3201. 找出有效子序列的最大长度 I

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3201. 找出有效子序列的最大长度 I
- 题意
- 思路
- 代码
- 拓展

3201. 找出有效子序列的最大长度 I

题意

给你一个整数数组 nums。

nums 的子序列 sub 的长度为 x ，如果其满足以下条件，则称其为有效子序列：

(sub[0] + sub[1]) % 2 == (sub[1] + sub[2]) % 2 == … == (sub[x - 2] + sub[x - 1]) % 2
返回 nums 的最长的有效子序列的长度。

一个子序列指的是从原数组中删除一些元素（也可以不删除任何元素），剩余元素保持原来顺序组成的新数组。

思路

贪心：最长有效子序列有三种可能：
- 全都是奇元素
- 全都是偶元素
- 奇偶元素交替
  代码注意：
- odd := sum(x & 1 for x in nums), 是海象表达式，赋值的同时返回值
- sum((x & 1) ^ (y & 1) for x, y in std.pairwise(nums)) 这句话非常的巧妙，首先获取每一对相邻的元素，然后 (x & 1) ^ (y & 1) 会比较 x 和 y 的奇偶性，算和是计算所有相邻元素对的奇偶性差异
  - 比如 2 1 3 3 4 6，生成 [(2, 1), (1, 3), (3, 3), (3, 4), (4, 6)]，其中只有 (2, 1), (3, 4) 是奇偶性差异的 pair
    - 当出现 (0, 1) 的时候，如果再出现一个 (0, 1)，期间一定会夹杂着一个 (1, 0) 对
  - 但显然这个是 (0, 1) (1, 0)，即使 1 和 3 不同，但是我们可以选择一个，那么序列就可以为 214 或者 234，长度为 2 + 1 = 3
  - 所以 1 + sum((x & 1) ^ (y & 1) for x, y in pairwise(nums)) 等价于 奇偶交替的最长子序列长度
dfs + 记忆化搜索
为了考虑所有可能的初始状态，代码调用 dfs 函数四次，分别对应四种不同的初始状态：

dfs(0, 0, 1)：从索引 0 开始，初始奇偶性为偶数，需要的奇偶性变化为奇
dfs(0, 0, 0)：从索引 0 开始，初始奇偶性为偶数，需要的奇偶性变化为偶
dfs(0, 1, 0)：从索引 0 开始，初始奇偶性为奇数，需要的奇偶性变化为偶
dfs(0, 1, 1)：从索引 0 开始，初始奇偶性为奇数，需要的奇偶性变化为奇

其实也是上面一样的思路，只不过用的dfs
lru_cache 类似 cache，可以实现记忆化搜索

代码

'''
Author: NEFU AB-IN
Date: 2024-06-30 10:30:18
FilePath: \LeetCode\CP404\b\b.py
LastEditTime: 2024-07-01 20:39:53
'''
# 3.8.19 import
from bisect import bisect_left, bisect_right
from collections import Counter, defaultdict, deque
from datetime import datetime, timedelta
from functools import lru_cache
from heapq import heapify, heappop, heappush, nlargest, nsmallest
from itertools import combinations, compress, permutations, starmap, tee
from math import ceil, fabs, floor, gcd, log, sqrt
from string import ascii_lowercase, ascii_uppercase
from sys import exit, setrecursionlimit, stdin, stdout
from typing import Any, Dict, Generic, List, TypeVar, UnionTYPE = TypeVar('TYPE')# Data structure
class SA(Generic[TYPE]):def __init__(self, x: TYPE, y: TYPE):self.x = xself.y = ydef __lt__(self, other: 'SA[TYPE]') -> bool:return self.x < other.xdef __eq__(self, other: 'SA[TYPE]') -> bool:return self.x == other.x and self.y == other.ydef __repr__(self) -> str:return f'SA(x={self.x}, y={self.y})'# Constants
N = int(2e5 + 10)  # If using AR, modify accordingly
M = int(20)  # If using AR, modify accordingly
INF = int(2e9)
OFFSET = int(100)# Set recursion limit
setrecursionlimit(INF)# Read
def input(): return stdin.readline().rstrip("\r\n")  # Remove when Mutiple data
def read(): return map(int, input().split())
def read_list(): return list(read())# Func
class std:letter_to_num = staticmethod(lambda x: ord(x.upper()) - 65)  # A -> 0num_to_letter = staticmethod(lambda x: ascii_uppercase[x])  # 0 -> Aarray = staticmethod(lambda x=0, size=N: [x] * size)array2d = staticmethod(lambda x=0, rows=N, cols=M: [std.array(x, cols) for _ in range(rows)])max = staticmethod(lambda a, b: a if a > b else b)min = staticmethod(lambda a, b: a if a < b else b)removeprefix = staticmethod(lambda s, prefix: s[len(prefix):] if s.startswith(prefix) else s)removesuffix = staticmethod(lambda s, suffix: s[:-len(suffix)] if s.endswith(suffix) else s)@staticmethoddef find(container: Union[List[TYPE], str], value: TYPE):"""Returns the index of value in container or -1 if value is not found."""if isinstance(container, list):try:return container.index(value)except ValueError:return -1elif isinstance(container, str):return container.find(value)@staticmethoddef pairwise(iterable):"""Return successive overlapping pairs taken from the input iterable."""a, b = tee(iterable)next(b, None)return zip(a, b)# ————————————————————— Division line ——————————————————————class Solution:def maximumLength(self, nums: List[int]) -> int:return max(odd := sum(x & 1 for x in nums),len(nums) - odd,1 + sum((x & 1) ^ (y & 1) for x, y in std.pairwise(nums)))def maximumLength(self, nums: List[int]) -> int:@lru_cache(None)def dfs(index, expected_parity, original_parity):# 递归终止条件：如果索引达到数组末尾，返回0if index == len(nums):return 0# 当前元素奇偶性符合预期if nums[index] % 2 == expected_parity:# 将当前元素加入子序列，并继续递归下一个元素# 更新expected_parity为(original_parity - expected_parity) % 2return 1 + dfs(index + 1, (original_parity - expected_parity) % 2, original_parity)else:# 当前元素不符合预期，不加入子序列，继续递归下一个元素return dfs(index + 1, expected_parity, original_parity)# 考虑所有可能的初始状态return max(dfs(0, 0, 1), dfs(0, 0, 0), dfs(0, 1, 0), dfs(0, 1, 1))

