Develop Reference

Checking if str1 can be re-arranged as str2 wth JavaScript

There are two strings called str1 and str2 and I'm trying to check if str1 can be re-arranged as str2.
FOR EXAMPLE: lets say str1 = "aabbcamaomsccdd" and str2="commas".
Is it possible to write the word "commas" out of "str1"
function scramble(str1, str2) {
let arr=[];
let str1arr = str1.split("");
let str2arr = str2.split("");
let j=0;
for(let i=0; i<str1.length; i++){
if(str1arr[i]==str2arr[j]){
arr.push(str1arr[i]);
str1arr=str1arr.splice(i,1);
j++;
i=0;
}
}if(arr.toString()===str2arr.toString()){
return true;
}else{
return false;
}
}
What I tried basically if str1arr[i]==str2arr[j] it will put the str1arr[i] value on a new array called arr and at the end it will compare str2 and the arr and return True or False.
The reason why I used str1arr=str1arr.splice(i,1); to delete the i after the match is because the for loop is reseting it self to check from the "i=0" each time i and j matches and that i would match with other duplicate letters (I hope thats what it does atleast :D).
It is an internet question and im not passing the tests. I only pass if the result is FALSE.
I want to know what I'm doing and thinking wrong here. Its not performance efficent too so any comment on that would be great too.

You could take arrays and sort them and check each character of the second string/array against the first one.
function compare([...a], [...b]) {
a.sort();
return b.sort().every((i => v => {
while (i < a.length && a[i] !== v) i++;
return a[i++] === v;
})(0));
}
console.log(compare("aabbcamaomsccdd", "commas")); // true
console.log(compare("aabbcamaomccdd", "commas")); // false

You should just check that both strings contain the same chars like so:
function scramble(str1, str2) {
var s1 = str1.split('');
var s2 = str2.split('');
var i;
for (i = 0; i < s2.length; i++) {
const idx = s1.indexOf(s2[i]);
if (idx === -1) {
return false;
}
s1.splice(idx, 1);
}
return s1.length === 0;
}
console.log(scramble('xcab1c', 'abxcc1'));

You could count the frequency of each character in your first string. Below I have used .reduce() to build an object with key-value pairs, where the key represents a character from your s1 string and the value is how many times it appears. You can then loop through the characters in s2 and check that every character appears in the frequency map. When you see a character you can subtract one from the value from the frequency object to signify that the character has been "used". If the .every() callback returns a falsy value (such as 0 for the value), then the result will be false, as your string can't be re-arranged:
const scramble = (s1, s2) => {
const s1Freq = [...s1].reduce((o, c) => ({...o, [c]: (o[c] || 0) +1}), {});
return [...s2].every(char => s1Freq[char]--);
}
console.log(scramble("aabbcamaomsccdd", "commas")); // true
console.log(scramble("abc321", "123")); // true
console.log(scramble("a3b2c11", "1231")); // true
console.log(scramble("a", "a")); // true
console.log(scramble("xyz", "xyt")); // false

How to get odd and even position characters from a string?

I'm trying to figure out how to remove every second character (starting from the first one) from a string in Javascript.
For example, the string "This is a test!" should become "hsi etTi sats!"
I also want to save every deleted character into another array.
I have tried using replace method and splice method, but wasn't able to get them to work properly. Mostly because replace only replaces the first character.
function encrypt(text, n) {
if (text === "NULL") return n;
if (n <= 0) return text;
var encArr = [];
var newString = text.split("");
var j = 0;
for (var i = 0; i < text.length; i += 2) {
encArr[j++] = text[i];
newString.splice(i, 1); // this line doesn't work properly
}
}

You could reduce the characters of the string and group them to separate arrays using the % operator. Use destructuring to get the 2D array returned to separate variables
let str = "This is a test!";
const [even, odd] = [...str].reduce((r,char,i) => (r[i%2].push(char), r), [[],[]])
console.log(odd.join(''))
console.log(even.join(''))
Using a for loop:
let str = "This is a test!",
odd = [],
even = [];
for (var i = 0; i < str.length; i++) {
i % 2 === 0
? even.push(str[i])
: odd.push(str[i])
}
console.log(odd.join(''))
console.log(even.join(''))

It would probably be easier to use a regular expression and .replace: capture two characters in separate capturing groups, add the first character to a string, and replace with the second character. Then, you'll have first half of the output you need in one string, and the second in another: just concatenate them together and return:
function encrypt(text) {
let removedText = '';
const replacedText1 = text.replace(/(.)(.)?/g, (_, firstChar, secondChar) => {
// in case the match was at the end of the string,
// and the string has an odd number of characters:
if (!secondChar) secondChar = '';
// remove the firstChar from the string, while adding it to removedText:
removedText += firstChar;
return secondChar;
});
return replacedText1 + removedText;
}
console.log(encrypt('This is a test!'));

Pretty simple with .reduce() to create the two arrays you seem to want.
function encrypt(text) {
return text.split("")
.reduce(({odd, even}, c, i) =>
i % 2 ? {odd: [...odd, c], even} : {odd, even: [...even, c]}
, {odd: [], even: []})
}
console.log(encrypt("This is a test!"));
They can be converted to strings by using .join("") if you desire.

