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Javascript includes and map together [duplicate]

I am supposed to write a program in JavaScript to find all the anagrams within a series of words provided. e.g.:
monk, konm, nkom, bbc, cbb, dell, ledl, llde
The output should be categorised into rows:
1. monk konm, nkom;
2. bbc cbb;
3. dell ledl, llde;
I already sorted them into alphabetical order and put them into an array. i.e.:
kmno kmno bbc bbc dell dell
However I am stuck in comparing and finding the matching anagram within the array.
Any help will be greatly appreciated.

Javascript objects are excellent for this purpose, since they are essentially key/value stores:
// Words to match
var words = ["dell", "ledl", "abc", "cba"];
// The output object
var anagrams = {};
for (var i in words) {
var word = words[i];
// sort the word like you've already described
var sorted = sortWord(word);
// If the key already exists, we just push
// the new word on the the array
if (anagrams[sorted] != null) {
anagrams[sorted].push(word);
}
// Otherwise we create an array with the word
// and insert it into the object
else {
anagrams[sorted] = [ word ];
}
}
// Output result
for (var sorted in anagrams) {
var words = anagrams[sorted];
var sep = ",";
var out = "";
for (var n in words) {
out += sep + words[n];
sep = "";
}
document.writeln(sorted + ": " + out + "<br />");
}

Here is my take:
var input = "monk, konm, bbc, cbb, dell, ledl";
var words = input.split(", ");
for (var i = 0; i < words.length; i++) {
var word = words[i];
var alphabetical = word.split("").sort().join("");
for (var j = 0; j < words.length; j++) {
if (i === j) {
continue;
}
var other = words[j];
if (alphabetical === other.split("").sort().join("")) {
console.log(word + " - " + other + " (" + i + ", " + j + ")");
}
}
}
where the output would be (the word, the match and the index of both):
monk - konm (0, 1)
konm - monk (1, 0)
bbc - cbb (2, 3)
cbb - bbc (3, 2)
dell - ledl (4, 5)
ledl - dell (5, 4)
To get the characters in the in alphabetical order, I used split("") ot get an array, called sort() and used join("") to get a string from the array.

Simple Solution
function anagrams(stringA, stringB) {
return cleanString(stringA) === cleanString(stringB);
}
function cleanString(str) {
return str.replace(/[^\w]/g).toLowerCase().split('').sort().join()
}
anagrams('monk','konm')
If it is anagrams function will return true otherwise false

I worked through a similar question to this today and wanted to share the results of my work. I was focused on just detecting the anagram so processing the list of words was not part of my exercise but this algorithm should provide a highly performant way to detect an anagram between two words.
function anagram(s1, s2){
if (s1.length !== s2.length) {
// not the same length, can't be anagram
return false;
}
if (s1 === s2) {
// same string must be anagram
return true;
}
var c = '',
i = 0,
limit = s1.length,
match = 0,
idx;
while(i < s1.length){
// chomp the next character
c = s1.substr(i++, 1);
// find it in the second string
idx = s2.indexOf(c);
if (idx > -1) {
// found it, add to the match
match++;
// assign the second string to remove the character we just matched
s2 = s2.substr(0, idx) + s2.substr(idx + 1);
} else {
// not found, not the same
return false;
}
}
return match === s1.length;
}
I think technically is can be solved like this:
function anagram(s1, s2){
return s1.split("").sort().join("") === s2.split("").sort().join("");
}
The reason I chose the earlier approach is that it is more performant for larger strings since you don't need to sort either string, convert to an array or loop through the entire string if any possible failure case is detected.

Probably not the most efficient way, but a clear way around using es6
function sortStrChars(str) {
if (!str) {
return;
}
str = str.split('');
str = str.sort();
str = str.join('');
return str;
}
const words = ["dell", "ledl", "abc", "cba", 'boo'];
function getGroupedAnagrams(words) {
const anagrams = {}; // {abc:[abc,cba], dell:[dell, ledl]}
words.forEach((word) => {
const sortedWord = sortStrChars(word);
if (anagrams[sortedWord]) {
return anagrams[sortedWord].push(word);
}
anagrams[sortedWord] = [word];
});
return anagrams;
}
const groupedAnagrams = getGroupedAnagrams(words);
for (const sortedWord in groupedAnagrams) {
console.log(groupedAnagrams[sortedWord].toString());
}

