Find longest substring of vowels - javascript

I have input where user types a string and I need to find longest chain of vowels in that string and render it out. I tried it by myself, but I am not sure how to do it. Here is my code:
vowelHandler = () => {
let input = document.getElementById('input2');
const hasVowel = /[aeiouAEIOU]/g;
let letters = [];
let vowels = '';
let count = 0;
let res = 0;
letters = input.value.match(/[aeiouAEIOU]/g);
if(hasVowel.test(letters)) {
for(let i = 0; i <= letters.length; i++) {
if(hasVowel.test(letters[i])) {
count++;
res = Math.max(res, count);
}else {
res = Math.max(res, count);
count = 0;
}
}
letters = letters.join(' ');
vowels = letters.substring(letters[res]);
document.getElementById('num2').innerHTML= vowels;
}else {
document.getElementById('num2').innerHTML= 'String does not contain vowels.';
}
}
<input type='text' id='input2'/>
<button onClick={this.vowelHandler}>CLICK</button>
<p id='num2'></p>

You could replace everything in the string that is not a vowel with a white space, split the string by white space, and check which of these substrings is the longest.
function getLongestVowelSubstring(s) {
const separateVowels = s.replace(/[^aeiou]/ig, ' ').split(' ');
const longest = separateVowels.reduce((longest, vowelGroup) => {
return vowelGroup.length > longest.length ? vowelGroup : longest;
}, '');
return longest;
}
console.log(getLongestVowelSubstring('get the looongest vowel substring'));

Try something like this:
vowelHandler = () => {
let input = document.getElementById('input2');
const hasVowel = /[aeiouAEIOU]+/g;
const letters = input.value.match(hasVowel);
if (letters) {
const longest = letters.reduce((acc, val) => acc.length > val.length ? acc : val, '');
document.getElementById('num2').innerHTML= longest;
} else {
document.getElementById('num2').innerHTML= 'String does not contain vowels.';
}
}

Use this working example:
let vowels = ['a', 'e', 'i', 'o', 'u']
function largestVowel(inputStr) {
let largestStr = '';
let tempVowelStr = '';
inputStr.split('').forEach(char => {
if (vowels.includes(char.toLowerCase())) {
tempVowelStr += char;
if (tempVowelStr.length > largestStr.length) {
largestStr = tempVowelStr;
}
} else {
tempVowelStr = '';
}
})
return (largestStr.length === 0) ? 'String does not contain vowels.' : largestStr;
}
console.log(largestVowel('ertyyfouigdg'));
console.log(largestVowel('sdfsfsf'));
console.log(largestVowel('sdeefsfiaeosf'));
Complexity: O(n)

Related

Is there a way I can merge all my functions and for-loop into one function since I need (str) to be given by the site

I need this to work without Hard-Coding (str), but instead given by the actual site. Is it even possible? Yes I understand my Order is unusual, However it works lol. Thanks for any assistance I may receive. With this function I need to receive an string of works and return the next increment of the letter ex(a=b, b=c, e=f). Also returns any vowels as UpperCased.
input: "hello3"
Output: Ifmmp3
input: "fun times!"
output: gvO Ujnft!
let str = ['f', 'u', 'n', ' ', '', 'i', 'm', 'e', 's', '!', 'n', 'n'];
function replaceLetter(letter) {
///above is fixed and cannot be edited
return letter.replace(/([a-zA-Z])[^a-zA-Z]*$/, function (nextLetter) {
var letter = nextLetter.charCodeAt(0);
switch (letter) {
case 90:
return 'A';
case 122:
return 'a';
default:
return String.fromCharCode(++letter);
}
});
}
var vowels = 'aeiou';
let emptyStr = [];
let finalStr = '';
let holder = '';
for (const i of str) {
[emptyStr.push(replaceLetter(i))].join().replaceAll(',', '');
}
for (let i = 0; i < emptyStr.length; i++) {
if (vowels.indexOf(emptyStr[i]) >= 0) {
finalStr += emptyStr[i].toUpperCase();
} else {
finalStr += emptyStr[i];
}
}
console.log(finalStr);```
var vowels = "aeiou";
let received_str = "hello"; // the dynamic string you'll recieve
function replaceLetters(string) {
///above is fixed and cannot be edited
let finalStr = "";
let str = [...string];
str.forEach((element) => {
let nextLetter = element.replace(/([a-zA-Z])[^a-zA-Z]*$/, function (
nextLetter
) {
var letter = nextLetter.charCodeAt(0);
switch (letter) {
case 90:
return "A";
case 122:
return "a";
default:
return String.fromCharCode(++letter);
}
});
if (nextLetter !== ",") {
if (vowels.indexOf(nextLetter) >= 0) {
finalStr += nextLetter.toUpperCase();
} else {
finalStr += nextLetter;
}
}
});
return finalStr;
}
console.log(replaceLetters(received_str));

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;
}

Find the first none-repeated character in a string, what are the mistakes here?