I think you were on the right track. What you missed is replace is using either a string or RegExp.
The replace() method returns a new string with some or all matches of a pattern replaced by a replacement. The pattern can be a string or a RegExp, and the replacement can be a string or a function to be called for each match. If pattern is a string, only the first occurrence will be replaced.
Source: String.prototype.replace()
If you are replacing a value (and not a regular expression), only the first instance of the value will be replaced. To replace all occurrences of a specified value, use the global (g) modifier
Source: JavaScript String replace() Method
So my suggestion would be to continue still with replace and pass the right RegExp to the function, I guess you can figure out from this example - this removes every second occurrence for char 't':
let count = 0;
let testString = 'test test test test';
console.log('original', testString);
// global modifier in RegExp
let result = testString.replace(/t/g, function (match) {
count++;
return (count % 2 === 0) ? '' : match;
});
console.log('removed', result);

like this?
var text = "This is a test!"
var result = ""
var rest = ""
for(var i = 0; i < text.length; i++){
if( (i%2) != 0 ){
result += text[i]
} else{
rest += text[i]
}
}
console.log(result+rest)

Maybe with split, filter and join:
const remaining = myString.split('').filter((char, i) => i % 2 !== 0).join('');
const deleted = myString.split('').filter((char, i) => i % 2 === 0).join('');

You could take an array and splice and push each second item to the end of the array.
function encrypt(string) {
var array = [...string],
i = 0,
l = array.length >> 1;
while (i <= l) array.push(array.splice(i++, 1)[0]);
return array.join('');
}
console.log(encrypt("This is a test!"));

function encrypt(text) {
text = text.split("");
var removed = []
var encrypted = text.filter((letter, index) => {
if(index % 2 == 0){
removed.push(letter)
return false;
}
return true
}).join("")
return {
full: encrypted + removed.join(""),
encrypted: encrypted,
removed: removed
}
}
console.log(encrypt("This is a test!"))
Splice does not work, because if you remove an element from an array in for loop indexes most probably will be wrong when removing another element.

I don't know how much you care about performance, but using regex is not very efficient.
Simple test for quite a long string shows that using filter function is on average about 3 times faster, which can make quite a difference when performed on very long strings or on many, many shorts ones.
function test(func, n){
var text = "";
for(var i = 0; i < n; ++i){
text += "a";
}
var start = new Date().getTime();
func(text);
var end = new Date().getTime();
var time = (end-start) / 1000.0;
console.log(func.name, " took ", time, " seconds")
return time;
}
function encryptREGEX(text) {
let removedText = '';
const replacedText1 = text.replace(/(.)(.)?/g, (_, firstChar, secondChar) => {
// in case the match was at the end of the string,
// and the string has an odd number of characters:
if (!secondChar) secondChar = '';
// remove the firstChar from the string, while adding it to removedText:
removedText += firstChar;
return secondChar;
});
return replacedText1 + removedText;
}
function encrypt(text) {
text = text.split("");
var removed = "";
var encrypted = text.filter((letter, index) => {
if(index % 2 == 0){
removed += letter;
return false;
}
return true
}).join("")
return encrypted + removed
}
var timeREGEX = test(encryptREGEX, 10000000);
var timeFilter = test(encrypt, 10000000);
console.log("Using filter is faster ", timeREGEX/timeFilter, " times")
Using actually an array for storing removed letters and then joining them is much more efficient, than using a string and concatenating letters to it.
I changed an array to string in filter solution to make it the same like in regex solution, so they are more comparable.

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'

How to write a character matching algorithm in JavaScript?

Given this input s1 = "dadxx" s2 = "ddxx" I'd expect the output to contain a bunch of a,b pairs wherever each character in s1 matched a character in s2 and vice versa (duplicates allowed). Among those pairs would be 0,0 because s1[0] and s2[0] are both equal to d.
The problem is that my output doesn't contain 2,1 even though s1[2] and s2[1] are both equal to d.
Can someone fix my algorithm or make a better one?
Here's a JSFiddle if it helps.
Here's my code:
// For each char, see if other string contains it
s1 = 'dadxx'
s2 = 'ddxx'
matchChars(s1,s2)
matchChars(s2,s1)
function matchChars(a,b) {
for (i = 0; i < a.length; i++) {
found = b.indexOf(a[i])
if (found >= 0) {
if (a===s1) console.log(i,found)
else console.log(found,i)
}
}
}

I believe the problem you're having is that you're only checking for a single match for s1[i] in s2 by using indexOf. That will find the first index of a matched value, not every index.
If you instead iterate through both strings and compare every character, you get the result I think you're trying to achieve.
// Define strings
s1 = 'dadxx'
s2 = 'ddxx'
matchChars(s1,s2)
matchChars(s2,s1)
function matchChars(a,b) {
// Convert strings to lower case for case insensitive matching
// Remove if case sensitive matching required
a = a.toLowerCase();
b = b.toLowerCase();
// Iterate through every letter in s1
for (i = 0; i < a.length; i++) {
// Iterate through every letter in s2
for (j = 0; j < b.length; j++) {
// Check if the letter in s1 matches letter in s2
if (a[i] === b[j]) {
// Changed per request of OP
(a === s1) ? console.log(i, j) : console.log(j, i);
// console.log([i, j]);
}
}
}
}
Working JSBin example: https://jsbin.com/wecijopohi/edit?js,console

