Develop Reference

Decipher any string that is passed

I am attempting this question on Codewars. I am not sure if am doing it right. Here is the question:
You are given a secret message you need to decipher. Here are the things you need to know to decipher it:
For each word:
the second and the last letter is switched (e.g. Hello becomes Holle)
the first letter is replaced by its character code (e.g. H becomes 72)
Note: there are no special characters used, only letters and spaces:
decipherThis('72olle 103doo 100ya'); // 'Hello good day'
decipherThis('82yade 115te 103o'); // 'Ready set go'
Now I have written this piece of code:
function decipherThis(str)
{
var msg = [];
msg.push(str.charCodeAt(0));
for (var i = 0; i<str.length; i++)
{
if (str[1] == true && str[1] != str[str.length])
{
msg.push(str[str.length]);
//str[1] = str[str.length]);
var news = str;
for (var j = 0; j<news.length; j++)
{
news[1] = news[news.length];
const newNew = delete news[0][1];
msg.push(newNew);
}
}
}
return msg;
};
var google = "hello"
decipherThis(google)
I am getting an error and I think I have created it for a single word. It does not understand words after blank spaces. Please help me fix this.
This is the error traceback:
Response received but no data was written to STDOUT or STDERR.

Please change characterCodeAt to charCodeAt and it will work.
characterCodeAt is not a valid method of a String class

The error is not Javascript related. It might occur because the decipherThis() function is returning an array instead of a string.
You could loop the original string backwards until the second character and concatonate it to the charcode of the first character like this:
function decipherThis(str)
{
var cipher = '',
words = str.split(' ');
for(var w = 0; w < words.length; w++) {
cipher += ' '+words[w].charCodeAt(0);
for(var i = words[w].length-1; i > 0; i--) {
cipher += words[w][i];
}
}
return cipher.trim();
}

How to make element in Array change its' place

I'm beginner in JS. I've tried to understand Caesar Cipher ROT13, but it was too complicated for me. So I've tried to write my own code. Here it is below:
function encrip() {
var alphabet = ["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 str = "Ni Hao";
var string = str.toUpperCase();
for (var i = 0; i < string.length; i++) {
for (var k = 0; k < alphabet.length; k++) {
if(string.charAt(i) == alphabet[k]) {
/* console.log(string.charAt(i) + ' ' + alphabet.indexOf(alphabet[k])); */
}
}
}
}
encrip();
But I am stuck. How to do:
1. Get value from var str and then access to var alphabet , after change each letter from var str value to next 3 from alphabet (var str each element's current position would be changed) For example: Input: Ni Hao ==> output: QL KDR
2. Create universal code, I mean, not only for changing position by 3, but when I give value '5', each element would be changed by next 5 positions from alphabet. So output can be changed when I change its' value
I hope I explained everything clearly. Thanks everyone in advance for help!!

you can use the following function to encrypt english words, the 1st parameter is the string to encrypt and the 2nd for shifting
function encryp(str,pos){
var alpha="ABCDEFGHIJKLMNOPQRSTUVWXYZ";
var strUC=str.toUpperCase();
var enc="";
for(var i=0;i<strUC.length;i++){
if(strUC.charAt(i)!=" "){
enc+=alpha.charAt((alpha.indexOf(strUC.charAt(i))+pos)%26)
}
else{
enc+=" "
}
// in your case pos=3
}
return enc;
}
console.log(encryp("NiHao",3));

You don't need two for loops to do this. Iterate over the input string and find the index of each character in the alphabet array, if found add the shift to it to get the encrypted character.
To handle overflow use the modulus operator to cycle through the array.
Also I assume that you are not going use any special symbols to do the encryption.
function encrip(string, shift) {
var alphabet = ["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"];
string = string.toUpperCase();
let arr = [];
for (var i = 0; i < string.length; i++) {
let char = alphabet.indexOf(string[i]) !== -1 ? alphabet[(alphabet.indexOf(string[i]) %26) + shift] : " ";
arr.push(char);
}
let encryp = arr.join("");
console.log(encryp);
return encryp;
}
encrip("Ni Hao", 3);