I had this question in an interview. Given an array of words ['cat', 'dog', 'tac', 'god', 'act'], return an array with all the anagrams grouped together. Makes sure the anagrams are unique.
var arr = ['cat', 'dog', 'tac', 'god', 'act'];
var allAnagrams = function(arr) {
var anagrams = {};
arr.forEach(function(str) {
var recurse = function(ana, str) {
if (str === '')
anagrams[ana] = 1;
for (var i = 0; i < str.length; i++)
recurse(ana + str[i], str.slice(0, i) + str.slice(i + 1));
};
recurse('', str);
});
return Object.keys(anagrams);
}
console.log(allAnagrams(arr));
//["cat", "cta", "act", "atc", "tca", "tac", "dog", "dgo", "odg", "ogd", "gdo", "god"]

Best and simple way to solve is using for loops and traversing it to each string and then store their result in object.
Here is the solution :-
function anagram(str1, str2) {
if (str1.length !== str2.length) {
return false;
}
const result = {};
for (let i=0;i<str1.length;i++) {
let char = str1[i];
result[char] = result[char] ? result[char] += 1 : result[char] = 1;
}
for (let i=0;i<str2.length;i++) {
let char = str2[i];
if (!result[char]) {
return false;
}
else {
result[char] = -1;
}
}
return true;
}
console.log(anagram('ronak','konar'));

I know this is an ancient post...but I just recently got nailed during an interview on this one. So, here is my 'new & improved' answer:
var AnagramStringMiningExample = function () {
/* Author: Dennis Baughn
* This has also been posted at:
* http://stackoverflow.com/questions/909449/anagrams-finder-in-javascript/5642437#5642437
* Free, private members of the closure and anonymous, innner function
* We will be building a hashtable for anagrams found, with the key
* being the alphabetical char sort (see sortCharArray())
* that the anagrams all have in common.
*/
var dHash = {};
var sortCharArray = function(word) {
return word.split("").sort().join("");
};
/* End free, private members for the closure and anonymous, innner function */
/* This goes through the dictionary entries.
* finds the anagrams (if any) for each word,
* and then populates them in the hashtable.
* Everything strictly local gets de-allocated
* so as not to pollute the closure with 'junk DNA'.
*/
(function() {
/* 'dictionary' referring to English dictionary entries. For a real
* English language dictionary, we could be looking at 20,000+ words, so
* an array instead of a string would be needed.
*/
var dictionaryEntries = "buddy,pan,nap,toot,toto,anestri,asterin,eranist,nastier,ratines,resiant,restain,retains,retinas,retsina,sainter,stainer,starnie,stearin";
/* This could probably be refactored better.
* It creates the actual hashtable entries. */
var populateDictionaryHash = function(keyword, newWord) {
var anagrams = dHash[keyword];
if (anagrams && anagrams.indexOf(newWord) < 0)
dHash[keyword] = (anagrams+','+newWord);
else dHash[keyword] = newWord;
};
var words = dictionaryEntries.split(",");
/* Old School answer, brute force
for (var i = words.length - 1; i >= 0; i--) {
var firstWord = words[i];
var sortedFirst = sortCharArray(firstWord);
for (var k = words.length - 1; k >= 0; k--) {
var secondWord = words[k];
if (i === k) continue;
var sortedSecond = sortCharArray(secondWord);
if (sortedFirst === sortedSecond)
populateDictionaryHash(sortedFirst, secondWord);
}
}/*
/*Better Method for JS, using JS Array.reduce(callback) with scope binding on callback function */
words.reduce(function (prev, cur, index, array) {
var sortedFirst = this.sortCharArray(prev);
var sortedSecond = this.sortCharArray(cur);
if (sortedFirst === sortedSecond) {
var anagrams = this.dHash[sortedFirst];
if (anagrams && anagrams.indexOf(cur) < 0)
this.dHash[sortedFirst] = (anagrams + ',' + cur);
else
this.dHash[sortedFirst] = prev + ','+ cur;
}
return cur;
}.bind(this));
}());
/* return in a nice, tightly-scoped closure the actual function
* to search for any anagrams for searchword provided in args and render results.
*/
return function(searchWord) {
var keyToSearch = sortCharArray(searchWord);
document.writeln('<p>');
if (dHash.hasOwnProperty(keyToSearch)) {
var anagrams = dHash[keyToSearch];
document.writeln(searchWord + ' is part of a collection of '+anagrams.split(',').length+' anagrams: ' + anagrams+'.');
} else document.writeln(searchWord + ' does not have anagrams.');
document.writeln('<\/p>');
};
};
Here is how it executes:
var checkForAnagrams = new AnagramStringMiningExample();
checkForAnagrams('toot');
checkForAnagrams('pan');
checkForAnagrams('retinas');
checkForAnagrams('buddy');
Here is the output of the above:
toot is part of a collection of 2
anagrams: toto,toot.
pan is part of a collection of 2
anagrams: nap,pan.
retinas is part of a collection of 14
anagrams:
stearin,anestri,asterin,eranist,nastier,ratines,resiant,restain,retains,retinas,retsina,sainter,stainer,starnie.
buddy does not have anagrams.

