):
    t, x, y = map(int, input().split())
    if t == 1:
        st.update(x - 1, y)
    elif t == 2:
        res.append(st.get_range(x - 1, y))
    else:
        res.append(st.max_right(x - 1, lambda z: z < y) + 1)

print('\n'.join(map(str, res)))

Submission status: Accepted

Solution code:
class SegmentTree:

    def __init__(self, n, op, e):
        """
        :param n: 要素数
        :param op: 二項演算
        :param e: 単位減
        """
        self.n = n
        self.op = op
        self.e = e
        self.size = 1 << (self.n - 1).bit_length()      # st[self.size + i] = array[i]
        self.tree = [self.e] * (self.size << 1)

    def built(self, array):
        """arrayを初期値とするセグメント木を構築"""
        for i in range(self.n):
            self.tree[self.size + i] = array[i]
        for i in range(self.size - 1, 0, -1):
            self.tree[i] = self.op(self.tree[i<<1], self.tree[(i<<1)|1])

    def update(self, i, x):
        """i 番目の要素を x に更新　(0-indexed) """
        i += self.size
        self.tree[i] = x
        while i > 1:
            i >>= 1
            self.tree[i] = self.op(self.tree[i<<1], self.tree[(i<<1)|1])

    def get(self, l, r):
        """ [l, r)の区間取得の結果を返す　(0-indexed) """
        l += self.size
        r += self.size
        res_l = self.e
        res_r = self.e
        while l < r:
            if l & 1:
                res_l = self.op(res_l, self.tree[l])
                l += 1
            if r & 1:
                r -= 1
                res_r = self.op(self.tree[r], res_r)
            l >>= 1
            r >>= 1
        return self.op(res_l, res_r)

    def max_right(self, l, f):
        """
        以下の条件を両方満たす r を(いずれか一つ)返す
            ・r = l or f(op(a[l], a[l + 1], ..., a[r - 1])) = true
            ・r = n or f(op(a[l], a[l + 1], ..., a[r])) = false
        """
        if l == self.n: return self.n
        l += self.size
        sm = self.e
        while True:
            while l % 2 == 0: l >>= 1
            if not f(self.op(sm, self.tree[l])):
                while l < self.size:
                    l = 2 * l
                    if f(self.op(sm, self.tree[l])):
                        sm = self.op(sm, self.tree[l])
                        l += 1
                return l - self.size
            sm = self.op(sm, self.tree[l])
            l += 1
            if (l & -l) == l: break
        return self.n

    def min_left(self, r, f):
        """
        以下の条件を両方満たす l を(いずれか一つ)返す
            ・l = r or f(op(a[l], a[l + 1], ..., a[r - 1])) = true
            ・l = 0 or f(op(a[l - 1], a[l], ..., a[r - 1])) = false
        """
        if r == 0: return 0
        r += self.size
        sm = self.e
        while True:
            r -= 1
            while r > 1 and (r % 2): r >>= 1
            if not f(self.op(self.tree[r], sm)):
                while r < self.size:
                    r = 2 * r + 1
                    if f(self.op(self.tree[r], sm)):
                        sm = self.op(self.tree[r], sm)
                        r -= 1
                return r + 1 - self.size
            sm = self.op(self.tree[r], sm)
            if (r & -r) == r: break
        return 0

    def __iter__(self):
        for a in self.tree[self.size:self.size+self.n]:
            yield a

    def __str__(self):
        return str(self.tree[self.size:self.size+self.n])

##################################################################################################################

N, Q = map(int, input().split())
A = list(map(int, input().split()))
e = 0
op = lambda x, y : x if x > y else y
st = SegmentTree(N, op, e)
st.built(A)
res = []

for _ in range(Q):
    t, x, y = map(int, input().split())
    if t == 1:
        st.update(x - 1, y)
    elif t == 2:
        res.append(st.get(x - 1, y))
    else:
        res.append(st.max_right(x - 1, lambda z: z < y) + 1)

print('\n'.join(map(str, res)))

