Develop Reference

How can I extract all contained characters in a String? [duplicate]

I have a string with repeated letters. I want letters that are repeated more than once to show only once.
Example input: aaabbbccc
Expected output: abc
I've tried to create the code myself, but so far my function has the following problems:
if the letter doesn't repeat, it's not shown (it should be)
if it's repeated once, it's show only once (i.e. aa shows a - correct)
if it's repeated twice, shows all (i.e. aaa shows aaa - should be a)
if it's repeated 3 times, it shows 6 (if aaaa it shows aaaaaa - should be a)
function unique_char(string) {
var unique = '';
var count = 0;
for (var i = 0; i < string.length; i++) {
for (var j = i+1; j < string.length; j++) {
if (string[i] == string[j]) {
count++;
unique += string[i];
}
}
}
return unique;
}
document.write(unique_char('aaabbbccc'));
The function must be with loop inside a loop; that's why the second for is inside the first.

Fill a Set with the characters and concatenate its unique entries:
function unique(str) {
return String.prototype.concat.call(...new Set(str));
}
console.log(unique('abc')); // "abc"
console.log(unique('abcabc')); // "abc"

Convert it to an array first, then use Josh Mc’s answer at How to get unique values in an array, and rejoin, like so:
var nonUnique = "ababdefegg";
var unique = Array.from(nonUnique).filter(function(item, i, ar){ return ar.indexOf(item) === i; }).join('');
All in one line. :-)

Too late may be but still my version of answer to this post:
function extractUniqCharacters(str){
var temp = {};
for(var oindex=0;oindex<str.length;oindex++){
temp[str.charAt(oindex)] = 0; //Assign any value
}
return Object.keys(temp).join("");
}

You can use a regular expression with a custom replacement function:
function unique_char(string) {
return string.replace(/(.)\1*/g, function(sequence, char) {
if (sequence.length == 1) // if the letter doesn't repeat
return ""; // its not shown
if (sequence.length == 2) // if its repeated once
return char; // its show only once (if aa shows a)
if (sequence.length == 3) // if its repeated twice
return sequence; // shows all(if aaa shows aaa)
if (sequence.length == 4) // if its repeated 3 times
return Array(7).join(char); // it shows 6( if aaaa shows aaaaaa)
// else ???
return sequence;
});
}

Using lodash:
_.uniq('aaabbbccc').join(''); // gives 'abc'

Per the actual question: "if the letter doesn't repeat its not shown"
function unique_char(str)
{
var obj = new Object();
for (var i = 0; i < str.length; i++)
{
var chr = str[i];
if (chr in obj)
{
obj[chr] += 1;
}
else
{
obj[chr] = 1;
}
}
var multiples = [];
for (key in obj)
{
// Remove this test if you just want unique chars
// But still keep the multiples.push(key)
if (obj[key] > 1)
{
multiples.push(key);
}
}
return multiples.join("");
}
var str = "aaabbbccc";
document.write(unique_char(str));

Your problem is that you are adding to unique every time you find the character in string. Really you should probably do something like this (since you specified the answer must be a nested for loop):
function unique_char(string){
var str_length=string.length;
var unique='';
for(var i=0; i<str_length; i++){
var foundIt = false;
for(var j=0; j<unique.length; j++){
if(string[i]==unique[j]){
foundIt = true;
break;
}
}
if(!foundIt){
unique+=string[i];
}
}
return unique;
}
document.write( unique_char('aaabbbccc'))
In this we only add the character found in string to unique if it isn't already there. This is really not an efficient way to do this at all ... but based on your requirements it should work.
I can't run this since I don't have anything handy to run JavaScript in ... but the theory in this method should work.

