LeetCode 2325. Decrypt message (map)

1. subject

Here is the string key and message , Represent an encryption key and an encrypted message respectively .
Decrypt message The steps are as follows ：

1. Use key in 26 English lowercase letters for the first time The order of occurrence is used as the letter in the replacement table The order .
2. Align the replacement table with the common English alphabet , Form a comparison table .
3. According to the comparison table Replace message Every letter in .
4. Space ' ' remain unchanged .

for example ,key = "happy boy"（ The actual encryption key will contain every letter in the alphabet At least once ）, Accordingly , You can get some comparison tables （'h' -> 'a'、'a' -> 'b'、'p' -> 'c'、'y' -> 'd'、'b' -> 'e'、'o' -> 'f'）.
Returns the decrypted message .

Example 1：

 Input ：key = "the quick brown fox jumps over the lazy dog", message = "vkbs bs t suepuv"
 Output ："this is a secret"
 explain ： The comparison table is shown in the above figure .
 extract  "the quick brown fox jumps over the lazy dog"  The first occurrence of each letter in can get a replacement table .

Example 2：

 Input ：key = "eljuxhpwnyrdgtqkviszcfmabo", message = "zwx hnfx lqantp mnoeius ycgk vcnjrdb"
 Output ："the five boxing wizards jump quickly"
 explain ： The comparison table is shown in the above figure .
 extract  "eljuxhpwnyrdgtqkviszcfmabo"  The first occurrence of each letter in can get a replacement table .
 
 Tips ：
26 <= key.length <= 2000
key  From lowercase English letters and  ' '  form 
key  Contains every character in the English alphabet （'a'  To  'z'） At least once 
1 <= message.length <= 2000
message  Consists of lowercase letters and  ' '  form 

source ： Power button （LeetCode） link ：https://leetcode.cn/problems/decode-the-message
Copyright belongs to the network . For commercial reprint, please contact the official authority , Non-commercial reprint please indicate the source .

2. Problem solving

• Imitate according to the meaning of the topic

class Solution:
    def decodeMessage(self, key: str, message: str) -> str:
        c = set()
        chars = [chr(ord('a')+i) for i in range(26)]
        idx = 0
        d = {
    }
        for x in key:
            if x not in c and x != ' ':
                c.add(x)
                d[x] = chars[idx]
                idx += 1
        return ''.join([d[x] if x in d else x for x in message])

32 ms 15.1 MB Python3

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