My solution to this old post:
// Words to match
var words = ["dell", "ledl", "abc", "cba"],
map = {};
//Normalize all the words
var normalizedWords = words.map( function( word ){
return word.split('').sort().join('');
});
//Create a map: normalizedWord -> real word(s)
normalizedWords.forEach( function ( normalizedWord, index){
map[normalizedWord] = map[normalizedWord] || [];
map[normalizedWord].push( words[index] );
});
//All entries in the map with an array with size > 1 are anagrams
Object.keys( map ).forEach( function( normalizedWord , index ){
var combinations = map[normalizedWord];
if( combinations.length > 1 ){
console.log( index + ". " + combinations.join(' ') );
}
});
Basically I normalize every word by sorting its characters so stackoverflow would be acefkloorstvw, build a map between normalized words and the original words, determine which normalized word has more than 1 word attached to it -> That's an anagram.

Maybe this?
function anagram (array) {
var organized = {};
for (var i = 0; i < array.length; i++) {
var word = array[i].split('').sort().join('');
if (!organized.hasOwnProperty(word)) {
organized[word] = [];
}
organized[word].push(array[i]);
}
return organized;
}
anagram(['kmno', 'okmn', 'omkn', 'dell', 'ledl', 'ok', 'ko']) // Example
It'd return something like
{
dell: ['dell', 'ledl'],
kmno: ['kmno', okmn', 'omkn'],
ko: ['ok', ko']
}
It's a simple version of what you wanted and certainly it could be improved avoiding duplicates for example.

My two cents.
This approach uses XOR on each character in both words. If the result is 0, then you have an anagram. This solution assumes case sensitivity.
let first = ['Sower', 'dad', 'drown', 'elbow']
let second = ['Swore', 'add', 'down', 'below']
// XOR all characters in both words
function isAnagram(first, second) {
// Word lengths must be equal for anagram to exist
if (first.length !== second.length) {
return false
}
let a = first.charCodeAt(0) ^ second.charCodeAt(0)
for (let i = 1; i < first.length; i++) {
a ^= first.charCodeAt(i) ^ second.charCodeAt(i)
}
// If a is 0 then both words have exact matching characters
return a ? false : true
}
// Check each pair of words for anagram match
for (let i = 0; i < first.length; i++) {
if (isAnagram(first[i], second[i])) {
console.log(`'${first[i]}' and '${second[i]}' are anagrams`)
} else {
console.log(`'${first[i]}' and '${second[i]}' are NOT anagrams`)
}
}

function isAnagram(str1, str2) {
var str1 = str1.toLowerCase();
var str2 = str2.toLowerCase();
if (str1 === str2)
return true;
var dict = {};
for(var i = 0; i < str1.length; i++) {
if (dict[str1[i]])
dict[str1[i]] = dict[str1[i]] + 1;
else
dict[str1[i]] = 1;
}
for(var j = 0; j < str2.length; j++) {
if (dict[str2[j]])
dict[str2[j]] = dict[str2[j]] - 1;
else
dict[str2[j]] = 1;
}
for (var key in dict) {
if (dict[key] !== 0)
return false;
}
return true;
}
console.log(isAnagram("hello", "olleh"));

I have an easy example
function isAnagram(strFirst, strSecond) {
if(strFirst.length != strSecond.length)
return false;
var tempString1 = strFirst.toLowerCase();
var tempString2 = strSecond.toLowerCase();
var matched = true ;
var cnt = 0;
while(tempString1.length){
if(tempString2.length < 1)
break;
if(tempString2.indexOf(tempString1[cnt]) > -1 )
tempString2 = tempString2.replace(tempString1[cnt],'');
else
return false;
cnt++;
}
return matched ;
}
Calling function will be isAnagram("Army",Mary);
Function will return true or false

let words = ["dell", "ledl","del", "abc", "cba", 'boo'];
//sort each item
function sortArray(data){
var r=data.split('').sort().join().replace(/,/g,'');
return r;
}
var groupObject={};
words.forEach((item)=>{
let sorteditem=sortArray(item);
//Check current item is in the groupObject or not.
//If not then add it as an array
//else push it to the object property
if(groupObject[sorteditem])
return groupObject[sorteditem].push(item);
groupObject[sorteditem]=[sorteditem];
});
//to print the result
for(i=0;i<Object.keys(groupObject).length;i++)
document.write(groupObject[Object.keys(groupObject)[i]] + "<br>");
/* groupObject value:
abc: (2) ["abc", "cba"]
boo: ["boo"]
del: ["del"]
dell: (2) ["dell", "ledl"]
OUTPUT:
------
dell,ledl
del
abc,cba
boo
*/