First of all, instead of your inner for loop scanning the whole alphabet array, you can use the built-in function indexOf:
alphabet.indexOf('K') // returns 10
Secondly, you'll want to build up your enciphered string in a separate variable. For each letter, get the index of that letter in the alphabet, add your cipher offset parameter to that index and add the resulting letter from the alphabet to your new string. An important step is that when you add to the index of the letter, you want to make sure the resulting index is within range for the alphabet array. You can do that using the % (modulo) operator, which will wrap high values back round to the start of the array. In full:
function encipher(input, offset) {
var alphabet = ["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 str = input.toUpperCase();
var result = '';
for (var i = 0; i < str.length; i++) {
letterIndex = alphabet.indexOf(str.charAt(i));
if (letterIndex === -1) {
result += str[i]; // if the letter isn't found in the alphabet, add it to the result unchanged
continue;
}
cipheredIndex = (letterIndex + offset) % alphabet.length; // wrap index to length of alphabet
result += alphabet[cipheredIndex];
}
console.log(result);
}
encipher('Ni Hao', 5); // output: 'SN MFT'

javascript string algorithm

Let's say I have a string variable called myString, and another string variable called myChar.
var myString = "batuhan"; // it's user input.
var myChar = "0"; // will be one character, always
What I need is, a function that returns all the combinations of myString and myChar.
Like:
"batuhan","batuha0n","batuh0an","batuh0a0n","batu0han","batu0ha0n","batu0h0an","batu0h0a0n","bat0uhan","bat0uha0n","bat0uh0an","bat0uh0a0n","bat0u0han","bat0u0ha0n","bat0u0h0an","bat0u0h0a0n","ba0tuhan","ba0tuha0n","ba0tuh0an","ba0tuh0a0n","ba0tu0han","ba0tu0ha0n","ba0tu0h0an","ba0tu0h0a0n","ba0t0uhan","ba0t0uha0n","ba0t0uh0an","ba0t0uh0a0n","ba0t0u0han","ba0t0u0ha0n","ba0t0u0h0an","ba0t0u0h0a0n","b0atuhan","b0atuha0n","b0atuh0an","b0atuh0a0n","b0atu0han","b0atu0ha0n","b0atu0h0an","b0atu0h0a0n","b0at0uhan","b0at0uha0n","b0at0uh0an","b0at0uh0a0n","b0at0u0han","b0at0u0ha0n","b0at0u0h0an","b0at0u0h0a0n","b0a0tuhan","b0a0tuha0n","b0a0tuh0an","b0a0tuh0a0n","b0a0tu0han","b0a0tu0ha0n","b0a0tu0h0an","b0a0tu0h0a0n","b0a0t0uhan","b0a0t0uha0n","b0a0t0uh0an","b0a0t0uh0a0n","b0a0t0u0han","b0a0t0u0ha0n","b0a0t0u0h0an","b0a0t0u0h0a0n"
Rules: myChar shouldn't follow myChar
How can I do that? Really my brain dead right now :/

It's possible to implement what you want using recursion.
// Example: allCombinations("abcd", "0") returns the array
// ["abcd", "abc0d", "ab0cd", "ab0c0d", "a0bcd", "a0bc0d", "a0b0cd", "a0b0c0d"]
function allCombinations(str, chr) {
if (str.length == 1)
return [str];
var arr = allCombinations(str.substring(1), chr);
var result = [];
var c = str.charAt(0);
for (var i = 0; i < arr.length; i++)
result.push(c + arr[i]);
for (var i = 0; i < arr.length; i++)
result.push(c + chr + arr[i]);
return result;
}

