LeetCode 솔루션 분류
[8/21] 936. Stamping The Sequence
본문
Hard
738139Add to ListShareYou are given two strings stamp
and target
. Initially, there is a string s
of length target.length
with all s[i] == '?'
.
In one turn, you can place stamp
over s
and replace every letter in the s
with the corresponding letter from stamp
.
- For example, if
stamp = "abc"
andtarget = "abcba"
, thens
is"?????"
initially. In one turn you can:- place
stamp
at index0
ofs
to obtain"abc??"
, - place
stamp
at index1
ofs
to obtain"?abc?"
, or - place
stamp
at index2
ofs
to obtain"??abc"
.
stamp
must be fully contained in the boundaries ofs
in order to stamp (i.e., you cannot placestamp
at index3
ofs
). - place
We want to convert s
to target
using at most 10 * target.length
turns.
Return an array of the index of the left-most letter being stamped at each turn. If we cannot obtain target
from s
within 10 * target.length
turns, return an empty array.
Example 1:
Input: stamp = "abc", target = "ababc" Output: [0,2] Explanation: Initially s = "?????". - Place stamp at index 0 to get "abc??". - Place stamp at index 2 to get "ababc". [1,0,2] would also be accepted as an answer, as well as some other answers.
Example 2:
Input: stamp = "abca", target = "aabcaca" Output: [3,0,1] Explanation: Initially s = "???????". - Place stamp at index 3 to get "???abca". - Place stamp at index 0 to get "abcabca". - Place stamp at index 1 to get "aabcaca".
Constraints:
1 <= stamp.length <= target.length <= 1000
stamp
andtarget
consist of lowercase English letters.
Accepted
29,418
Submissions
51,998
관련자료
-
링크
댓글 1
학부유학생님의 댓글
- 익명
- 작성일
Runtime: 81 ms, faster than 97.01% of Python3 online submissions for Stamping The Sequence.
Memory Usage: 15.4 MB, less than 37.31% of Python3 online submissions for Stamping The Sequence.
Memory Usage: 15.4 MB, less than 37.31% of Python3 online submissions for Stamping The Sequence.
class Solution:
def movesToStamp(self, stamp: str, target: str) -> List[int]:
res = []
combs = set()
for i in range(len(stamp)):
for j in range(len(stamp)-i):
combs.add('#'*i + stamp[i:len(stamp)-j] + '#'*j)
# print(combs)
finished = '#'*len(target)
while target!=finished:
found = False
for i in range(len(target)-len(stamp),-1,-1):
if target[i:i+len(stamp)] in combs:
target = target[:i] + '#'*len(stamp) + target[i+len(stamp):]
res.append(i)
found = True
if not found: return []
return res[::-1]