Submission status: Accepted

Solution code:
"""
セグメント木
func：二項演算の関数（モノイドである必要あり）
e：単位元（モノイドにおける）
update, find, bisect, 全てにおいて, 1-indexとなっている。
(大抵の場合、Atcoderの問題文の表記のままの値を、メソッドに代入すれば良い)
"""
class SegmentTree:
    def __init__(self, n, func, e, arrange=None):
        self.init(n)
        self.func = func
        self.e = e
        self.make_arrange(arrange)
    
    def init(self, n):
        self.inf = pow(2, 32)
        self.n = n
        self.N = 1
        while self.N < self.n:
            self.N *= 2
        self.size = self.N * 2
    
    def make_arrange(self, arrange):
        self.set_arrange(arrange)
        self.construct(arrange)

    def set_arrange(self, arrange):
        if arrange == None:
            self.segment = [self.e]*(self.size)
            return
        self.segment = [0]*(self.N) + arrange + [self.e]*(self.size-self.N-self.n)
    
    def construct(self, arrange):
        if arrange == None:
            return
        for i in range(self.N-1, 0, -1):
            self.segment[i] = self.func(self.segment[2*i], self.segment[2*i+1])
    
    def update(self, i, x):
        i += (self.N - 1)
        self.segment[i] = x
        while i > 1:
            i = i//2
            self.segment[i] = self.func(self.segment[2*i], self.segment[2*i+1])
    

    def count(self, l, r):
        res = self.e
        l += self.N-1
        r += self.N
        while r > l:
            if l & 1:
                res = self.func(res, self.segment[l])
                l += 1
            if r & 1:
                r -= 1
                res = self.func(res, self.segment[r])
            l >>= 1
            r >>= 1
        return res
    

    def bisect_sub(self, a, b, k, l, r, x):
        if r <= a or b <= l:
            return b+1
        if self.segment[k] < x:
            return b+1
        if k >= self.N:
            return r
        
        find_l = self.bisect_sub(a, b, 2*k, l, (l+r)//2, x)
        if find_l <= b:
            return find_l

        find_r = self.bisect_sub(a, b, 2*k+1, (l+r)//2, r, x)
        return find_r

    def bisect(self, l, r, x):
        return self.bisect_sub(l-1, r, 1, 0, self.size-self.N, x)
    

def main():
    n, q = map(int, input().split())
    p = list(map(int, input().split()))
    seg = SegmentTree(n, max, 0, arrange=p)

    res = []
    for i in range(q):
        a, b, c = list(map(int, input().split()))
        if a == 1:
            seg.update(b, c)
        elif a == 2:
            ans = seg.count(b, c)
            res.append(ans)
        else:
            ans = seg.bisect(b, n, c)
            res.append(ans)

    print(*res, sep="\n")
            

    
if __name__ == "__main__":
    main()

Submission status: Accepted

Solution code:
mod = 1000000007
eps = 10**-9
 
 
def main():
    import sys
    input = sys.stdin.buffer.readline
    
    #by tamatoさん
    class SegmentTree:
        def __init__(self, A, initialize=True, segfunc=min, ident=2000000000):
            self.N = len(A)
            self.LV = (self.N - 1).bit_length()
            self.N0 = 1 << self.LV
            self.segfunc = segfunc
            self.ident = ident
            if initialize:
                self.data = [self.ident] * self.N0 + A + [self.ident] * (self.N0 - self.N)
                for i in range(self.N0 - 1, 0, -1):
                    self.data[i] = segfunc(self.data[i * 2], self.data[i * 2 + 1])
            else:
                self.data = [self.ident] * (self.N0 * 2)
 
        def update(self, i, x):
            i += self.N0 - 1
            self.data[i] = x
            for _ in range(self.LV):
                i >>= 1
                self.data[i] = self.segfunc(self.data[i * 2], self.data[i * 2 + 1])
 
        # open interval [l, r)
        def query(self, l, r):
            l += self.N0 - 1
            r += self.N0 - 1
            ret_l = self.ident
            ret_r = self.ident
            while l < r:
                if l & 1:
                    ret_l = self.segfunc(ret_l, self.data[l])
                    l += 1
                if r & 1:
                    ret_r = self.segfunc(self.data[r - 1], ret_r)
                    r -= 1
                l >>= 1
                r >>= 1
            return self.segfunc(ret_l, ret_r)
 