Try this if duplicate characters have to be displayed once, i.e.,
for i/p: aaabbbccc o/p: abc
var str="aaabbbccc";
Array.prototype.map.call(str,
(obj,i)=>{
if(str.indexOf(obj,i+1)==-1 ){
return obj;
}
}
).join("");
//output: "abc"
And try this if only unique characters(String Bombarding Algo) have to be displayed, add another "and" condition to remove the characters which came more than once and display only unique characters, i.e.,
for i/p: aabbbkaha o/p: kh
var str="aabbbkaha";
Array.prototype.map.call(str,
(obj,i)=>{
if(str.indexOf(obj,i+1)==-1 && str.lastIndexOf(obj,i-1)==-1){ // another and condition
return obj;
}
}
).join("");
//output: "kh"

<script>
uniqueString = "";
alert("Displays the number of a specific character in user entered string and then finds the number of unique characters:");
function countChar(testString, lookFor) {
var charCounter = 0;
document.write("Looking at this string:<br>");
for (pos = 0; pos < testString.length; pos++) {
if (testString.charAt(pos) == lookFor) {
charCounter += 1;
document.write("<B>" + lookFor + "</B>");
} else
document.write(testString.charAt(pos));
}
document.write("<br><br>");
return charCounter;
}
function findNumberOfUniqueChar(testString) {
var numChar = 0,
uniqueChar = 0;
for (pos = 0; pos < testString.length; pos++) {
var newLookFor = "";
for (pos2 = 0; pos2 <= pos; pos2++) {
if (testString.charAt(pos) == testString.charAt(pos2)) {
numChar += 1;
}
}
if (numChar == 1) {
uniqueChar += 1;
uniqueString = uniqueString + " " + testString.charAt(pos)
}
numChar = 0;
}
return uniqueChar;
}
var testString = prompt("Give me a string of characters to check", "");
var lookFor = "startvalue";
while (lookFor.length > 1) {
if (lookFor != "startvalue")
alert("Please select only one character");
lookFor = prompt(testString + "\n\nWhat should character should I look for?", "");
}
document.write("I found " + countChar(testString, lookFor) + " of the<b> " + lookFor + "</B> character");
document.write("<br><br>I counted the following " + findNumberOfUniqueChar(testString) + " unique character(s):");
document.write("<br>" + uniqueString)
</script>

Here is the simplest function to do that
function remove(text)
{
var unique= "";
for(var i = 0; i < text.length; i++)
{
if(unique.indexOf(text.charAt(i)) < 0)
{
unique += text.charAt(i);
}
}
return unique;
}

The one line solution will be to use Set. const chars = [...new Set(s.split(''))];

If you want to return values in an array, you can use this function below.
const getUniqueChar = (str) => Array.from(str)
.filter((item, index, arr) => arr.slice(index + 1).indexOf(item) === -1);
console.log(getUniqueChar("aaabbbccc"));
Alternatively, you can use the Set constructor.
const getUniqueChar = (str) => new Set(str);
console.log(getUniqueChar("aaabbbccc"));

Here is the simplest function to do that pt. 2
const showUniqChars = (text) => {
let uniqChars = "";
for (const char of text) {
if (!uniqChars.includes(char))
uniqChars += char;
}
return uniqChars;
};

const countUnique = (s1, s2) => new Set(s1 + s2).size
a shorter way based on #le_m answer

let unique=myArray.filter((item,index,array)=>array.indexOf(item)===index)

print like this * "a" -> "a1" * "aabbbaa" -> "a2b3a2"

