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", thensis"?????"initially. In one turn you can:- place stampat index0ofsto obtain"abc??",
- place stampat index1ofsto obtain"?abc?", or
- place stampat index2ofsto obtain"??abc".
 stampmust be fully contained in the boundaries ofsin order to stamp (i.e., you cannot placestampat index3ofs).
- 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
- stampand- targetconsist 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] 
								 
							







