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Javascript find the most repetitive character occurrence from the string

Let's say we have this string:
BBBBAAAABBAAAAAACCCCCBDDDDEEEEEEE,FFF
As you can see, here B is occurring 4 times at first but B is also present before DDDD.
Similarly, A is occurring 4 times at the beginning and later 6 times.
I want the expected output if I am searching B it should 4 times as the max occurrence B is 4. However if I am searching A then it should return 6 because the most occurrence for A is 6.
Here is my code I tried:
function checkRepeatativeString(str) {
let hashMap = {};
let seen = new Set();
let counter = 1;
let maxValue = 1;
let isPreviousValueSame = false;
let isNextValueSame = true;
for (let i = 0; i < str.length; i++) {
/**
* is previous value same
*/
if (str[i] == str[i-1]) {
isPreviousValueSame = true;
}
/**
* is next value same
*/
if (str[i] == str[i+1]) {
isNextValueSame = true;
}
if (seen.has(str[i]) && isPreviousValueSame) {
hashMap[str[i]][0]++;
hashMap[str[i]][1]++;
isPreviousValueSame = false;
} else if(seen.has(str[i]) && !isNextValueSame) {
maxValue = Math.max(hashMap[str[i]][1], maxValue);
counter = 0;
hashMap[str[i]] = [counter, maxValue];
} else {
maxValue = Math.max(maxValue, counter);
seen.add(str[i]);
hashMap[str[i]] = [counter, maxValue];
isPreviousValueSame = false;
}
}
return hashMap;
}
let str = "BBBBAAAABBAAAAAACCCCCBDDDDEEEEEEE,FFF";
console.log(checkRepeatativeString(str));
This code is working but if you look for B, I am getting stuck at the beginning of value.
My program returns out for B:
B: [ 1, 1 ]
^ ^
Inside array, 1 is a counter which scans the string and second 1 in array is a max value which should return the output. However my program is returning 1 for B. I am expecting 4 as max value.
Help would be appreciated~

Quick and dirty.
function maxConsecutiveCharacters(check, haystack) {
if(check.length !== 1) return false;
let result = 0;
let buffer = 0;
for(let i = 0; i < haystack.length; i++) {
if(haystack[i] === check) {
buffer++;
}
else {
if(buffer > result) {
result = buffer;
}
buffer = 0;
}
if(buffer > result) {
result = buffer;
}
}
return result;
}

That looks overly complicated. Consider approaching the problem from a different angle - first split up the string into segments of repeating characters, and group them into an object based on the length of the longest substring for a given character.
const checkRepeatativeString = (str) => {
const longestCounts = {};
for (const consecutive of (str.match(/(.)\1*/g) || [])) {
const char = consecutive[0];
longestCounts[char] = Math.max(
longestCounts[char] || 0, // Use the existing value in the object if it exists and is higher
consecutive.length // Otherwise, use the length of the string iterated over
);
}
return longestCounts;
};
let str = "BBBBAAAABBAAAAAACCCCCBDDDDEEEEEEE,FFF";
console.log(checkRepeatativeString(str));
Simpler code often means less surface area for bugs.

Return String as Sorted Blocks - Codewars Challenge - JavaScript

I've been trying to solve this codewars challenge. The idea is to return the string, rearranged according to its hierarchy, or separated into chunks according to the repeating character.
You will receive a string consisting of lowercase letters, uppercase letters and digits as input. Your task is to return this string as blocks separated by dashes ("-"). The elements of a block should be sorted with respect to the hierarchy listed below, and each block cannot contain multiple instances of the same character.
The hierarchy is:
lowercase letters (a - z), in alphabetic order
uppercase letters (A - Z), in alphabetic order
digits (0 - 9), in ascending order
Examples
"21AxBz" -> "xzAB12"
since input does not contain repeating characters, you only need 1 block
"abacad" -> "abcd-a-a"
character "a" repeats 3 times, thus 3 blocks are needed
"" -> ""
an empty input should result in an empty output
What I've tried actually works for the given test cases:
describe("Sample tests", () => {
it("Tests", () => {
assert.equal(blocks("21AxBz"), "xzAB12");
assert.equal(blocks("abacad"), "abcd-a-a");
assert.equal(blocks(""), "");
});
});
But fails when there are any repeating characters, besides in the test cases:
function repeatingChar(str){
const result = [];
const strArr = str.toLowerCase().split("").sort().join("").match(/(.)\1+/g);
if (strArr != null) {
strArr.forEach((elem) => {
result.push(elem[0]);
});
}
return result;
}
function blocks(s) {
if (s.length === 0) {
return '';
}
//if string does not contain repeating characters
if (!/(.).*\1/.test(s) === true) {
let lower = s.match(/[a-z]/g).join('');
let upper = s.match(/[A-Z]/g).join('');
let numbers = s.match(/[\d]/g).sort().join('');
return lower + upper + numbers;
}
//if string contains repeating characters
if (!/(.).*\1/.test(s) === false) {
let repeatChar = (repeatingChar(s)[0]);
let repeatRegex = new RegExp(repeatingChar(s)[0]);
let repeatCount = s.match(/[repeatRegex]/gi).length;
let nonAChars = s.match(/[^a]/gi).join('');
function getPosition(string, subString, index) {
return s.split(repeatChar, index).join(repeatChar).length;
}
let index = getPosition(s, repeatChar, 2);
// console.log('indexxxxx', index);
return s.slice(0, index) + nonAChars.slice(1) + ('-' + repeatChar).repeat(repeatCount - 1);
}
}
console.log(blocks("abacad"));
And actually, I'm not sure what's wrong with it, because I don't know how to unlock any other tests on Codewars.
You can see that what I'm trying to do, is find the repeating character, get all characters that are not the repeating character, and slice the string from starting point up until the 2 instance of the repeating character, and then add on the remaining repeating characters at the end, separated by dashes.
Any other suggestions for how to do this?