I am new to js.
can you tell me how to print like this * "a" -> "a1" * "aabbbaa" -> "a2b3a2"
i tried with hash map but test cases failing.
providing my code below.
i am not good in hash map.
can you tell me how to solve with hash map so that in future I can fix it my self.
not sure what data structure to use for this one.
providing my code below.
const _ = require("underscore");
const rle = ( input ) => {
console.log("input--->" + input);
//var someString ="aaa";
var someString = input;
var arr = someString.split("");
var numberCount = {};
for(var i=0; i< arr.length; i++) {
var alphabet = arr[i];
if(numberCount[alphabet]){
numberCount[alphabet] = numberCount[alphabet] + 1;
}
else{
numberCount[alphabet] = 1;
}
}
console.log("a:" + numberCount['a'], "b:" + numberCount['b']);
}
/**
* boolean doTestsPass()
* Returns true if all the tests pass. Otherwise returns false.
*/
/**
* Returns true if all tests pass; otherwise, returns false.
*/
const doTestsPass = () => {
const VALID_COMBOS = {"aaa": "a3", "aaabbc":"a3b2c1"};
let testPassed = true;
_.forEach(VALID_COMBOS, function(value, key) {
console.log(key, rle(key));
if (value !== rle(key)) {
testPassed = false;
}
});
return testPassed;
}
/**
* Main execution entry.
*/
if(doTestsPass())
{
console.log("All tests pass!");
}
else
{
console.log("There are test failures.");
}

You could
match groups of characters,
get the character and the count and
join it to a string.
function runLengthEncoding(string) {
return string
.match(/(.)\1*/g) // keep same characters in a single string
.map(s => s[0] + s.length) // take first character of string and length
.join(''); // create string of array
}
console.log(['a', 'aaa', 'aaabbc'].map(runLengthEncoding));
This is a bit more understandable version which iterates the given string and count the characters. If a different character is found, the last character and count is added to the result string.
At the end, a check is made, to prevent counting of empty strings and the last character cound is added to the result.
function runLengthEncoding(string) {
var result = '',
i,
count = 0,
character = string[0];
for (i = 0; i < string.length; i++) {
if (character === string[i]) {
count++;
continue;
}
result += character + count;
character = string[i];
count = 1;
}
if (count) {
result += character + count;
}
return result;
}
console.log(['', 'a', 'aaa', 'aaabbc'].map(runLengthEncoding));

You can reduce the array into a multidimensional array. map and join the array to convert to string.
const rle = (input) => {
return input.split("").reduce((c, v) => {
if (c[c.length - 1] && c[c.length - 1][0] === v) c[c.length - 1][1]++;
else c.push([v, 1]);
return c;
}, []).map(o => o.join('')).join('');
}
console.log(rle("a"));
console.log(rle("aabbbaa"));
console.log(rle("aaaaaa"));

Your function rle doesn't return a result.
Also note, this implementation may pass the test cases you wrote, but not the examples you mentioned in your question: for the string "aabbaa" this will produce "a4b2", not " a2b2a2" .

A simpler solution:
function runLengthEncoding(str) {
let out = "";
for (let i = 0; i < str.length; ++i) {
let temp = str[i];
let count = 1;
while (i < str.length && str[i+1] == temp) {
++count;
++i;
}
out += temp + count;
} // end-for
return out;
}
console.log(runLengthEncoding("a"));
console.log(runLengthEncoding("aabbbaa"));
console.log(runLengthEncoding("aaaaaa"));

Javascript Cannot call method 'indexOf' of undefined at order

I need to make a function that will sort a string. Each word in the String will contain a single number. Numbers can be from 1 to 9(no 0).
For example an input: "is2 Thi1s T4est 3a" the function should return "Thi1s is2 3a T4est".
My code is :
function order(words)
{
// ...
if(words == '')
{
return words;
}
var all_words = words.split(" ");
var checked_words = new Array();
var joined_words = "";
for(i = 1; i <= 9; i++)
{
//console.log(checked_words);
//checked_words[i-1] = all_words;
for(j = 1;j <= all_words.length; j++)
{
if(all_words[i-1].indexOf(i) != -1)
{
checked_words.push(all_words[i-1]);
if(i == (all_words.length))
{
joined_words = checked_words.join(" ");
return joined_words;
}
}
}
}
}
The problem is it kept showing "TypeError: Cannot call method 'indexOf' of undefined at order". Please help thanks!