Compare string length, if not equal, return false
Create character Hashmap which stores count of character in strA e.g. Hello --> {H: 1, e: 1, l: 2, o: 1}
Loop over the second string and lookup the current character in Hashmap. If not exist, return false, else decrement the value by 1
If none of the above return falsy, it must be an anagram
Time complexity: O(n)
function isAnagram(strA: string, strB: string): boolean {
const strALength = strA.length;
const strBLength = strB.length;
const charMap = new Map<string, number>();
if (strALength !== strBLength) {
return false;
}
for (let i = 0; i < strALength; i += 1) {
const current = strA[i];
charMap.set(current, (charMap.get(current) || 0) + 1);
}
for (let i = 0; i < strBLength; i += 1) {
const current = strB[i];
if (!charMap.get(current)) {
return false;
}
charMap.set(current, charMap.get(current) - 1);
}
return true;
}

function findAnagram(str1, str2) {
let mappedstr1 = {}, mappedstr2 = {};
for (let item of str1) {
mappedstr1[item] = (mappedstr1[item] || 0) + 1;
}
for (let item2 of str2) {
mappedstr2[item2] = (mappedstr2[item2] || 0) + 1;
}
for (let key in mappedstr1) {
if (!mappedstr2[key]) {
return false;
}
if (mappedstr1[key] !== mappedstr2[key]) {
return false;
}
}
return true;
}
console.log(findAnagram("hello", "hlleo"));

Another example only for comparing 2 strings for an anagram.
function anagram(str1, str2) {
if (str1.length !== str2.length) {
return false;
} else {
if (
str1.toLowerCase().split("").sort().join("") ===
str2.toLowerCase().split("").sort().join("")
) {
return "Anagram";
} else {
return "Not Anagram";
}
}
}
console.log(anagram("hello", "olleh"));
console.log(anagram("ronak", "konar"));

const str1 ="1123451"
const str2 = "2341151"
function anagram(str1,str2) {
let count = 0;
if (str1.length!==str2.length) { return false;}
for(i1=0;i1<str1.length; i1++) {
for (i2=0;i2<str2.length; i2++) {
if (str1[i1]===str2[i2]){
count++;
break;
}
}
}
if (count===str1.length) { return true}
}
anagram(str1,str2)

Another solution for isAnagram using reduce
const checkAnagram = (orig, test) => {
return orig.length === test.length
&& orig.split('').reduce(
(acc, item) => {
let index = acc.indexOf(item);
if (index >= 0) {
acc.splice(index, 1);
return acc;
}
throw new Error('Not an anagram');
},
test.split('')
).length === 0;
};
const isAnagram = (tester, orig, test) => {
try {
return tester(orig, test);
} catch (e) {
return false;
}
}
console.log(isAnagram(checkAnagram, '867443', '473846'));
console.log(isAnagram(checkAnagram, '867443', '473846'));
console.log(isAnagram(checkAnagram, '867443', '475846'));

var check=true;
var str="cleartrip";
var str1="tripclear";
if(str.length!=str1.length){
console.log("Not an anagram");
check=false;
}
console.log(str.split("").sort());
console.log("----------"+str.split("").sort().join(''));
if(check){
if((str.split("").sort().join(''))===((str1.split("").sort().join('')))){
console.log("Anagram")
}
else{
console.log("not a anagram");
}
}

Here is my solution which addresses a test case where the input strings which are not anagrams, can be removed from the output. Hence the output contains only the anagram strings. Hope this is helpful.
/**
* Anagram Finder
* #params {array} wordArray
* #return {object}
*/
function filterAnagram(wordArray) {
let outHash = {};
for ([index, word] of wordArray.entries()) {
let w = word.split("").sort().join("");
outHash[w] = !outHash[w] ? [word] : outHash[w].concat(word);
}
let filteredObject = Object.keys(outHash).reduce(function(r, e) {
if (Object.values(outHash).filter(v => v.length > 1).includes(outHash[e])) r[e] = outHash[e]
return r;
}, {});
return filteredObject;
}
console.log(filterAnagram(['monk', 'yzx','konm', 'aaa', 'ledl', 'bbc', 'cbb', 'dell', 'onkm']));

i have recently faced this in the coding interview, here is my solution.
function group_anagrams(arr) {
let sortedArr = arr.map(item => item.split('').sort().join(''));
let setArr = new Set(sortedArr);
let reducedObj = {};
for (let setItem of setArr) {
let indexArr = sortedArr.reduce((acc, cur, index) => {
if (setItem === cur) {
acc.push(index);
}
return acc;
}, []);
reducedObj[setItem] = indexArr;
}
let finalArr = [];
for (let reduceItem in reducedObj) {
finalArr.push(reducedObj[reduceItem].map(item => arr[item]));
}
return finalArr;
}
group_anagrams(['car','cra','rca', 'cheese','ab','ba']);
output will be like
[
["car", "cra", "rca"],
["cheese"],
["ab", "ba"]
]