        # return smallest i(l <= i < r) s.t. check(A[i]) == True
        def binsearch(self, l, r, check):
            if not check(self.query(l, r)):
                return r
            l += self.N0 - 1
            val = self.ident
            while True:
                if check(self.segfunc(val, self.data[l])):
                    break
                if l & 1:
                    val = self.segfunc(val, self.data[l])
                    l += 1
                l >>= 1
            while l < self.N0:
                newval = self.segfunc(val, self.data[l * 2])
                if not check(newval):
                    val = newval
                    l = (l << 1) + 1
                else:
                    l <<= 1
            return l - self.N0 + 1
 
    def check(val):
        return val >= v

    n,q = map(int,input().split())
    morty = list(map(int,input().split()))
    stree = SegmentTree(morty,segfunc=max,ident=-2000000000)

    for _ in range(q):
        t,x,v = map(int,input().split())
        if t == 1:
            stree.update(x, v)
        elif t==2:
            print(stree.query(x,v+1))
        else:
            print(stree.binsearch(x,n+1, check))
if __name__ == '__main__':
    main()

Submission status: Accepted

Solution code:
import sys
def input():
	return sys.stdin.buffer.readline()[:-1]

class SegTree():
	def __init__(self, a, func, idem):
		#the number of nodes is 2n-1
		self.n = 1 << len(a).bit_length()
		self.size = len(a)
		self.func = func
		self.idem = idem
		self.node = [idem] * (2*self.n-1)
		for i in range(len(a)+self.n-2, -1, -1):
			if i >= self.n-1:
				self.node[i] = a[i-self.n+1]
			else:
				self.node[i] = self.func(self.node[2*i+1], self.node[2*i+2])

	def get(self, x):
		return self.node[x+self.n-1]

	def set(self, x, val):
		x += self.n-1
		self.node[x] = val
		while x > 0:
			x = (x-1)>>1
			self.node[x] = self.func(self.node[2*x+1], self.node[2*x+2])
		return

	#[l, r)
	def prod(self, l, r):
		L, R = l+self.n, r+self.n
		s_l, s_r = self.idem, self.idem
		while L<R:
			if R & 1:
				R -= 1
				s_r = self.func(self.node[R-1], s_r)
			if L & 1:
				s_l = self.func(s_l, self.node[L-1])
				L += 1
			L >>= 1
			R >>= 1
		return self.func(s_l, s_r)

	#maximum of r (within [l, n)) which satisfies val(prod(l, r))
	def max_right(self, l, val):
		if l == self.size:
			return self.size
		L = l + self.n
		s = self.idem
		while True:
			while (L % 2 == 0):
				L >>= 1
			if not val(self.func(s, self.node[L-1])):
				while L < self.n:
					L <<= 1
					if val(self.func(s, self.node[L-1])):
						s = self.func(s, self.node[L-1])
						L += 1
				return L - self.n
			s = self.func(s, self.node[L-1])
			L += 1
			if (L & -L) == L:
				return self.size

	#minimum of l (within [0, r)) which satisfies val(prod(l, r))
	def min_left(self, r, val):
		if r == 0:
			return 0
		R = r + self.n
		s = self.idem
		while True:
			R -= 1
			while R > 1 and R % 2:
				R >>= 1
			if not val(self.func(self.node[R-1], s)):
				while R < self.n:
					R = (R << 1) + 1
					if val(self.func(self.node[R-1], s)):
						s = self.func(self.node[R-1], s)
						R -= 1
				return R + 1 - self.n
			s = self.func(self.node[R-1], s)
			if (R & -R) == R:
				return 0


n, q = map(int, input().split())
a = list(map(int, input().split()))
st = SegTree(a, max, 0)
for _ in range(q):
	t, x, y = map(int, input().split())
	if t == 1:
		st.set(x-1, y)
	elif t == 2:
		print(st.prod(x-1, y))
	else:
		print(st.max_right(x-1, lambda z: z < y) + 1)

Submission status: Accepted

Solution code:
from typing import Callable, List, TypeVar

S = TypeVar("S")


class SegmentTree:
    """Segment Tree
    References:
        https://github.com/atcoder/ac-library/blob/master/atcoder/segtree.hpp
    """