I've been trying to figure out what that first loop is doing and I finally realised it's checking for the numbers in the string. You shouldn't do it that way because it's 1) weird, 2) non-peformant if it worked.
Instead use one loop and a regex to find the number in the string.
for (var i = 0; i < all_words.length; i++) {
// take the first match and coerce it to an integer
var n = +all_words[i].match(/\d+/)[0];
checked_words[n] = all_words[i];
}
// checked_words has an undefined element at index 0;
// slice takes a copy of the array from index 1
return checked_words.slice(1).join(' ');
DEMO
Alternatively you could use reduce.
var joined_words = str.split(' ').reduce(function (p, c) {
var n = +c.match(/\d+/)[0];
p[n] = c;
return p;
}, []).slice(1).join(' ');
DEMO
You're not even tied to the numbers 0-9. As long as you filter out the undefined elements, you're golden...
var str = 'is2 Thi1s T4est 3a Ye12s!';
var out = str.split(' ').reduce(function (p, c) {
var n = +c.match(/\d+/)[0];
p[n] = c;
return p;
}, []).slice(1).filter(function (el) {
return el !== undefined;
}).join(' ');
DEMO

I need to get the last two value after dot using javascript split

I am getting the results like demo.in,demo.co.in,demo.tv,demo.org.in
I need to split the extension separately using JavaScript split function
var vsp = i.split(".");
this is my code I will get the result as
demo,in demo,co,in
but I need to get the extension separately

Working fiddle(demo version)
var values = [
"demo.in",
"demo.co.in",
"demo.tv","demo.org"
];
var results = [];
// iterate through the values
for (var i = 0, len = values.length; i < len; i++) {
// Split the parts on every dot.
var parts = values[i].split(".");
// Remove the first part (before the first dot).
parts = parts.slice(1, parts.length);
// Join the results together
results.push(parts.join("."));
};
console.dir(results); // all done
// Nicely display the values for the OP:
for (var i = 0, len = results.length; i < len; i++) {
document.body.innerHTML += (i + 1) + ": " + results[i] + "<br />";
};

I have no idea what you want, so here's some functions to cover the likely cases:
var s = 'demo.co.in'
// Return everything before the first '.'
function getExtension(s) {
var m = s.match(/^[^.]+/);
return m? m[0] : '';
}
alert(getExtension(s)); // demo
// Return everything after the last '.'
function getExtension2(s) {
var m = s.match(/[^.]+$/);
return m? m[0] : '';
}
alert(getExtension2(s)); // in
// Return everything after the first '.'
function getExtension3(s) {
return s.replace(/^[^.]+\./, '');
}
alert(getExtension3(s)); // co.in

I could not understand exactly .. "the extension" . You can try like below code
var urls = "demo,demo.in,my.demo.co.in,demo.tv,demo.org.in"
.split(',');
var splited = urls.reduce( function( o, n ){
var parts = n.split( '.' );
o[ n ] = (function(){
var ext = [];
while( !(parts.length == 1 || ext.length == 2) ) {
ext.unshift( parts.pop() );
};
return ext.join('.');
}());
return o;
}, {} );
console.log( JSON.stringify( splited ) );
which prints
{
"demo":"",
"demo.in":"in",
"my.demo.co.in":"co.in",
"demo.tv":"tv",
"demo.org.in":"org.in"
}
process result using
for( var i in splited ) {
console.log( i, splited[i]);
}

Try this:
var vsp = i.split(".");
for(var i=0; i< vsp.length; i++){
if(i !== 0){ //leaving the first match
// do something with vsp[i]
}
}
I hope you are not considering www also. :). If so, then keep i>1 instead of i !== 0.