You say duplicates are allowed but not required. I'm submitting this as a more modern approach, not as a correction to the accepted solution, which looks good to me. https://jsfiddle.net/avc705zr/3/
match = (a, b) => {
let re, match, matches = []
a.split('').forEach((l, i) => {
re = new RegExp(l, 'g')
while ((match = re.exec(b)) != null) {
matches.push([i, match.index])
}
})
return matches
}
However, in my experience when you actually need functionality like this, you only need one of the strings to exhausted. In other words, you are looking for matches in string 2 of all instances in string 1 -- which is to say, unique characters in string 1. So a modification which might come up in the real world might instead be like:
Array.prototype.unique = function() {
return this.filter(function (value, index, self) {
return self.indexOf(value) === index;
});
}
match = (a, b) => {
let re, match, matches = []
a.split('').unique().forEach(l => {
re = new RegExp(l, 'g')
while ((match = re.exec(b)) != null) {
matches.push([l, match.index])
}
})
return matches
}

How do you sort letters in JavaScript, with capital and lowercase letters combined?

I'm working on a JavaScript (jQuery's OK too if this needs it, but I doubt it will) function to alphabetize a string of letters. Let's say the string that I want to sort is: "ACBacb".
My code as of now is this:
var string='ACBacb';
alert(string.split('').sort().join(''));
This returns ABCabc. I can see why that happens, but that is not the format that I am looking for. Is there a way that I can sort it by putting the same letters next to each other, capital letter first? So when I put in ACBacb, I get AaBbCc?

Array.sort can have a sort function as optional argument.
What about sorting the string first (ACBacbA becomes AABCabc), and then sorting it case-insensitive:
function case_insensitive_comp(strA, strB) {
return strA.toLowerCase().localeCompare(strB.toLowerCase());
}
var str = 'ACBacbA';
// split the string in chunks
str = str.split("");
// sorting
str = str.sort();
str = str.sort( case_insensitive_comp )
// concatenate the chunks in one string
str = str.join("");
alert(str);
As per Felix suggestion, the first sort function can be omitted and merged in the second one. First, do a case-insensitive comparison between both characters. If they are equal, check their case-sensitive equivalents. Return -1 or 1 for a difference and zero for equality.
function compare(strA, strB) {
var icmp = strA.toLowerCase().localeCompare(strB.toLowerCase());
if (icmp != 0) {
// spotted a difference when considering the locale
return icmp;
}
// no difference found when considering locale, let's see whether
// capitalization matters
if (strA > strB) {
return 1;
} else if (strA < strB) {
return -1;
} else {
// the characters are equal.
return 0;
}
}
var str = 'ACBacbA';
str = str.split('');
str = str.sort( compare );
str = str.join('');

You can pass a custom comparison function to Array.sort()
The already given answers are right so far that you have to use a custom comparison function. However you have to add an extra step to sort capital letters before lower case once:
function cmp(x,y) {
if(x.toLowerCase() !== y.toLowerCase()) {
x = x.toLowerCase();
y = y.toLowerCase();
}
return x > y ? 1 : (x < y ? -1 : 0);
// or
// return x.localeCompare(y);
}
If the letters are the same, the originals have to be compared, not the lower case versions. The upper case letter is always "larger" than the lower case version.
DEMO (based on #Matt Ball's version)

a working example http://jsfiddle.net/uGwZ3/
var string='ACBacb';
alert(string.split('').sort(caseInsensitiveSort).join(''));
function caseInsensitiveSort(a, b)
{
var ret = 0;
a = a.toLowerCase();b = b.toLowerCase();
if(a > b)
ret = 1;
if(a < b)
ret = -1;
return ret;
}

Use a custom sort function like this:
function customSortfunc(a,b){
var lca = a.toLowerCase(), lcb = b.toLowerCase();
return lca > lcb ? 1 : lca < lcb ? -1 : 0;
}
var string='ACBacb';
alert(string.split('').sort(customSortfunc).join(''));
You can read more about the sort function here: https://developer.mozilla.org/en/JavaScript/Reference/Global_Objects/Array/sort
Beware: if you use localeCompare like other answer suggests "u" and "ü" will be sorted together as the same letter, since it disregards all diacritics.

basic example for another replace, combined with lowercase :D
<button onclick="myFunction('U')">Try it</button>
<p id="demo"></p>
<script>
function myFunction(val) {
var str = "HELLO WORLD!";
var res = str.toLowerCase().split("o");
var elem = document.getElementById("demo").innerHTML
for(i = 0; i < res.length; i++){
(i > 0)?elem += val + res[i]:elem += res[i];
}
}
</script>