My solution has more code, but it avoids using .sort(), so I think this solution has less time complexity. Instead it makes a hash out of every word and compares the hashes:
const wordToHash = word => {
const hash = {};
// Make all lower case and remove spaces
[...word.toLowerCase().replace(/ /g, '')].forEach(letter => hash[letter] ? hash[letter] += 1 : hash[letter] = 1);
return hash;
}
const hashesEqual = (obj1, obj2) => {
const keys1 = Object.keys(obj1), keys2 = Object.keys(obj2);
let match = true;
if(keys1.length !== keys2.length) return false;
for(const key in keys1) { if(obj1[key] !== obj2[key]) match = false; break; }
return match;
}
const checkAnagrams = (word1, word2) => {
const hash1 = wordToHash(word1), hash2 = wordToHash(word2);
return hashesEqual(hash1, hash2);
}
console.log( checkAnagrams("Dormitory", "Dirty room") );

/*This is good option since
logic is easy,
deals with duplicate data,
Code to check anagram in an array,
shows results in appropriate manner,
function check can be separately used for comparing string in this regards with all benefits mentioned above.
*/
var words = ["deuoll", "ellduo", "abc","dcr","frt", "bu","cba","aadl","bca","elduo","bac","acb","ub","eldou","ellduo","ert","tre"];
var counter=1;
var ele=[];
function check(str1,str2)
{
if(str2=="")
return false;
if(str1.length!=str2.length)
return false;
var r1=[...(new Set (str1.split('').sort()))];
var r2=[...(new Set (str2.split('').sort()))];
var flag=true;
r1.forEach((item,index)=>
{
if(r2.indexOf(item)!=index)
{ flag=false;}
});
return flag;
}
var anagram=function ()
{
for(var i=0;i<words.length && counter!=words.length ;i++)
{
if(words[i]!="")
{
document.write("<br>"+words[i]+":");
counter++;
}
for(var j=i+1;j<words.length && counter !=words.length+1;j++)
{
if(check(words[i],words[j]))
{
ele=words[j];
document.write(words[j]+"&nbsp");
words[j]="";
counter++;
}
}
}
}
anagram();

If you just need count of anagrams
const removeDuplicatesAndSort = [...new Set(yourString.split(', '))].map(word => word.split('').sort().join())
const numberOfAnagrams = removeDuplicatesAndSort.length - [...new Set(removeDuplicatesAndSort)].length

function isAnagram(str1, str2){
let count = 0;
if (str1.length !== str2.length) {
return false;
} else {
let val1 = str1.toLowerCase().split("").sort();
let val2 = str2.toLowerCase().split("").sort();
for (let i = 0; i < val2.length; i++) {
if (val1[i] === val2[i]) {
count++;
}
}
if (count == str1.length) {
return true;
}
}
return false;
}
console.log(isAnagram("cristian", "Cristina"))

function findAnagrams (str, arr){
let newStr = "";
let output = [];
for (let i = 0; i < arr.length; i++) {
for (let j = 0; j < arr[i].length; j++) {
for (let k = 0; k < str.length; k++) {
if (str[k] === arr[i][j] && str.length === arr[i].length) {
newStr += arr[i][j];
}
}
} if(newStr.length === str.length){
output.push(newStr);
newStr = "";
}
}
return output;
}

const getAnagrams = (...args) => {
const anagrams = {};
args.forEach((arg) => {
const letters = arg.split("").sort().join("");
if (anagrams[letters]) {
anagrams[letters].push(arg);
} else {
anagrams[letters] = [arg];
}
});
return Object.values(anagrams);
}

function isAnagaram(str1, str2){
if(str1.length!== str2.length){
return false;
}
var obj1 = {};
var obj2 = {};
for(var arg of str1){
obj1[arg] = (obj1[arg] || 0 ) + 1 ;
}
for(var arg of str2){
obj2[arg] = (obj2[arg] || 0 ) + 1 ;
}
for( var key in obj1){
if(obj1[key] !== obj2[key]){
return false;
}
}
return true;
}
console.log(isAnagaram('texttwisttime' , 'timetwisttext'));

let validAnagram = (firstString, secondString) => {
if (firstString.length !== secondString.length) {
return false;
}
let secondStringArr = secondString.split('');
for (var char of firstString) {
charIndexInSecondString = secondString.indexOf(char);
if (charIndexInSecondString === -1) {
return false;
}
secondString = secondString.replace(char, '');
}
return true;
}

