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I am doing codewars challenge Replace With Alphabet Position, but i don't understand it

The challenge says given a string, replace every letter with its position in the alphabet.
For 4 hours i couldn't find a solution. I tried regex(don't really understand how does it works yet). And i tried to find a solution, but i don't understand it fully
The code is this
function alphabetPosition(text) {
var result = "";
for (var i = 0; i < text.length; i++){
var code = text.toUpperCase().charCodeAt(i)
if (code > 64 && code < 91) {
result += (code - 64) + " "
};
}
return result.slice(0, result.length-1);
}
I don't understand after var code = ......
Why doing this
if (code > 64 && code < 91) {
result += (code - 64) + " "
};

for (var i = 0; i < text.length; i++)
This simply loops over every letter in the text parameter.
var code = text.toUpperCase().charCodeAt(i)
This begins to define the letter's alphabet position. It basically says: "for every character in text, return its character code charCodeAt(i).
if (code > 64 && code < 91) {
result += (code - 64) + " "
};
Just what the first comment says, the character codes for uppercase letters A thru Z start at 65 (A) and end at 90 (Z). Therefore, we only want the codes for letters A thru Z (e.g., we don't care about codes for spaces or quotation marks).
Since we want A to equal 1, we must subtract 64 from the character code
(65 - 64 = 1, likewise for B, 66 - 64 = 2, etc).
result += (code - 64) + " "
This portion adds our filtered letter codes to the result and adds a space for readability. It does it for every character code filtered by the if statement.
return result.slice(0, result.length-1);
Then, we simply return the entire string and subtract the unnecessary space at the end.
Note: you could also simply return result.trim() to remove the whitespace at the end.

Alternatives to escape(string) in JavaScript

When changing to TypeScript I'm not allowed to use escape(string) anymore because it's deprecated. The reason I still use it is that the alternatives encodeURI and encodeURIComponent give a different results.
var s = "Å"
console.log(escape(s));
console.log(encodeURI(s));
console.log(encodeURIComponent(s));
I don't use this for URLs, but for a CSV export.
What are other alternatives that will give me the same result as escape(string)?

In EcmaScript spec there is algorithm:
Call ToString(string).
Compute the number of characters in Result(1).
Let R be the empty string.
Let k be 0.
If k equals Result(2), return R.
Get the character at position k within Result(1).
If Result(6) is one of the 69 nonblank ASCII characters ABCDEFGHIJKLMNOPQRSTUVWXYZ
abcdefghijklmnopqrstuvwxyz 0123456789 #*_+-./, go to step 14.
Compute the 16-bit unsigned integer that is the Unicode character encoding of Result(6).
If Result(8), is less than 256, go to step 12.
Let S be a string containing six characters “%uwxyz” where wxyz are four hexadecimal digits encoding the
value of Result(8).
Go to step 15.
Let S be a string containing three characters “%xy” where xy are two hexadecimal digits encoding the
value of Result(8).
Go to step 15.
Let S be a string containing the single character Result(6).
Let R be a new string value computed by concatenating the previous value of R and S.
Increase k by 1.
Go to step 5.
which can be coded like this:
(function(global) {
var allowed = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789#*_+-./,';
global.escapeString = function(str) {
str = str.toString();
var len = str.length, R = '', k = 0, S, chr, ord;
while(k < len) {
chr = str[k];
if (allowed.indexOf(chr) != -1) {
S = chr;
} else {
ord = str.charCodeAt(k);
if (ord < 256) {
S = '%' + ("00" + ord.toString(16)).toUpperCase().slice(-2);
} else {
S = '%u' + ("0000" + ord.toString(16)).toUpperCase().slice(-4);
}
}
R += S;
k++;
}
return R;
};
})(typeof window == 'undefined' ? global : window);

Permute string until it matches some input?

I've looked this up online without much results because it's quite hard to describe in a few words.
Basically, I need to have a function, say puntil which takes the argument string. Basically, the function permutes until the string is equal to the argument.
For example if you run puntil('ab') it should do inside the function:
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
aa
ab !! MATCH
Another example, for puntil('abcd') it will do
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
aa
ab
ac
ad
ae
af
ag
ah
ai
aj
ak
al
am
an
ao
ap
aq
ar
as
at
au
av
aw
ax
ay
az
... etc etc ..
until it matches abcd.
Basically an infinite permutation until it matches.
Any ideas?

