3291. 形成目标字符串需要的最少字符串数 I

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3291. 形成目标字符串需要的最少字符串数 I

题意

给你一个字符串数组 words 和一个字符串 target。

如果字符串 x 是 words 中 任意 字符串的前缀，则认为 x 是一个 有效 字符串。

现计划通过 连接 有效字符串形成 target ，请你计算并返回需要连接的 最少 字符串数量。如果无法通过这种方式形成 target，则返回 -1。

思路

1. Trie + 记忆化搜索
   首先就是把所有的word全部放进trie，然后进行记忆化搜索即可
   从target的初始下标开始，去trie中匹配，如果到j匹配上了，然后就dfs(j + 1)，如果匹配不上，直接break
   记录每个状态的能被最少几个前缀组成，每次用之后的dfs的值更新这个值即可

代码

# 3.8.9 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# 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 Std:class TrieNode:"""TrieNode class can quickly process string prefixes, a common feature used in applications like autocomplete and spell checking."""_sid_cnt = 0  # sid counter, representing string index starting from 0def __init__(self):"""Initialize children dictionary and cost. The trie tree is a 26-ary tree."""self.children_ = {}self._sid = -1  # Unique ID for the node, -1 if not assigneddef add(self, word: str) -> int:"""Add a word to the trie with the associated cost and return a unique ID."""node = selffor c in word:if c not in node.children_:node.children_[c] = Std.TrieNode()node = node.children_[c]return node._sid# ————————————————————— Division line ——————————————————————class Solution:def minValidStrings(self, words: List[str], target: str) -> int:n = len(target)trie = Std.TrieNode()for word in words:trie.add(word)@lru_cache(None)def dfs(i: int) -> int:nonlocal trieif i == n:return 0res = INFnode = triefor j in range(i, n):c = target[j]if c not in node.children_:breaksub_res = dfs(j + 1)res = Math.min(res, sub_res + 1)node = node.children_[c]return resresult = dfs(0)dfs.cache_clear()return result if result != INF else -1