Recursive function to return each character in Input

i am trying to use recursion to return each character in a string. However, the output is not
//We define a function with input parameter.
function countCharInString(string) {
//vi Define an empty objec
const result = {};
//we loop through the length of string
for (let i = 0; i < string.length; i++) {
//create another variable for each element in string
const ch = string[i];
//BASE CASE: if string is empty, return Object with nothing
if (!result[ch]) {
return result[ch]=0;
} else {
//RECURSION: 1 plus whatever the length of the substring from the next character onwards is
return countCharInString(result[ch] + 1)
}
}
}
console.log(countCharInString("Vi skal tælle bogstaver"))
the output should be the following:
var result = {
l : 3,
a : 2,
e : 2,
s : 2,
t : 2,
v : 2,
b: 1,
i : 1,
k : 1,
o : 1,
r : 1,
æ : 1
};

i would suggest to do it with a simple reduce like so
var inputString = 'donald duck';
var result = inputString.split('').reduce((acc, char, index) => {
if (acc[char] !== undefined) {
acc[char] = acc[char] + 1;
}
else {
acc = { ...acc, [char]: 1 }
}
return acc
}, {})
see: https://jsfiddle.net/yswu91zh/21/

Only recursion would not give you the output that you are asking for. After recursively counting character you have to sort it by frequency and then by character. I have excluded a bunch of punctuation with space from counting, if you want exclude more just add it to the punctuation string. You have to use String.prototype.localeCompare() method to compare the characters. This method compares two strings in the current locale. As you are using Danish language you have to specify locale as da.
const punctuations = '.,:;!? ';
const countCharInString = (str, p = {}) => {
if (str.length === 0) return p;
const key = str[0].toLowerCase();
if (!punctuations.includes(key)) {
if (!p[key]) p[key] = 1;
else p[key] += 1;
}
return countCharInString(str.slice(1), p);
};
const cmp = (x, y) => {
if (x[1] === y[1]) {
return x[0].localeCompare(y[0], 'da');
}
return x[1] < y[1] ? 1 : -1;
};
const ret = Object.fromEntries(
Object.entries(countCharInString('Vi skal tælle bogstaver')).sort(cmp)
);
console.log(ret);

Removing consecutive duplicate characters

I am trying to remove consecutive duplicate characters from the string. So the input of aabbcde should output as cde.
I wrote the following code, but I am getting the incorrect output. I could not understand why.
var a = "aabbccddeef"
var ncopy = a;
let leftPointer = 0;
let rightPointer = 1;
let posList = [];
while (true) {
if (a[leftPointer] == a[rightPointer]) {
ncopy = ncopy.replace(a.slice(leftPointer, rightPointer + 1), '')
leftPointer += 2
rightPointer = leftPointer + 1;
if (leftPointer >= a.length || rightPointer >= a.length) {
break;
}
} else {
leftPointer++;
rightPointer++;
}
}
console.log(ncopy);

You can do this with a simple regex:
'aabbbcde'.replace(/(.)\1+/g, '')
Result:
'cde'
Snippet:
var a = 'aabbbcde';
var result = a.replace(/(.)\1+/g, '');
console.log(result);

'aabbccde'
.split('')
.reduce(
(acc, cur) => {
if (acc.indexOf(cur) !== -1) {
return acc.slice(acc.length);
}
return [...acc, cur];
}
)
.join('')

You can do it with a following recursive function:
const input = 'aabbccddeeffg';
function getResult(str) {
const newStr = str.match(/(.)\1*/g).filter(str => !((str.length % 2 === 0) && (str.split('').every((char,i,arr) => char === arr[0])))).join('');
return newStr === str ? newStr : getResult(newStr)
}
const result = getResult(input);
console.log(result);

counting duplicate arrays within an array in javascript

I have an array of arrays as follows:
[[3, 4], [1, 2], [3, 4]]
I wish to create a new array of arrays that has no duplicates, and has a count of the number of occurrences of each element in the first array:
[[3,4,2], [1,2,1]]
here is what I have so far:
var alreadyAdded = 0;
dataset.forEach(function(data) {
From = data[0];
To = data[1];
index = 0;
newDataSet.forEach(function(newdata) {
newFrom = newData[0];
newTo = newData[1];
// check if the point we are looking for is already added to the new array
if ((From == newFrom) && (To == newTo)) {
// if it is, increment the count for that pair
var count = newData[2];
var newCount = count + 1;
newDataSet[index] = [newFrom, newTo, newCount];
test = "reached here";
alreadyAdded = 1;
}
index++;
});
// the pair was not already added to the new dataset, add it
if (alreadyAdded == 0) {
newDataSet.push([From, To, 1]);
}
// reset alreadyAdded variable
alreadyAdded = 0;
});
I am very new to Javascript, can someone help explain to me what I'm doing wrong? I'm sure there is a more concise way of doing this, however I wasn't able to find an example in javascript that dealt with duplicate array of arrays.