Here is the fiddle
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 output = "";
var match = "cccc"; //<----------- match here
//This is your main function
function permutate() {
var d = 0; // d for depth
while (true) {
//Your main alphabet iteration
for (var i=0; i<alphabet.length; i++){
//First level
if (d === 0) {
console.log(alphabet[i])
output = alphabet[i];
}
else
iterator(alphabet[i], d); //Call iterator for d > 0
if (output === match)
return;
}
d++; //Increase the depth
}
}
//This function runs n depths of iterations
function iterator(s, depth){
if (depth === 0)
return;
for (var i=0; i<alphabet.length; i++){
if (depth > 1)
iterator(s + alphabet[i], depth - 1)
else {
console.log(s + alphabet[i]);
output = s + alphabet[i];
}
if (output === match)
return;
}
};
Explanation:
Your program needs to traverse a tree of alphabet like this
[a]
-[a]
-[a]
-[a]...
-[b]...
[b] ...
-[b] -
-[a]...
-[b]...
[b] - ...
[c] - ...
This could have easily been done through a conventional recursive function if not for the requirement that you need to finish each depth first.
So we need a special iterator(s, depth) function which can perform number of nested iterations (depth) requested.
So the main function can call the iterator with increasing depths (d++).
That's all!!
Warning: This is a prototype only. This can be optimized and improved. It uses global variables for the ease of demonstrating. Your real program should avoid globals. Also I recommend calling the iterator() inside setTimeout if your match word is too long.
The n depths can only be limited by your resources.

Fiddle here
function next(charArray, rightBound){
if(!rightBound){
rightBound = charArray.length;
}
var oldValue = charArray[rightBound-1];
var newValue = nextCharacter(charArray[rightBound-1]);
charArray[rightBound-1] = newValue;
if(newValue < oldValue){
if(rightBound > 1){
next(charArray, rightBound-1);
}
else{
charArray.push('a');
}
}
return charArray;
}
function nextCharacter(char){
if(char === 'z'){
return 'a'
}
else{
return String.fromCharCode(char.charCodeAt(0) + 1)
}
}
function permuteUntil(word){
var charArray = ['a'];
var wordChain = ['a'];
while(next(charArray).join('') !== word){
wordChain.push(charArray.join(''));
}
wordChain.push(word);
return wordChain.join(', ');
}
alert(permuteUntil('ab'));

What OP is asking is a bit ambiguous, so I'll post for both the things (that I doubt) OP is asking.
First, the question can be, what will be the position of input string in the infinite permutation of alphabets (which I see as more legit question, I've given the reason later). This can be done in the following manner:
Taking an example (input = dhca). So, all strings of 1 to 3 characters length will come before this string. So, add: 26^1 + 26^2 + 26^3 to the answer. Then, 1st character is d, which means, following the dictionary, if 1st character is a | b | c, all characters past that are valid. So, add 3 * 26^3 to the answer. Now, say 1st character is d. Then, we can have all characters from a to g (7) and last 2 characters can be anything. So, add 7 * 26^2 to the answer. Going on in this way, we get the answer as:
26^1 + 26^2 + 26^3 + (3 * 26^3) + (7 * 26^2) + (2 * 26^1) + (0) + 1
= 75791
OK. Now the second thing, which I think OP is actually asking (to print all strings before we get a match). Now, why I think this is unfeasible is because if we have input as zzzzz (5 characters long) we need to print 26^1 + 26^2 + 26^3 + 26^4 + 26^5 strings, which is 12356630. So, for this part, I assume max length of input string is 5 (And definitely no more) because for 6 character length string, we need to print ~321272406 strings --> NOT POSSIBLE.
So, a simple solution to this can be:
Create an array of size 27 as: arr[] = {'', 'a', 'b', ..., 'y', 'z'}. 1st character is null.
Write 5 (max string length) nested loops from 0 to 26 (inclusive) and add it to dummy string and print it. Something like.
for i from 0 to 26
String a = "" + arr[i]
for j from 0 to 26
String b = a + arr[j]
for k from 0 to 26
String c = b + arr[k]
for l from 0 to 26
String d = c + arr[l]
for m from 0 to 26
String e = d + arr[m]
print e
if(e equals input string)
break from all loops //use some flag, goto statement etc.

