Decode the Message Leetcode Python Solution

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admin July 4, 2022

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 key and message, which represent a cipher key and a secret message, respectively. The steps to decode message are as follows:

  1. Use the first appearance of all 26 lowercase English letters in key as the order of the substitution table.
  2. Align the substitution table with the regular English alphabet.
  3. Each letter in message is then substituted using the table.
  4. Spaces ' ' are transformed to themselves.

Return the decoded message.

 

Example 1:

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".

Example 2:

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".

 

Constraints:

Source: Leetcode

Solution:

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.

Python solution:

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)

 

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