Powered by:NEFU AB-IN
文章目录
3113. 边界元素是最大值的子数组数目
题意
给你一个 正 整数数组 nums 。
请你求出 nums 中有多少个子数组,满足子数组中 第一个 和 最后一个 元素都是这个子数组中的 最大 值。
思路
- st表
找每个数x出现的下标,并判断当x和前一个x之间的最大值等于x时,记录下标;如果不是x,说明区间断了,那么就统计之前记录的下标,结果为n * (n + 1) // 2
,n为下标数组长度,并清空数组,加入新下标。最后将所有的非空下标数组都做一遍统计 - 单调栈
https://leetcode.cn/problems/find-the-number-of-subarrays-where-boundary-elements-are-maximum/solutions/2738894/on-dan-diao-zhan-pythonjavacgo-by-endles-y00d/
利用单调栈(处理两边比自己大或小的第一个元素)去清理掉垃圾数据,维护一个底大顶小的单调栈,记录元素及其出现次数。对栈顶元素判断大小
代码
python
'''
Author: NEFU AB-IN
Date: 2024-08-01 19:55:47
FilePath: \LeetCode\3113\3113.py
LastEditTime: 2024-08-01 20:49:33
'''
# 3.8.19 import
import random
from collections import Counter, defaultdict, deque
from datetime import datetime, timedelta
from functools import lru_cache, reduce
from heapq import heapify, heappop, heappush, nlargest, nsmallest
from itertools import combinations, compress, permutations, starmap, tee
from math import ceil, comb, fabs, floor, gcd, hypot, log, perm, sqrt
from string import ascii_lowercase, ascii_uppercase
from sys import exit, setrecursionlimit, stdin
from typing import Any, Callable, Dict, List, Optional, Tuple, TypeVar, Union
# Constants
TYPE = TypeVar('TYPE')
N = int(2e5 + 10)
M = int(20)
INF = int(1e12)
OFFSET = int(100)
MOD = int(1e9 + 7)
# Set recursion limit
setrecursionlimit(int(2e9))
class Arr:
array = staticmethod(lambda x=0, size=N: [x() if callable(x) else x for _ in range(size)])
array2d = staticmethod(lambda x=0, rows=N, cols=M: [Arr.array(x, cols) for _ in range(rows)])
graph = staticmethod(lambda size=N: [[] for _ in range(size)])
class Math:
max = staticmethod(lambda a, b: a if a > b else b)
min = staticmethod(lambda a, b: a if a < b else b)
class IO:
input = staticmethod(lambda: stdin.readline().rstrip("\r\n"))
read = staticmethod(lambda: map(int, IO.input().split()))
read_list = staticmethod(lambda: list(IO.read()))
class Std:
class SparseTable:
def __init__(self, data: list, func=lambda x, y: x | y):
"""Initialize the Sparse Table with the given data and function."""
self.func = func
self.st = [list(data)]
i, n = 1, len(self.st[0])
while 2 * i <= n:
pre = self.st[-1]
self.st.append([func(pre[j], pre[j + i]) for j in range(n - 2 * i + 1)])
i <<= 1
def query(self, begin: int, end: int):
"""Query the combined result over the interval [begin, end]."""
lg = (end - begin + 1).bit_length() - 1
return self.func(self.st[lg][begin], self.st[lg][end - (1 << lg) + 1])
# --------------------------------------------------------------- Division line ------------------------------------------------------------------
class Solution:
def numberOfSubarrays(self, nums: List[int]) -> int:
st = Std.SparseTable(nums, Math.max)
index_list_dict = defaultdict(list)
res = 0
for i, num in enumerate(nums):
list_ = index_list_dict[num]
if list_:
pre_id = list_[-1]
if st.query(pre_id, i) == num:
list_.append(i)
else:
res += len(list_) * (len(list_) + 1) // 2
list_.clear()
list_.append(i)
else:
list_.append(i)
for _, list_ in index_list_dict.items():
res += len(list_) * (len(list_) + 1) // 2
return res
python
class Solution:
def numberOfSubarrays(self, nums: List[int]) -> int:
ans = len(nums)
st = [[inf, 0]] # 无穷大哨兵
for x in nums:
while x > st[-1][0]:
st.pop()
if x == st[-1][0]:
ans += st[-1][1]
st[-1][1] += 1
else:
st.append([x, 1])
return ans