I'm doing some exercises. the question is to find the first none-repeated character in a string.
My idea is: turn string to array. assign array[0] to a new variable, and remove this array[0] from array. Check if this new array contain this variable, if not, return this variable. Else, use filter to remove same value elements and get a new array. repeat the process. the code as following.
const NoneReChar = (str) => {
let tempArr = str.split('');
let start = tempArr[0];
while (true) {
tempArr.shift();
if (!tempArr.includes(start)) {
return start;
} else {
tempArr.filter(char => char !== start);
start = tempArr[0];
}
}
}
console.log(NoneReChar("aaaabbbeccc"))
I was expect output 'e', but I keep getting 'a'...where are the mistakes I made here?
The Array.filter() method doesn't mutate the original array. You need to assign the result of filter to tempArr:
tempArr = tempArr.filter(char => char !== start);
Example:
const NoneReChar = (str) => {
let tempArr = str.split('');
let start = tempArr[0];
while (true) {
tempArr.shift();
if (!tempArr.includes(start)) {
return start;
} else {
tempArr = tempArr.filter(char => char !== start);
start = tempArr[0];
}
}
}
console.log(NoneReChar("aaaabbbeccc"))
However, you don't handle the not found case. To handle it instead of true, the while clause should stop when the array is empty:
const NoneReChar = (str) => {
let tempArr = str.split('');
let start = tempArr[0];
while (tempArr.length) {
tempArr.shift();
if (!tempArr.includes(start)) {
return start;
} else {
tempArr = tempArr.filter(char => char !== start);
start = tempArr[0];
}
}
return null;
}
console.log(NoneReChar("aabbcc"))
Another option is comparing the length of the array before and after filtering. If the length is the same, the item was not repeated:
const NoneReChar = (str) => {
let tempArr = str.split('');
while (tempArr.length) {
const [start, ...rest] = tempArr; // take the 1st item and the rest
tempArr = rest.filter(char => char !== start); // filter out start
if(tempArr.length === rest.length) { // check current and previous arrays, and if the length still matches, start didn't appear again
return start;
}
}
return null;
}
console.log(NoneReChar("aabzbcc"))
const NoneReChar = (str) => {
const tempArr = str.split('');
let result = 'Not Found';
for (let index = 0; index < tempArr.length; index++) {
const firstIndex = tempArr.indexOf(tempArr[index]);
const lastIndex = tempArr.lastIndexOf(tempArr[index]);
if (firstIndex === lastIndex) {
result = tempArr[index];
break;
}
}
return result;
}
console.log(NoneReChar("aaaabbbeccc"));

Determine groups in a string

I am trying to determine groups in a string.
For example "AAABBCCCCD" should have ["AAA", "BB", "CCCC", "D"] as groups based on the pattern in the string.
Here is my simple function to do that:
const patternGroup = (str) => {
let cursor = str[0]
let groups = []
let currentGroup = ""
for (let i = 0; i < str.length; i++) {
let ch = str[i]
if (ch === cursor) {
currentGroup += ch
}
else {
groups.push(currentGroup)
currentGroup = ch
cursor = ch
}
}
if (currentGroup !== "") {
groups.push(currentGroup)
}
return groups
}
It works as intented but I am looking for a simpler function, maybe using map/reduce or regex. Any ideas?
you can simply use RegExp:
var input = "AAABBCCCCD";
const res = input.match(/([^])(\1*)/g);
console.log(res)
Similar to your implementation:
const patternGroupWithReduce = (str) => {
let currentGroup = ""
return str.split("").reduce((acc, cur, i) => {
currentGroup += cur
if (str[i + 1] !== cur) {
acc.push(currentGroup)
currentGroup = ""
}
return acc
}, [])
}
const patternGroupWithRegex = (str) => {
return str.match(/(.)\1*/g)
}
RegEx /(.)\1/g* uses the back reference for the capturing group.
\1* matches the same char as recently captured one as many times as possible.
Here is the visualization for it:
You could use
(.)\1*
As in:
let string = "AAABBCCCCD";
let rx = /(.)\1*/g;
let parts = string.match(rx);
console.log(parts);
See a demo for the expression on regex101.com.

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);

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