Competitive programming is essential and today we are going to solve the round bounded in circle leetcode problem. This problem appeared in amazon coding interview.
The question states that:-
You are given the strings
message, which represent a cipher key and a secret message, respectively. The steps to decode
message are as follows:
keyas the order of the substitution table.
messageis then substituted using the table.
' 'are transformed to themselves.
key = "happy boy"(actual key would have at least one instance of each letter in the alphabet), we have the partial substitution table of (
'h' -> 'a',
'a' -> 'b',
'p' -> 'c',
'y' -> 'd',
'b' -> 'e',
'o' -> 'f').
Return the decoded message.
Input: key = "the quick brown fox jumps over the lazy dog", message = "vkbs bs t suepuv" Output: "this is a secret" Explanation: The diagram above shows the substitution table. It is obtained by taking the first appearance of each letter in "the quick brown fox jumps over the lazy dog".
Input: key = "eljuxhpwnyrdgtqkviszcfmabo", message = "zwx hnfx lqantp mnoeius ycgk vcnjrdb" Output: "the five boxing wizards jump quickly" Explanation: The diagram above shows the substitution table. It is obtained by taking the first appearance of each letter in "eljuxhpwnyrdgtqkviszcfmabo".
26 <= key.length <= 2000
keyconsists of lowercase English letters and
keycontains every letter in the English alphabet (
'z') at least once.
1 <= message.length <= 2000
messageconsists of lowercase English letters and
So, Here first we will create a list of the alphabet key in order for the list index plus 97 to behave like the ASCII value of the list key value.
then we will run the loop for each index of message find its index and then plus 97 to their index find the chr accourding to the ASCII value keep concatenate with ans and return after the loop finishes.
class Solution: def decodeMessage(self, key: str, message: str) -> str: li =  for i in key: if i != ' ' and i not in li: li.append(i) ans = '' for i in message: if i==' ': ans+=' ' else: ans+=chr(97+li.index(i)) return ans
Time complexity: O(n)
Space complexity: O(26)