拓展

当2变为k时，可以用dp解决

'''
Author: NEFU AB-IN
Date: 2024-06-30 10:30:18
FilePath: \LeetCode\CP404\c\c.py
LastEditTime: 2024-07-01 22:18:53
'''
# 3.8.19 import
from bisect import bisect_left, bisect_right
from collections import Counter, defaultdict, deque
from datetime import datetime, timedelta
from functools import lru_cache
from heapq import heapify, heappop, heappush, nlargest, nsmallest
from itertools import combinations, compress, permutations, starmap, tee
from math import ceil, fabs, floor, gcd, log, sqrt
from string import ascii_lowercase, ascii_uppercase
from sys import exit, setrecursionlimit, stdin, stdout
from typing import Any, Dict, Generic, List, TypeVar, UnionTYPE = TypeVar('TYPE')# Data structure
class SA(Generic[TYPE]):def __init__(self, x: TYPE, y: TYPE):self.x = xself.y = ydef __lt__(self, other: 'SA[TYPE]') -> bool:return self.x < other.xdef __eq__(self, other: 'SA[TYPE]') -> bool:return self.x == other.x and self.y == other.ydef __repr__(self) -> str:return f'SA(x={self.x}, y={self.y})'# Constants
N = int(2e5 + 10)  # If using AR, modify accordingly
M = int(20)  # If using AR, modify accordingly
INF = int(2e9)
OFFSET = int(100)# Set recursion limit
setrecursionlimit(INF)# Read
def input(): return stdin.readline().rstrip("\r\n")  # Remove when Mutiple data
def read(): return map(int, input().split())
def read_list(): return list(read())# Func
class std:letter_to_num = staticmethod(lambda x: ord(x.upper()) - 65)  # A -> 0num_to_letter = staticmethod(lambda x: ascii_uppercase[x])  # 0 -> Aarray = staticmethod(lambda x=0, size=N: [x] * size)array2d = staticmethod(lambda x=0, rows=N, cols=M: [std.array(x, cols) for _ in range(rows)])max = staticmethod(lambda a, b: a if a > b else b)min = staticmethod(lambda a, b: a if a < b else b)removeprefix = staticmethod(lambda s, prefix: s[len(prefix):] if s.startswith(prefix) else s)removesuffix = staticmethod(lambda s, suffix: s[:-len(suffix)] if s.endswith(suffix) else s)@staticmethoddef find(container: Union[List[TYPE], str], value: TYPE):"""Returns the index of value in container or -1 if value is not found."""if isinstance(container, list):try:return container.index(value)except ValueError:return -1elif isinstance(container, str):return container.find(value)@staticmethoddef pairwise(iterable):"""Return successive overlapping pairs taken from the input iterable."""a, b = tee(iterable)next(b, None)return zip(a, b)# ————————————————————— Division line ——————————————————————class Solution:def maximumLength(self, nums: List[int], k: int) -> int:n = len(nums)dp = std.array2d(0, n, k + OFFSET)if n <= 2:return nfor j in range(k):dp[0][j] = 1res = 0for i in range(n):for j in range(i):dp[i][(nums[i] + nums[j]) % k] = std.max(dp[i][(nums[i] + nums[j]) % k], dp[j][(nums[i] + nums[j]) % k] + 1)res = std.max(res, max(dp[i]))return res