You asked for a more elegant solution, here's a simple function that converts integers into lowercase strings of characters allowing you to easily iterate through strings.
function toWord(val, first, out) {
if(first == 1)
out = String.fromCharCode(97 + val % 26) + out;
else
out = String.fromCharCode(97 + (val-1) % 26) + out;
if(val % 26 == 0 && first == 0) {
val -= 26;
}
else {
val -= val %26;
}
val = val / 26;
if(val != 0)
return toWord(val, 0, out);
else
return out;
}
It's by no means perfect, but it's short and simple. When calling the function set val to be the integer you want to convert, first as 1, and out as "".
For example the following will apply yourFunction to the first 10,000 lowercase strings
for(int i=0; i<10000; i++) {
youFunction(toWord(i, 1, ""));
}

So you need to always start incrementing from a? Since they are char values you can easily do this with the following construct:
Note that this is a java solution :)
public static char[] increment(char[] arr, int pos) {
// get current position char
char currentChar = arr[pos];
// increment it by one
currentChar++;
// if it is at the end of it's range
boolean overflow = false;
if (currentChar > 'Z') {
overflow = true;
currentChar = 'A';
}
// always update current char
arr[pos] = currentChar;
if (overflow) {
if (pos == 0) {
// resize array and add new character
char[] newArr = new char[arr.length + 1];
System.arraycopy(arr, 0, newArr, 0, arr.length);
newArr[arr.length] = 'A';
arr = newArr;
} else {
// overflowed, increment one position back
arr = increment(arr, pos - 1);
}
}
return arr;
}
public static void main(String[] args) {
final String stringToUse = "A";
final String stringToSearch = "ABCD";
char[] use = stringToUse.toCharArray();
for (;;) {
final String result = new String(use);
System.out.println("Candidate: " + result);
if (stringToSearch.equals(result)) {
System.out.println("Found!");
break;
}
use = increment(use, use.length - 1);
}
}

create all possible variations of a string with inserted character

I'm trying to take the variable email and create all possible combinations with a "." in it like so:
Results
andrew
andre.w
andr.ew
andr.e.w
and.rew
and.re.w
and.r.ew
and.r.e.w
an.drew
an.dre.w
an.dr.ew
an.dr.e.w
an.d.rew
an.d.re.w
an.d.r.ew
an.d.r.e.w
a.ndrew
a.ndre.w
a.ndr.ew
a.ndr.e.w
a.nd.rew
a.nd.re.w
a.nd.r.ew
a.nd.r.e.w
a.n.drew
a.n.dre.w
a.n.dr.ew
a.n.dr.e.w
a.n.d.rew
a.n.d.re.w
a.n.d.r.ew
a.n.d.r.e.w
I'm not sure how to do about doing this exactly. I know how to use a loop to go over each character, but as far as the rest goes I'm stumped. I was looking at substr, slice and few other functions but couldn't get anything working.
Code
var email = "andrew";
for (var i = 0; i < email.length; i++) {
console.log( email[i] + "." );
}

That's easy:
var str = 'andrew';
var results = [],
bin;
for (var i = 0; i < Math.pow(2, str.length - 1); ++i) {
bin = i.toString(2).split('').reverse().join('');
results.push(str.replace(/./g, function(letter, index) {
if (bin.charAt(index) == 1) {
letter += '.';
}
return letter;
}));
}
console.log(results);
Demo: http://jsfiddle.net/9qLY6/
Short description:
For 'abc' string there are 2 positions for a dot character: between a and b; b and c. These 2 positions might be presented as a digits of a binary number. All the possible combinations in this case are:
00
01
10
11
If you treat 1 as - . there, and 0 as no . there - you can just iterate over 2^(n-1) numbers and put . if the corresponding bit is set.

If you're interested in a recursive solution like Dinesh mentioned, here's some code to get you started.
function withPeriods(str, prev) {
prev = prev || '';
if(!str || str.length == 0) {
return prev ? [prev] : [];
} else if(str.length == 1) {
return [prev + str];
} else {
var c = str.charAt(0);
var newStr = str.slice(1);
return withPeriods(newStr, prev+c).concat(withPeriods(newStr, prev+c+'.'));
}
}
The idea here is that you are working your way through the string, keeping the current result in the 'prev' variable. If the string is length 0 or 1, there's nothing left to do. Otherwise, you need consider two options: one where you take a character from 'str' and add it to 'prev', and one where you do that but also add a '.'

If you think about it, you need to either insert a dot, or not insert one, at every possible location in the string (between any two characters). A funky way to do this is to realize that if you have n characters, there are n-1 places. If you wrote the combinations of period = 1 and no period = 0, then you can write all possible solutions as a 2^n-1 binary sequence. Showing this for a four letter word "word":
000 word
001 wor.d
010 wo.rd
011 wo.r.d
100 w.ord
101 w.or.d
110 w.o.rd
111 w.o.r.d
In pseudo code (can't test JS syntax right now):
n = strlen( email );
combinations = 1 << n - 1; // left shift operation
for i = 0 to combinations - 1:
dot = 1
for j = 0 to n:
print email[j];
if dot & i:
print '.'
dot << 1;
Can you take it from here?