You may or may not have noticed this but this is basically counting in binary. If we define bit 0 to be the absence of myChar and bit 1 to be the presence of myChar, then the following sequence:
var myString = ".....";
var myChar = "1";
var sequence = [
".....1",
"....1.",
"....1.1",
"...1.."
];
is basically counting from 1 to 4 in binary:
var sequence = [
0b0000001,
0b0000010,
0b0000011,
0b0000100
];
Therefore, all you need is a for loop to count up to the bit amount of the length of the string plus 1 (because the position at the end of the string is also legal):
var len = Math.pow(2,myString.length+1);
for (var x = 0; x < len; x++) {
// x in binary is all the possible combinations
// now use the "1" bits in x to modify the string:
// Convert myString to array for easy processing:
var arr = myString.split('');
arr.push(""); // last position;
for (var i = myString.length; i >= 0; i--) {
if ((x >> i) & 0x01) { // check if bit at position i is 1
arr[i] = myChar + arr[i];
}
}
console.log(arr.join('')); // print out one combination
}
Of course, this works only for small strings of up to 31 characters. For larger strings you'd need to do the binary counting using things other than numbers. Doing it in a string form is one option. Another option is to use a bigint library such as BigInteger.js to do the counting.

Word Array to String

how to do this in Javascript or Jquery?
Please suggest in 2 steps:
1.- Word Array to Single Byte Array.
2.- Byte Array to String.
Maybe this can help:
function hex2a(hex) {
var str = '';
for (var i = 0; i < hex.length; i += 2)
str += String.fromCharCode(parseInt(hex.substr(i, 2), 16));
return str;
}

What you are trying to achieve is already implemented in CryptoJS. From the documentation:
You can convert a WordArray object to other formats by explicitly calling the toString method and passing an encoder.
var hash = CryptoJS.SHA256("Message");
alert(hash.toString(CryptoJS.enc.Base64));
alert(hash.toString(CryptoJS.enc.Hex));
Honestly I have no idea why you want to implement that yourself... But if you absolutely need to do it "manually" in the 2 steps you mentioned, you could try something like this:
function wordToByteArray(wordArray) {
var byteArray = [], word, i, j;
for (i = 0; i < wordArray.length; ++i) {
word = wordArray[i];
for (j = 3; j >= 0; --j) {
byteArray.push((word >> 8 * j) & 0xFF);
}
}
return byteArray;
}
function byteArrayToString(byteArray) {
var str = "", i;
for (i = 0; i < byteArray.length; ++i) {
str += escape(String.fromCharCode(byteArray[i]));
}
return str;
}
var hash = CryptoJS.SHA256("Message");
var byteArray = wordToByteArray(hash.words);
alert(byteArrayToString(byteArray));
The wordToByteArray function should work perfectly, but be aware that byteArrayToString will produce weird results in almost any case. I don't know much about encodings, but ASCII only uses 7 bits so you won't get ASCII chars when trying to encode an entire byte. So I added the escape function to at least be able to display all those strange chars you might get. ;)
I'd recommend you use the functions CryptoJS has already implemented or just use the byte array (without converting it to string) for your analysis.

JavaScript strings outside of the BMP

BMP being Basic Multilingual Plane
According to JavaScript: the Good Parts:
JavaScript was built at a time when Unicode was a 16-bit character set, so all characters in JavaScript are 16 bits wide.
This leads me to believe that JavaScript uses UCS-2 (not UTF-16!) and can only handle characters up to U+FFFF.
Further investigation confirms this:
> String.fromCharCode(0x20001);
The fromCharCode method seems to only use the lowest 16 bits when returning the Unicode character. Trying to get U+20001 (CJK unified ideograph 20001) instead returns U+0001.
Question: is it at all possible to handle post-BMP characters in JavaScript?
2011-07-31: slide twelve from Unicode Support Shootout: The Good, The Bad, & the (mostly) Ugly covers issues related to this quite well:

Depends what you mean by ‘support’. You can certainly put non-UCS-2 characters in a JS string using surrogates, and browsers will display them if they can.
But, each item in a JS string is a separate UTF-16 code unit. There is no language-level support for handling full characters: all the standard String members (length, split, slice etc) all deal with code units not characters, so will quite happily split surrogate pairs or hold invalid surrogate sequences.
If you want surrogate-aware methods, I'm afraid you're going to have to start writing them yourself! For example:
String.prototype.getCodePointLength= function() {
return this.length-this.split(/[\uD800-\uDBFF][\uDC00-\uDFFF]/g).length+1;
};
String.fromCodePoint= function() {
var chars= Array.prototype.slice.call(arguments);
for (var i= chars.length; i-->0;) {
var n = chars[i]-0x10000;
if (n>=0)
chars.splice(i, 1, 0xD800+(n>>10), 0xDC00+(n&0x3FF));
}
return String.fromCharCode.apply(null, chars);
};

I came to the same conclusion as bobince. If you want to work with strings containing unicode characters outside of the BMP, you have to reimplement javascript's String methods. This is because javascript counts characters as each 16-bit code value. Symbols outside of the BMP need two code values to be represented. You therefore run into a case where some symbols count as two characters and some count only as one.
I've reimplemented the following methods to treat each unicode code point as a single character: .length, .charCodeAt, .fromCharCode, .charAt, .indexOf, .lastIndexOf, .splice, and .split.
You can check it out on jsfiddle: http://jsfiddle.net/Y89Du/
Here's the code without comments. I tested it, but it may still have errors. Comments are welcome.
if (!String.prototype.ucLength) {
String.prototype.ucLength = function() {
// this solution was taken from
// http://stackoverflow.com/questions/3744721/javascript-strings-outside-of-the-bmp
return this.length - this.split(/[\uD800-\uDBFF][\uDC00-\uDFFF]/g).length + 1;
};
}
if (!String.prototype.codePointAt) {
String.prototype.codePointAt = function (ucPos) {
if (isNaN(ucPos)){
ucPos = 0;
}
var str = String(this);
var codePoint = null;
var pairFound = false;
var ucIndex = -1;
var i = 0;
while (i < str.length){
ucIndex += 1;
var code = str.charCodeAt(i);
var next = str.charCodeAt(i + 1);
pairFound = (0xD800 <= code && code <= 0xDBFF && 0xDC00 <= next && next <= 0xDFFF);
if (ucIndex == ucPos){
codePoint = pairFound ? ((code - 0xD800) * 0x400) + (next - 0xDC00) + 0x10000 : code;
break;
} else{
i += pairFound ? 2 : 1;
}
}
return codePoint;
};
}
if (!String.fromCodePoint) {
String.fromCodePoint = function () {
var strChars = [], codePoint, offset, codeValues, i;
for (i = 0; i < arguments.length; ++i) {
codePoint = arguments[i];
offset = codePoint - 0x10000;
if (codePoint > 0xFFFF){
codeValues = [0xD800 + (offset >> 10), 0xDC00 + (offset & 0x3FF)];
} else{
codeValues = [codePoint];
}
strChars.push(String.fromCharCode.apply(null, codeValues));
}
return strChars.join("");
};
}
if (!String.prototype.ucCharAt) {
String.prototype.ucCharAt = function (ucIndex) {
var str = String(this);
var codePoint = str.codePointAt(ucIndex);
var ucChar = String.fromCodePoint(codePoint);
return ucChar;
};
}
if (!String.prototype.ucIndexOf) {
String.prototype.ucIndexOf = function (searchStr, ucStart) {
if (isNaN(ucStart)){
ucStart = 0;
}
if (ucStart < 0){
ucStart = 0;
}
var str = String(this);
var strUCLength = str.ucLength();
searchStr = String(searchStr);
var ucSearchLength = searchStr.ucLength();
var i = ucStart;
while (i < strUCLength){
var ucSlice = str.ucSlice(i,i+ucSearchLength);
if (ucSlice == searchStr){
return i;
}
i++;
}
return -1;
};
}
if (!String.prototype.ucLastIndexOf) {
String.prototype.ucLastIndexOf = function (searchStr, ucStart) {
var str = String(this);
var strUCLength = str.ucLength();
if (isNaN(ucStart)){
ucStart = strUCLength - 1;
}
if (ucStart >= strUCLength){
ucStart = strUCLength - 1;
}
searchStr = String(searchStr);
var ucSearchLength = searchStr.ucLength();
var i = ucStart;
while (i >= 0){
var ucSlice = str.ucSlice(i,i+ucSearchLength);
if (ucSlice == searchStr){
return i;
}
i--;
}
return -1;
};
}
if (!String.prototype.ucSlice) {
String.prototype.ucSlice = function (ucStart, ucStop) {
var str = String(this);
var strUCLength = str.ucLength();
if (isNaN(ucStart)){
ucStart = 0;
}
if (ucStart < 0){
ucStart = strUCLength + ucStart;
if (ucStart < 0){ ucStart = 0;}
}
if (typeof(ucStop) == 'undefined'){
ucStop = strUCLength - 1;
}
if (ucStop < 0){
ucStop = strUCLength + ucStop;
if (ucStop < 0){ ucStop = 0;}
}
var ucChars = [];
var i = ucStart;
while (i < ucStop){
ucChars.push(str.ucCharAt(i));
i++;
}
return ucChars.join("");
};
}
if (!String.prototype.ucSplit) {
String.prototype.ucSplit = function (delimeter, limit) {
var str = String(this);
var strUCLength = str.ucLength();
var ucChars = [];
if (delimeter == ''){
for (var i = 0; i < strUCLength; i++){
ucChars.push(str.ucCharAt(i));
}
ucChars = ucChars.slice(0, 0 + limit);
} else{
ucChars = str.split(delimeter, limit);
}
return ucChars;
};
}