    __slots__ = ["e", "op", "_n", "_size", "tree"]

    def __init__(self, a: List[S], e: S, op: Callable[[S, S], S]) -> None:
        self.e = e
        self.op = op
        self._n = len(a)
        self._size = 1 << (self._n - 1).bit_length()

        self.tree = [e] * self._size + a + [e] * (self._size - self._n)
        for i in range(self._size - 1, 0, -1):
            self._update(i)

    def _update(self, k: int) -> None:
        """Update the value of a[k]."""
        self.tree[k] = self.op(self.tree[2 * k], self.tree[2 * k + 1])

    def set(self, k: int, x: S) -> None:
        """Assign x to a[k] in O(log n)."""
        assert 0 <= k < self._n

        k += self._size
        self.tree[k] = x
        while k:
            k >>= 1
            self._update(k)

    def get(self, k: int) -> S:
        """Return a[k] in O(1)."""
        assert 0 <= k < self._n
        return self.tree[k + self._size]

    def prod(self, l: int, r: int) -> S:
        """Return op(a[l], ..., a[r - 1]). Return e, if l == r.
        Complexity: O(log n)
        """
        assert 0 <= l <= r <= self._n

        sml, smr = self.e, self.e
        l += self._size
        r += self._size

        while l < r:
            if l & 1:
                sml = self.op(sml, self.tree[l])
                l += 1
            if r & 1:
                r -= 1
                smr = self.op(self.tree[r], smr)
            l >>= 1
            r >>= 1
        return self.op(sml, smr)

    def prod_all(self) -> S:
        """Return op(a[0], ..., a[n - 1]. Return e if n == 0.
        Complexity: O(1)
        """
        return self.tree[1]

    def max_right(self, l: int, f: Callable[[S], bool]) -> int:
        """
        Return an index r satisfying both:
            1. r = l or f(op(a[l], a[l + 1], ..., a[r - 1])) = true
            2. r = n or f(op(a[l], a[l + 1], ..., a[r])) = false.

        If f is monotone, this is the maximum r satisfying:
            f(op(a[l], a[l + 1], ..., a[r - 1])) = true.

        Complexity: O(log n)
        """
        assert 0 <= l <= self._n
        assert f(self.e)

        if l == self._n:
            return self._n

        l += self._size
        sm = self.e

        while True:
            while not l & 1:
                l >>= 1

            if not f(self.op(sm, self.tree[l])):
                while l < self._size:
                    l *= 2
                    if f(self.op(sm, self.tree[l])):
                        sm = self.op(sm, self.tree[l])
                        l += 1
                return l - self._size

            sm = self.op(sm, self.tree[l])
            l += 1

            if (l & -l) == l:
                break

        return self._n

    def min_left(self, r: int, f: Callable[[S], bool]) -> int:
        """
        Return an index l satisfying both:
            1. l = r or f(op(a[l], a[l + 1], ..., a[r - 1])) = true
            2. l = 0 or f(op(a[l - 1], a[l + 1], ..., a[r - 1])) = false.
        If f is monotone, this is the minimum l satisfying:
            f(op(a[l], a[l + 1], ..., a[r - 1])) = true.

        Complexity: O(log n)
        """
        assert 0 <= r <= self._n
        assert f(self.e)

        if not r:
            return 0

        r += self._size
        sm = self.e

        while True:
            r -= 1
            while r > 1 and r & 1:
                r >>= 1

            if not f(self.op(self.tree[r], sm)):
                while r < self._size:
                    r = 2 * r + 1
                    if f(self.op(self.tree[r], sm)):
                        sm = self.op(self.tree[r], sm)
                        r -= 1
                return r + 1 - self._size

            if (r & -r) == r:
                break

        return 0


def practice2_j():
    # https://atcoder.jp/contests/practice2/tasks/practice2_j
    import sys

    readline = sys.stdin.readline

    N, Q = map(int, readline().split())
    (*A,) = map(int, readline().split())
    tree = SegmentTree(A, -1, max)
    res = []
    for _ in range(Q):
        t, x, y = map(int, readline().split())
        if t == 1:
            tree.set(x - 1, y)
        elif t == 2:
            res.append(tree.prod(x - 1, y))
        else:
            res.append(tree.max_right(x - 1, lambda n: n < y) + 1)
    print("\n".join(map(str, res)))


if __name__ == "__main__":
    practice2_j()

Submission status: Accepted

Solution code:
from typing import Callable, List, TypeVar

S = TypeVar("S")


class SegmentTree:
    """Segment Tree
    References:
        https://github.com/atcoder/ac-library/blob/master/atcoder/segtree.hpp
    """