You might take a recursive approach to this problem. Maybe you can use the base case as a string with 2 characters.

String that contains all ascii characters

I want to create a string in JavaScript that contains all ascii characters. How can I do this?

var s = ' !"#$%&\'()*+,-./0123456789:;<=>?#ABCDEFGHIJKLMNOPQRSTUVWXYZ[\\]^_`abcdefghijklmnopqrstuvwxyz{|}~';

My javascript is a bit rusty, but something like this:
s = '';
for( var i = 32; i <= 126; i++ )
{
s += String.fromCharCode( i );
}
Not sure if the range is correct though.
Edit:
Seems it should be 32 to 127 then. Adjusted.
Edit 2:
Since char 127 isn't a printable character either, we'll have to narrow it down to 32 <= c <= 126, in stead of 32 <= c <= 127.

Just loop the character codes and convert each to a character:
var s = '';
for (var i=32; i<=127;i++) s += String.fromCharCode(i);

Just wanted to put this here for reference. (takes about 13/100 to 26/100 of a ms on my computer to generate).
var allAsciiPrintables = JSON.stringify((Array.from(Array(126 + 32).keys()).slice(32).map((item) => {
return String.fromCharCode(item);
})).join(''));
Decomposed:
var allAsciiPrintables = (function() {
/* ArrayIterator */
var result = Array(126 + 32).keys();
/* [0, 126 + 32] */
result = Array.from(result);
/* [32, 126 + 32] */
result = result.slice(32);
/* transform each item from Number to its ASCII as String. */
result = result.map((item) => {
return String.fromCharCode(item);
});
/* convert from array of each string[1] to a single string */
result = result.join('');
/* create an escaped string so you can replace this code with the string
to avoid having to calculate this on each time the program runs */
result = JSON.stringify(result);
/* return the string */
return result;
})();
The most efficient solution(if you do want to generate the whole set each time the script runs, is probably)(takes around 3/100-35/100 of a millisecond on my computer to generate).
var allAsciiPrintables = (() => {
var result = new Array(126-32);
for (var i = 32; i <= 126; ++i) {
result[i - 32] = (String.fromCharCode(i));
}
return JSON.stringify(result.join(''));
})();
strangely, this is only 3-10 times slower than assigning the string literal directly(with backticks to tell javascript to avoid most backslash parsing).
var x;
var t;
t = performance.now();
x = '!\"#$%&\'()*+,-./0123456789:;<=>?#ABCDEFGHIJKLMNOPQRSTUVWXYZ[\\]^_`abcdefghijklmnopqrstuvwxyz{|}~';
t = performance.now() - t;
console.log(t);
.

This is a version written in python. Gives all ASCII characters in order as a single string.
all_ascii = ''.join(chr(k) for k in range(128)) # 7 bits
all_chars = ''.join(chr(k) for k in range(256)) # 8 bits
printable_ascii = ''.join(chr(k) for k in range(128) if len(repr(chr(k))) == 3)
>>> print(printable_ascii)
' !"#$%&\'()*+,-./0123456789:;<=>?#ABCDEFGHIJKLMNOPQRSTUVWXYZ[]^_`abcdefghijklmnopqrstuvwxyz{|}~'
The last string here, printable_ascii contains only those characters that contain no escapes (i.e. have length == 1). The chars like: \x05, \x06 or \t, \n which does not have its own glyph in your system's font, are filtered out.
len(repr(chr(k))) == 3 includes 2 quotes that come from repr call.

Without doing several appends:
var s = Array.apply(null, Array(127-32))
.map(function(x,i) {
return String.fromCharCode(i+32);
}).join("");
document.write(s);

Here is an ES6 one liner:
asciiChars = Array.from({ length: 95 }, (e, i) => String.fromCharCode(i + 32)).join('');
console.log(asciiChars)

let str = '';// empty string declear
for( var i = 32; i <= 126; i++ )
{
str = str + String.fromCharCode( i ); /* this method received one integer and
convert it into a ascii characters and store it str variable one by one by using
string concatenation method. The loop start for 32 and end 126 */
}

Here is a version in coffeescript
require 'fluentnode'
all_Ascii = ->
(String.fromCharCode(c) for c in [0..255])
describe 'all Ascii', ->
it 'all_Ascii', ->
all_Ascii.assert_Is_Function()
all_Ascii().assert_Size_Is 256
all_Ascii()[0x41].assert_Is 'A'
all_Ascii()[66 ].assert_Is 'B'
all_Ascii()[50 ].assert_Is '2'
all_Ascii()[150 ].assert_Is String.fromCharCode(150)