贴纸拼单词
https://www.lintcode.com/problem/1081/?utm_source=sc-libao-ql
描述
给出N种不同类型的贴纸。 每个贴纸上都写有一个小写英文单词。
通过裁剪贴纸上的所有字母并重排序来拼出字符串target
。
每种贴纸可以使用多次,假定每种贴纸数量无限。
拼出target
最少需要多少张贴纸?如果不可能拼成,则返回-1。
stickers
的长度范围为[1, 50]
.stickers
仅由小写英文字母组成(没有撇号'
).target
的长度范围为[1, 15]
,而且由小写字母组成。在所有的测试用例中,单词是从1000个最常见的美国英文单词中任意选择的,目标单词则是由两个任意单词拼接而成。
时间限制可能比以往更有挑战性,期望的时间界是平均35毫秒50个测试样例。
样例
样例 1:
输入:["with", "example", "science"], "thehat"输出:3解释:使用两张"with"和一张"example",经过裁剪和重排字母,可以得到"thehat"。这也是所需要的最少贴纸数。
样例 2:
输入:["notice", "possible"], "basicbasic"输出:-1解释:无法拼成"basicbasic"。
解题思路:
dfs+memo, 遍历可用的stickers,可重复使用sticker。从target去掉重叠字母,得到remaining word,继续dfs.
status: 剩余的单词,利用当前单词remaining_counter = Counter(target)-Counter(sticker), 得到剩余的{remaining_chars: count}, 组合成新单词,注意这里char可能有重复,所以需要char*count
new_remaining = ''.join(char * new_remaining_counter[char] for char in new_remaining_counter)
出口:remaining word为空,返回0
注意
@lru_cache(None)
```python
from typing import (
List,
)
from collections import Counter
from functools import lru_cache # Import lru_cache
class Solution:
"""
@param stickers: a string array
@param target: a string
@return: the minimum number of stickers that you need to spell out the target
"""
def min_stickers(self, stickers: List[str], target: str) -> int:
# Write your code here
sticker_counters = [Counter(i) for i in stickers]
@lru_cache(None)
def dp(remaining: str) -> int:
if not remaining:
return 0
min_stickers = float('inf')
remaining_counter = Counter(remaining)
for sticker_counter in sticker_counters:
if sticker_counter[remaining[0]] == 0:
continue
new_remaining_counter = remaining_counter-sticker_counter #需要char* 为了防止漏掉重复的char
new_remaining = ''.join(char * new_remaining_counter[char] for char in new_remaining_counter)
min_stickers = min(min_stickers,1+dp(new_remaining))
return min_stickers
result = dp(target)
return result if result != float('inf') else -1
```
也可以用BFS+bit做,用target字符位置是否设置作为bit status, 存入q。 然后遍历sticker,设置target的字符位。如果当前状态已经遍历过则略过,否则存入q。当target所有位置都遍历过,则直接返回。
因为bfs层遍历,从使用一个sticker, 二个stikcer, 三个sticker。。。先到的status就是使用sticker最少的。
```python
from typing import (
List,
)
from collections import deque,Counter
class Solution:
"""
@param stickers: a string array
@param target: a string
@return: the minimum number of stickers that you need to spell out the target
"""
def min_stickers(self, stickers: List[str], target: str) -> int:
# Write your code here
q = deque([0]) #should be a list not a set {}
dist = {0:0}
n = len(target)
final_state = 1<<n # no empty space, NOT 1 >> n
while q:
now = q.popleft()
for sticker in stickers:
sticker_counter = Counter(sticker)
state = now #should initialize to now
for i,c in enumerate(target):
if not now&1<<i and sticker_counter[c] > 0:
state |= 1<<i
sticker_counter[c] -= 1
if state in dist:
continue
q.append(state)
dist[state] = dist[now] + 1
if state == final_state -1:
return dist[state]
return -1
```
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