There are several ways to do it (probably none as easy as it should be). You could define a function like this:
function mySplit(str, delim) {
var idx = (str.indexOf(delim) == -1) ? str.indexOf(delim) : str.length;
return [str.substr(0,idx), str.substr(idx)];
}
And then call it like: var sp = mySplit(i, ".");

You can also use lastIndexOf, which returns the location of the last . and from there you can get the rest of the string using substring.
function getExtension(hostName) {
var extension = null;
if(hostName && hostName.length > 0 && hostName.indexOf(".") !== -1) {
extension = hostName.substring(hostName.lastIndexOf(".") + 1);
}
return extension;
}
From there, you can use this function in a loop to get the extensions of many host names.
Edit Just noticed the "last two values" part in the title :) Thanks #rab

Javascript Function to split and return a value from a string

I am trying to grab a certain value. I am new to javascript and I can't figure out why this is not working.
If I parse "kid_2" I should get "kostas". Instead of "Kostas" I always get "02-23-2000". So I must have a logic problem in the loop but I am really stuck.
function getold_val(fieldname,str){
var chunks=str.split("||");
var allchunks = chunks.length-1;
for(k=0;k<allchunks;k++){
var n=str.indexOf(fieldname);
alert(chunks[k]);
if(n>0){
var chunkd=chunks[k].split("::");
alert(chunkd);
return chunkd[1];
}
}
}
var test = getold_val('kid_2','date_1::02-23-2000||date_2::06-06-1990||kid_1::George||kid_2::Kostas||');
alert(test);

A regex may be a little more appealing. Here's a fiddle:
function getValue(source, key){
return (new RegExp("(^|\\|)" + key + "::([^$\\|]+)", "i").exec(source) || {})[2];
}
getValue("date_1::02-23-2000||date_2::06-06-1990||kid_1::George||kid_2::Kostas||","kid_2");
But if you want something a little more involved, you can parse that string into a dictionary like so (fiddle):
function splitToDictionary(val, fieldDelimiter, valueDelimiter){
var dict = {},
fields = val.split(fieldDelimiter),
kvp;
for (var i = 0; i < fields.length; i++) {
if (fields[i] !== "") {
kvp = fields[i].split(valueDelimiter);
dict[kvp[0]] = kvp[1];
}
}
return dict;
}
var dict = splitToDictionary("date_1::02-23-2000||date_2::06-06-1990||kid_1::George||kid_2::Kostas||","||","::");
console.log(dict["date_1"]);
console.log(dict["date_2"]);
console.log(dict["kid_1"]);
console.log(dict["kid_2"]);​

This works, here's my fiddle.
function getold_val(fieldname,str) {
var chunks = str.split('||');
for(var i = 0; i < chunks.length-1; i++) {
if(chunks[i].indexOf(fieldname) >= 0) {
return(chunks[i].substring(fieldname.length+2));
}
}
}
alert(getold_val('kid_2', 'date_1::02-23-2000||date_2::06-06-1990||kid_1::George||kid_2::Kostas||'));
The issue with your code was (as #slebetman noticed as well) the fact that a string index can be 0 because it starts exactly in the first letter.
The code is almost the same as yours, I just didn't use the second .split('::') because I felt a .substring(...) would be easier.

There are two bugs. The first error is in the indexOf call:
var n = str.indexOf(fieldname);
This will always return a value greater than or equal to 0 since the field exists in the string. What you should be doing is:
var n = chunks[k].indexOf(fieldname);
The second error is in your if statement. It should be:
if(n >= 0) {
...
}
or
if(n > -1) {
...
}
The substring you are looking for could very well be the at the beginning of the string, in which case its index is 0. indexOf returns -1 if it cannot find what you're looking for.
That being said, here's a better way to do what you're trying to do:
function getold_val(fieldName, str) {
var keyValuePairs = str.split("||");
var returnValue = null;
if(/||$/.match(str)) {
keyValuePairs = keyValuePairs.slice(0, keyValuePairs.length - 1);
}
var found = false;
var i = 0;
while(i < keyValuePairs.length && !found) {
var keyValuePair = keyValuePairs[i].split("::");
var key = keyValuePair[0];
var value = keyValuePair[1];
if(fieldName === key) {
returnValue = value;
found = true;
}
i++;
}
return returnValue;
}