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)

I have a string with counts and alphabets in ordered manner i have to print the alphabets according to their count

var input = `2
6
z2k1o2
6
m2v1p2`
var newInput = input.split("\n")
//console.log(newInput.length)
var input_arr = input.trim().split("\n")
var n = Number(input_arr[0])
//console.log(input_arr)
for (var i = 1; i < input_arr.length; i = i + 2) {
var length = Number(input_arr[i])
var string = input_arr[i + 1].trim()
}
//console.log(string)
var newstring = string
//console.log(newstring)
var alpha = []
var num = []
for (i = 1; i < string.length; i += 2) {
num.push(string[i])
}
var newnum = num.map(Number)
//console.log(newnum)
for (i = 0; i < string.length; i += 2) {
alpha.push(string[i])
}
var newalpha = (alpha)
//console.log(newalpha)
var answer = []
for (i = 0; i < newnum.length; i++) {
for (j = 0; j < newnum[i]; j++) {
answer.push(newalpha[i])
}
}
console.log(answer.join(""))
Here I'm getting only one output can you please explain why And do you like share any other approach for this problem.
This is the input z2k1o2 and the output should be zzkoo
The input will follow this patter of alphabets ans counts..

I'd do this with a regular expression that captures pairs of (character, number) and uses the function-replacement mode of .replace() to generate the replacement string.
> "a2b1c2".replace(/([a-z])([0-9]+)/ig, (_, a, b) => a.repeat(+b))
"aabcc"
>"pos2es2".replace(/([a-z])([0-9]+)/ig, (_, a, b) => a.repeat(+b))
"possess"

Here's where you're running into your error:
for (var i = 1; i < input_arr.length; i = i + 2) {
var length = Number(input_arr[i])
var string = input_arr[i + 1].trim()
}
You're going through the entire input array and saving each of the lengths and strings, but you're overwriting the length and the string each time you read it - only the last length and last string are saved, and so only the last length and last string are processed / printed.

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"));

Perform a merge on two strings

I'm trying to build a collaborative doc editor and implement operational transformation. Imagine we have a string that is manipulated simultaneously by 2 users. They can only add characters, not remove them. We want to incorporate both of their changes.
The original string is: catspider
The first user does this: cat<span id>spider</span>
The second user does this: c<span id>atspi</span>der
I'm trying to write a function that will produce: c<span id>at<span id>spi</span>der</span>
The function I've written is close, but it produces c<span id>at<span i</span>d>spider</span> codepen here
String.prototype.splice = function(start, newSubStr) {
return this.slice(0, start) + newSubStr + this.slice(start);
};
function merge(saved, working, requested) {
if (!saved || !working || !requested) {
return false;
}
var diffSavedWorking = createDiff(working, saved);
var diffRequestedWorking = createDiff(working, requested);
var newStr = working;
for (var i = 0; i < Math.max(diffRequestedWorking.length, diffSavedWorking.length); i++) {
//splice does an insert `before` -- we will assume that the saved document characters
//should always appear before the requested document characters in this merger operation
//so we first insert requested and then saved, which means that the final string will have the
//original characters first.
if (diffRequestedWorking[i]) {
newStr = newStr.splice(i, diffRequestedWorking[i]);
//we need to update the merge arrays by the number of
//inserted characters.
var length = diffRequestedWorking[i].length;
insertNatX(diffSavedWorking, length, i + 1);
insertNatX(diffRequestedWorking, length, i + 1);
}
if (diffSavedWorking[i]) {
newStr = newStr.splice(i, diffSavedWorking[i]);
//we need to update the merge arrays by the number of
//inserted characters.
var length = diffSavedWorking[i].length;
insertNatX(diffSavedWorking, length, i + 1);
insertNatX(diffRequestedWorking, length, i + 1);
}
}
return newStr;
}
//arr1 should be the shorter array.
//returns inserted characters at their
//insertion index.
function createDiff(arr1, arr2) {
var diff = [];
var j = 0;
for (var i = 0; i < arr1.length; i++) {
diff[i] = "";
while (arr2[j] !== arr1[i]) {
diff[i] += arr2[j];
j++;
}
j++;
}
var remainder = arr2.substr(j);
if (remainder) diff[i] = remainder;
return diff;
}
function insertNatX(arr, length, pos) {
for (var j = 0; j < length; j++) {
arr.splice(pos, 0, "");
}
}
var saved = 'cat<span id>spider</span>';
var working = 'catspider';
var requested = 'c<span id>atspi</span>der';
console.log(merge(saved, working, requested));
Would appreciate any thoughts on a better / simpler way to achieve this.

