Daily Challenge #235 - Reversing a Process

#challenge

Suppose we know the process A by which a string s has been coded to a string r.

Implement a function that will decode r to get back the original string s.

Explanation of the known process A:
data: a string s composed of lowercase letters from a to z and a positive integer num
we know there is a correspondence between abcde...uvwxyz and 0, 1, 2 ..., 23, 24, 25 : 0 <-> a, 1 <-> b ...
If c is a character of s whose corresponding number is x, apply to x the function f: x-> f(x) = num * x % 26, then find ch the corresponding character of f(x).
Accumulate all these ch in a string r.
concatenate num and r and return the result.

code("mer", 6015) -> "6015ekx"
m <-> 12, 12 * 6015 % 26 == 4, 4 <-> e
e <-> 4, 4 * 6015 % 26 == 10, 10 <-> k
r <-> 17, 17 * 6015 % 26 == 23, 23 <-> x
We get "ekx" so the answer is: "6015ekx"

Task
A string s has been coded to a string r by the above process (A). Write a function decode(r) to get back s whenever it is possible.

Indeed it can happen that the decoding is impossible when positive integer num has not been correctly chosen. In that case return "Impossible to decode".

Example:
decode("6015ekx") -> "mer"
decode("5057aan") -> "Impossible to decode"

Tests:
decode("1273409kuqhkoynvvknsdwljantzkpnmfgf")
decode("761328qockcouoqmoayqwmkkic")

Good luck!

This challenge comes from g964 on CodeWars. Thank you to CodeWars, who has licensed redistribution of this challenge under the 2-Clause BSD License!

Want to propose a challenge idea for a future post? Email yo+challenge@dev.to with your suggestions!

Top comments (2)

'' • Apr 30 '20 • Edited

"Not correcly chosen" should be clarified.

ES6


  const alphabet = "abcdefghijklmnopqrstuvwxyz"

  const decode = (encoded) => {
    let number = parseInt(encoded.split(/\D/)[0])
    let text = encoded.substring(("" + number).length)
    let result = ""
    for(let i=0;i<text.length;i++) {
        let encodedIndex = alphabet.indexOf(text[i])
        for(let j=0;j<26;j++) {
            if((j*number - encodedIndex) % 26 === 0) {
                result += alphabet[j]
                break;
            }
        }
    }

    return result;
}