Depending on how large the dataset is that you're iterating over I'd be cautious of looping over it so many times. You can avoid having to do that by creating an 'index' for each element in the original dataset and then using it to reference the elements in your grouping. This is the approach that I took when I solved the problem. You can see it here on jsfiddle. I used Array.prototype.reduce to create an object literal which contained the grouping of elements from the original dataset. Then I iterated over it's keys to create the final grouping.
var dataSet = [[3,4], [1,2], [3,4]],
grouping = [],
counts,
keys,
current;
counts = dataSet.reduce(function(acc, elem) {
var key = elem[0] + ':' + elem[1];
if (!acc.hasOwnProperty(key)) {
acc[key] = {elem: elem, count: 0}
}
acc[key].count += 1;
return acc;
}, {});
keys = Object.keys(counts);
for (var i = 0, l = keys.length; i < l; i++) {
current = counts[keys[i]];
current.elem.push(current.count);
grouping.push(current.elem);
}
console.log(grouping);

Assuming order of sub array items matters, assuming that your sub arrays could be of variable length and could contain items other than numbers, here is a fairly generic way to approach the problem. Requires ECMA5 compatibility as it stands, but would not be hard to make it work on ECMA3.
Javascript
// Create shortcuts for prototype methods
var toClass = Object.prototype.toString.call.bind(Object.prototype.toString),
aSlice = Array.prototype.slice.call.bind(Array.prototype.slice);
// A generic deepEqual defined by commonjs
// http://wiki.commonjs.org/wiki/Unit_Testing/1.0
function deepEqual(a, b) {
if (a === b) {
return true;
}
if (toClass(a) === '[object Date]' && toClass(b) === '[object Date]') {
return a.getTime() === b.getTime();
}
if (toClass(a) === '[object RegExp]' && toClass(b) === '[object RegExp]') {
return a.toString() === b.toString();
}
if (a && typeof a !== 'object' && b && typeof b !== 'object') {
return a == b;
}
if (a.prototype !== b.prototype) {
return false;
}
if (toClass(a) === '[object Arguments]') {
if (toClass(b) !== '[object Arguments]') {
return false;
}
return deepEqual(aSlice(a), aSlice(b));
}
var ka,
kb,
length,
index,
it;
try {
ka = Object.keys(a);
kb = Object.keys(b);
} catch (eDE) {
return false;
}
length = ka.length;
if (length !== kb.length) {
if (Array.isArray(a) && Array.isArray(b)) {
if (a.length !== b.length) {
return false;
}
} else {
return false;
}
} else {
ka.sort();
kb.sort();
for (index = 0; index < length; index += 1) {
if (ka[index] !== kb[index]) {
return false;
}
}
}
for (index = 0; index < length; index += 1) {
it = ka[index];
if (!deepEqual(a[it], b[it])) {
return false;
}
}
return true;
};
// Recursive function for counting arrays as specified
// a must be an array of arrays
// dupsArray is used to keep count when recursing
function countDups(a, dupsArray) {
dupsArray = Array.isArray(dupsArray) ? dupsArray : [];
var copy,
current,
count;
if (a.length) {
copy = a.slice();
current = copy.pop();
count = 1;
copy = copy.filter(function (item) {
var isEqual = deepEqual(current, item);
if (isEqual) {
count += 1;
}
return !isEqual;
});
current.push(count);
dupsArray.push(current);
if (copy.length) {
countDups(copy, dupsArray);
}
}
return dupsArray;
}
var x = [
[3, 4],
[1, 2],
[3, 4]
];
console.log(JSON.stringify(countDups(x)));
Output
[[3,4,2],[1,2,1]]
on jsFiddle