Google-Spreadsheet Scripts

I have a column of cells in a particular sheet of Google Spreadsheet document.
This column references multiple values in another sheet using the built-in JOIN command:
=JOIN(", ",Regular!B3,Regular!B9,Regular!B10,Regular!B11,Regular!B12,Regular!B13,Regular!B14)
typical output for each such cell is a list of integers that are comma-separated, f.ex:
2, 5, 10, 12, 13
Some cells use ranges like this:
=JOIN(", ",Regular!B3:B9)
I want to lock these cells in the formula as such: Regular!$B$3,Regular!$B:$9...
Right now I want each reference to lock both column and row, but a solution that lets me pick row, column or both is a better solution.
1) I haven't found a way to do this without using a custom script - have I missed something?
2) My custom script solution is unfinished:
function eachCellInRange(range, op) {
var numRows = range.getNumRows();
var numCols = range.getNumColumns();
for (var i = 1; i <= numRows; i++) {
for (var j = 1; j <= numCols; j++) {
op(range.getCell(i,j), i, j);
}
}
};
function lockCell(cell, row, col) {
var formula = cell.getFormula();
if(formula) {
var startIdx = formula.indexOf('(');
if(startIdx > 0) {
//!! REGEX HERE !! //
cell.setValue(formula);
}
}
}
function lockRows() {
var range = SpreadsheetApp.getActiveSpreadsheet().getActiveSheet().getActiveRange();
eachCellInRange(range, lockCell);
};
I need to make a regex pattern that will identify the B3, B9... parts of the formula and change them to $B$3, $B$9... but also not break in the B1:B8 case
Currently all references are prefixed with SheetName! (e.g. Regular!B9:B20), in the future some may not be, so the most general solution is preferred.

I'm not sure whether this is what you're looking for but I would replace the little bit you currently have:
if(formula) {
var startIdx = formula.indexOf('(');
if(startIdx > 0) {
//!! REGEX HERE !! //
cell.setValue(formula);
}
}
by
if(formula.substring(0,6) == "=JOIN(") {
formula = formula.replace(/([A-Z]+(?=[0-9]))/g, function($1) {
return "$" +$1 + "$";
});
alert(formula);
// cell.setValue(formula);
}
Which ensures that the formula is a JOIN formula.
Also, I'm not that familiar with JS, but I put it in JSFiddle to see how it goes.
Warning: This will fail if your sheet names have alphanumeric characters (mix of letters and digits).

Using #Jerry's useful answer, I was able to suit it to my needs:
function eachCellInRange(range, op) {
var numRows = range.getNumRows();
var numCols = range.getNumColumns();
for (var i = 1; i <= numRows; i++) {
for (var j = 1; j <= numCols; j++) {
op(range.getCell(i,j), i, j);
}
}
};
var lockOn = 1, lockOff = -1, lockNop = 0,
lockChar = '$', lockEmpty = '';
function lock2char(newLock, curLock) {
if(newLock == lockNop) newLock = curLock;
return (newLock > lockNop) ? lockChar : lockEmpty;
}
function bool2lock(boolValue) {
return (boolValue) ? lockOn : lockOff;
}
function lockCell(lockCol, lockRow, cell, row, col) {
var formula = cell.getFormula();
if(formula) {
var startIdx = formula.indexOf('(');
if(startIdx > 0) {
var newFormula = formula.replace(/([A-Z|\$]+(?=[0-9]))/g, function(part) {
var prefix = lock2char(lockCol, (part.charAt(0) == lockChar));
var suffix = lock2char(lockRow, (part.charAt(part.length -1) == lockChar));
part = part.replace(/\$/g, '');
return prefix + part + suffix;
});
cell.setFormula(newFormula);
}
}
}
function lockRows() {
var range = SpreadsheetApp.getActiveSpreadsheet().getActiveSheet().getActiveRange();
eachCellInRange(range, lockCell.bind(this, lockOff, lockOn));
};