More recent JavaScript engines have String.fromCodePoint.
const ideograph = String.fromCodePoint( 0x20001 ); // outside the BMP
Also a code-point iterator, which gets you the code-point length.
function countCodePoints( str )
{
const i = str[Symbol.iterator]();
let count = 0;
while( !i.next().done ) ++count;
return count;
}
console.log( ideograph.length ); // gives '2'
console.log( countCodePoints(ideograph) ); // '1'

Yes, you can. Although support to non-BMP characters directly in source documents is optional according to the ECMAScript standard, modern browsers let you use them. Naturally, the document encoding must be properly declared, and for most practical purposes you would need to use the UTF-8 encoding. Moreover, you need an editor that can handle UTF-8, and you need some input method(s); see e.g. my Full Unicode Input utility.
Using suitable tools and settings, you can write var foo = '𠀁'.
The non-BMP characters will be internally represented as surrogate pairs, so each non-BMP character counts as 2 in the string length.

Using for (c of this) instruction, one can make various computations on a string that contains non-BMP characters. For instance, to compute the string length, and to get the nth character of the string:
String.prototype.magicLength = function()
{
var c, k;
k = 0;
for (c of this) // iterate each char of this
{
k++;
}
return k;
}
String.prototype.magicCharAt = function(n)
{
var c, k;
k = 0;
for (c of this) // iterate each char of this
{
if (k == n) return c + "";
k++;
}
return "";
}

This old topic has now a simple solution in ES6:
Split characters into an array
simple version
[..."😴😄😃⛔🎠🚓🚇"] // ["😴", "😄", "😃", "⛔", "🎠", "🚓", "🚇"]
Then having each one separated you can handle them easily for most common cases.
Credit: DownGoat
Full solution
To overcome special emojis as the one in the comment, one can search for the connection charecter (char code 8205 in UTF-16) and make some modifications. Here is how:
let myStr = "👩‍👩‍👧‍👧😃𝌆"
let arr = [...myStr]
for (i = arr.length-1; i--; i>= 0) {
if (arr[i].charCodeAt(0) == 8205) { // special combination character
arr[i-1] += arr[i] + arr[i+1]; // combine them back to a single emoji
arr.splice(i, 2)
}
}
console.log(arr.length) //3
Haven't found a case where this doesn't work. Comment if you do.
To conclude
it seems that JS uses the 8205 char code to represent UCS-2 characters as a UTF-16 combinations.