After fixing a typo I tried your solution in the debugger; it works!
Fixed the inner forEach-loop variable name to match case. Also some var-keywords added.
var alreadyAdded = 0;
dataset.forEach(function (data) {
var From = data[0];
var To = data[1];
var index = 0;
newDataSet.forEach(function (newData) {
var newFrom = newData[0];
var newTo = newData[1];
// check if the point we are looking for is already added to the new array
if ((From == newFrom) && (To == newTo)) {
// if it is, increment the count for that pair
var count = newData[2];
var newCount = count + 1;
newDataSet[index] = [newFrom, newTo, newCount];
test = "reached here";
alreadyAdded = 1;
}
index++;
});
// the pair was not already added to the new dataset, add it
if (alreadyAdded == 0) {
newDataSet.push([From, To, 1]);
}
// reset alreadyAdded variable
alreadyAdded = 0;
});

const x = [[3, 4], [1, 2], [3, 4]];
const with_duplicate_count = [
...x
.map(JSON.stringify)
.reduce( (acc, v) => acc.set(v, (acc.get(v) || 0) + 1), new Map() )
.entries()
].map(([k, v]) => JSON.parse(k).concat(v));
console.log(with_duplicate_count);

Spliting string twice with 2 separators in javascript

Let's say I need to split the string a.b.c.d#.e.f.g.h#.i.j.k.l with separator as # and then ".".
str = a.b.c.d#.e.f.g.h#.i.j.k.l
res = str.split("#")
res[0] will store a.b.c.d when I use split for the 1st time .
I need to split this again and save the data.
can anyone help ?

I think the simplest way to do it is using a regex:
var str = "a.b.c.d#.e.f.g.h#.i.j.k.l";
var res = str.split(/[.#]/);

If I may, if you have to split a string with character a then character b, the simplest in my mind would be : string.split('a').join('b').split('b')

I know this post is a bit old, but came across it, and fell like there was a better, perhaps more modern, solution. Excluding comments, this is a neat solution to this problem, and is a bit more flexible. It uses the reduce prototype, which works pretty well. This can be modified to output an object with key/value pairs as well.
const input = 'a.b.c#.e.f.g#.h.i.j'
const firstDelimiter = '#';
const secondDelimiter = '.';
// This will prevent sub arrays from containing empty values, i.e. ['','e','f','g']
const cleanInput = input.replace(/#./g, '#');
//First split by the firstDelimiter
const output = cleanInput.split(firstDelimiter).reduce( (newArr, element, i) => {
// By this point, you will have an array like ['a.b.c', 'e.f.g', 'h.i.j']
// Each element is passed into the callback, into the element variable
let subArr = element.split(secondDelimiter); // split that element by the second delimiter, created another array like ['a', 'b', 'c'], etc.
console.log(i, newArr)
// newArr is the accumulator, and will maintain it's values
newArr[i] = subArr;
return newArr;
}, []); //It's important to include the [] for the initial value of the accumulator
console.log('final:', output);

If you want to split by '#' and then for each item split by '.'
Input:'a.b.c.d#.e.f.g.h#.i.j.k'
Output:[ a b c d e f g h i j k]
var str = 'a.b.c.d#.e.f.g.h#.i.j.k.l';
var ar = [];
var sp = str.split('#');
for (var i = 0; i < sp.length; i++) {
var sub = sp[i].split('.');
for (var j = 0; j < sub.length; j++) {
ar.push(sub[j]);
}
}
alert(ar);

Here's a function I made that splits an array:
function splitArray(array, delimiter) {
let returnValue = [];
for(let i = 0; i < array.length; i++) {
array[i] = array[i].split(delimiter);
array[i].forEach(elem => {
returnValue.push(elem);
});
};
return returnValue;
};
In your case, use it like this:
str = "a.b.c.d#.e.f.g.h#.i.j.k.l";
console.log( splitArray (str.split ("#"), "." ) );

You can use the explode function similar to php
// example 1: explode(' ', 'Kevin van Zonneveld');
// returns 1: {0: 'Kevin', 1: 'van', 2: 'Zonneveld'}
function explode(delimiter, string, limit) {
// discuss at: http://phpjs.org/functions/explode/
// original by: Kevin van Zonneveld (http://kevin.vanzonneveld.net)
if (arguments.length < 2 || typeof delimiter === 'undefined' || typeof string === 'undefined') return null;
if (delimiter === '' || delimiter === false || delimiter === null) return false;
if (typeof delimiter === 'function' || typeof delimiter === 'object' || typeof string === 'function' || typeof string ===
'object') {
return {
0: ''
};
}
if (delimiter === true) delimiter = '1';
// Here we go...
delimiter += '';
string += '';
var s = string.split(delimiter);
if (typeof limit === 'undefined') return s;
// Support for limit
if (limit === 0) limit = 1;
// Positive limit
if (limit > 0) {
if (limit >= s.length) return s;
return s.slice(0, limit - 1)
.concat([s.slice(limit - 1)
.join(delimiter)
]);
}
// Negative limit
if (-limit >= s.length) return [];
s.splice(s.length + limit);
return s;
}
For more functions visit http://phpjs.org/