    __slots__ = ["e", "op", "_n", "_size", "tree"]

    def __init__(self, a: List[S], e: S, op: Callable[[S, S], S]) -> None:
        self.e = e
        self.op = op
        self._n = len(a)
        self._size = 1 << (self._n - 1).bit_length()

        self.tree = [e] * self._size + a + [e] * (self._size - self._n)
        for i in range(self._size - 1, 0, -1):
            self._update(i)

    def _update(self, k: int) -> None:
        """Update the value of a[k]."""
        self.tree[k] = self.op(self.tree[2 * k], self.tree[2 * k + 1])

    def set(self, k: int, x: S) -> None:
        """Assign x to a[k] in O(log n)."""
        assert 0 <= k < self._n

        k += self._size
        self.tree[k] = x
        while k:
            k >>= 1
            self._update(k)

    def get(self, k: int) -> S:
        """Return a[k] in O(1)."""
        assert 0 <= k < self._n
        return self.tree[k + self._size]

    def prod(self, l: int, r: int) -> S:
        """Return op(a[l], ..., a[r - 1]). Return e, if l == r.
        Complexity: O(log n)
        """
        assert 0 <= l <= r <= self._n

        sml, smr = self.e, self.e
        l += self._size
        r += self._size

        while l < r:
            if l & 1:
                sml = self.op(sml, self.tree[l])
                l += 1
            if r & 1:
                r -= 1
                smr = self.op(self.tree[r], smr)
            l >>= 1
            r >>= 1
        return self.op(sml, smr)

    def prod_all(self) -> S:
        """Return op(a[0], ..., a[n - 1]. Return e if n == 0.
        Complexity: O(1)
        """
        return self.tree[1]

    def max_right(self, l: int, f: Callable[[S], bool]) -> int:
        """
        Return an index r satisfying both:
            1. r = l or f(op(a[l], a[l + 1], ..., a[r - 1])) = true
            2. r = n or f(op(a[l], a[l + 1], ..., a[r])) = false.

        If f is monotone, this is the maximum r satisfying:
            f(op(a[l], a[l + 1], ..., a[r - 1])) = true.

        Complexity: O(log n)
        """
        assert 0 <= l <= self._n
        assert f(self.e)

        if l == self._n:
            return self._n

        l += self._size
        sm = self.e

        while True:
            while not l & 1:
                l >>= 1

            if not f(self.op(sm, self.tree[l])):
                while l < self._size:
                    l *= 2
                    if f(self.op(sm, self.tree[l])):
                        sm = self.op(sm, self.tree[l])
                        l += 1
                return l - self._size

            sm = self.op(sm, self.tree[l])
            l += 1

            if (l & -l) == l:
                break

        return self._n

    def min_left(self, r: int, f: Callable[[S], bool]) -> int:
        """
        Return an index l satisfying both:
            1. l = r or f(op(a[l], a[l + 1], ..., a[r - 1])) = true
            2. l = 0 or f(op(a[l - 1], a[l + 1], ..., a[r - 1])) = false.
        If f is monotone, this is the minimum l satisfying:
            f(op(a[l], a[l + 1], ..., a[r - 1])) = true.

        Complexity: O(log n)
        """
        assert 0 <= r <= self._n
        assert f(self.e)

        if not r:
            return 0

        r += self._size
        sm = self.e

        while True:
            r -= 1
            while r > 1 and r & 1:
                r >>= 1

            if not f(self.op(self.tree[r], sm)):
                while r < self._size:
                    r = 2 * r + 1
                    if f(self.op(self.tree[r], sm)):
                        sm = self.op(self.tree[r], sm)
                        r -= 1
                return r + 1 - self._size

            if (r & -r) == r:
                break

        return 0


# https://atcoder.jp/contests/practice2/tasks/practice2_j
import sys

readline = sys.stdin.buffer.readline

N, Q = map(int, readline().split())
(*A,) = map(int, readline().split())
tree = SegmentTree(A, -1, max)
res = []
for _ in range(Q):
    t, x, y = map(int, readline().split())
    if t == 1:
        tree.set(x - 1, y)
    elif t == 2:
        res.append(tree.prod(x - 1, y))
    else:
        res.append(tree.max_right(x - 1, lambda n: n < y) + 1)
print("\n".join(map(str, res)))