Reputation: 493
I have worked on this for many hours trying to solve it on my own but I came to the conclusion that I need help after several hours of hitting a dead end. I know very close but can't get rid of the spaces between words. here is my code:
function rot13(encodedStr) {
var codeArr = encodedStr.split("");
var decodedArr = [];
var letters = ["A", "B", "C", "D", "E", "F","G", "H", "I", "J", "K", "L", "M", "N", "O", "P", "Q", "R", "S", "T", "U", "V", "W", "X", "Y", "Z"];
var space = " ";
var location = 0;
var temp = [];
for(var i = 0; i<codeArr.length; i++){
if(letters[i] !== " "){
location = letters.indexOf(codeArr[i]);
decodedArr.push(letters[(location+13)%26]);
}else
decodedArr.push(letters[i]);
}
return decodedArr.join("");
}
rot13("SERR CVMMN!");
This is supposed to return "FREE PIZZA!"
Upvotes: 0
Views: 337
Reputation: 4007
Lets mark all letters as uppercase to avoid ambiguity between rot13('SERR CVMMN!');
and rot13('SeRR cvMmn!');
var codeArr = encodedStr.toUpperCase().split(""); // sanitize input.
Non alpha characters should be passed through. Space is not a special case
if (/[A-Z]/.test(codeArr[i])) { // if the encoded character is an alphabet character
// decode the character and push it on the decodedArr
} else {
decodedArr.push(codeArr[i]); // let things like spaces and !
// pass through to the output.
}
I think your logic is a little backwards. Instead of looping over the alphabet why not loop over the encoded text and decode sequentially?
for (var i = 0; i < codeArr.length; i += 1) {
if (/[A-Z]/.test(codeArr[i])) { // if the encoded character is an alphabet character
// decode the character and push it on the decodedArr
// the decoded character is 13 ahead of the current character in base 26.
var ACode = 'A'.charCodeAt(); // 65
var letter = codeArr[i];
var letterCode = letter.charCodeAt() - ACode; // ASCII code
letterCode = (letterCode + 13) % 26;
decodedArr.push(letters[letterCode]);
} else {
decodedArr.push(codeArr[i]); // let things like spaces and !
// pass through to the output.
}
}
Upvotes: 3