For funzies, here's how I would have approached the problem:
const isLower = new RegExp('[a-z]');
const isUpper = new RegExp('[A-Z]');
const isDigit = new RegExp('[0-9]');
const isDigitOrUpper = new RegExp('[0-9A-Z]');
const isDigitOrLower = new RegExp('[0-9a-z]');
const isLowerOrUpper = new RegExp('[a-zA-Z]');
function lowerUpperNumber(a, b)
{
if(isLower.test(a) && isDigitOrUpper.test(b))
{
return -1;
}
else if(isUpper.test(a) && isDigitOrLower.test(b))
{
if(isDigit.test(b))
{
return -1;
}
else if(isLower.test(b))
{
return 1;
}
}
else if(isDigit.test(a) && isLowerOrUpper.test(b))
{
return 1;
}
else if(a > b)
{
return 1;
}
else if(a < b)
{
return -1;
}
return 0;
}
function makeBlocks(input)
{
let sortedInput = input.split('');
sortedInput.sort(lowerUpperNumber);
let output = '';
let blocks = [];
for(let c of sortedInput)
{
let inserted = false;
for(let block of blocks)
{
if(block.indexOf(c) === -1)
{
inserted = true;
block.push(c);
break;
}
}
if(!inserted)
{
blocks.push([c]);
}
}
output = blocks.map(block => block.join('')).join('-');
return output;
}
console.log(makeBlocks('21AxBz'));
console.log(makeBlocks('abacad'));
console.log(makeBlocks('Any9Old4String22With7Numbers'));
console.log(makeBlocks(''));

The first obvious error I can see is let repeatCount = s.match(/[repeatRegex]/gi).length;. What you really want to do is:
let repeatRegex = new RegExp(repeatingChar(s)[0], 'g');
let repeatCount = s.match(repeatRegex).length;
The next is that you only look at one of the repeating characters, and not all of them, so you won't get blocks of the correct form, so you'll need to loop over them.
let repeatedChars = repeatingChar(s);
for(let c of repeatedChars)
{
//figure out blocks
}
When you're building the block, you've decided to focus on everything that's not "a". I'm guessing that's not what you originally wrote, but some debugging code, to work on that one sample input.
If I understand your desire correctly, you want to take all the non-repeating characters and smoosh them together, then take the first instance of the first repeating character and stuff that on the front and then cram the remaining instances of the repeating character on the back, separated by -.
The problem here is that the first repeating character might not be the one that should be first in the result. Essentially you got lucky with the repeating character being a.
Fixing up your code, I would create an array and build the blocks up individually, then join them all together at the end.
let repeatedChars = repeatingChar(s);
let blocks = []
for(let c of repeatedChars)
{
let repeatRegex = new RegExp(c, 'g');
let repeatCount = s.match(repeatRegex).length;
for(let i = 1; i <= repeatCount; i++)
{
if(blocks.length < i)
{
let newBlock = [c];
blocks.push(newBlock);
}
else
{
block[i - 1].push(c);
}
}
}
let tailBlocks = blocks.map(block => block.join('')).join('-');
However, this leaves me with a problem of how to build the final string with the non-repeating characters included, all in the right order.
So, to start with, let's make the initial string. To do so we'll need a custom sort function (sorry, it's pretty verbose. If only we could use regular ASCII ordering):
function lowerUpperNumber(a, b)
{
if(a.match(/[a-z]/) && b.match(/[A-Z0-9]/))
{
return -1;
}
else if(a.match(/[A-Z]/) && (b.match(/[0-9]/) || b.match(/[a-z]/)))
{
if(b.match(/[0-9]/))
{
return -1;
}
else if(b.match(/[a-z]/))
{
return 1;
}
}
else if(a.match(/[0-9]/) && b.match(/[a-zA-Z]/))
{
return 1;
}
else if(a > b)
{
return 1;
}
else if(a < b)
{
return -1;
}
return 0;
}
Then create the head of the final output:
let firstBlock = [...(new Set(s))].sort(lowerUpperNumber);
The Set creates a set of unique elements, i.e. no repeats.
Because we've created the head string, when creating the blocks of repeated characters, we'll need one fewer than the above loop gives us, so we'll be using s.match(repeatRegex).length-1.
I get the desire to short circuit the complicated bit and return quickly when there are no repeated characters, but I'm going to remove that bit for brevity, and also I don't want to deal with undefined values (for example try '123' as your input).
Let's put it all together:
function lowerUpperNumber(a, b)
{
if(a.match(/[a-z]/) && b.match(/[A-Z0-9]/))
{
return -1;
}
else if(a.match(/[A-Z]/) && (b.match(/[0-9]/) || b.match(/[a-z]/)))
{
if(b.match(/[0-9]/))
{
return -1;
}
else if(b.match(/[a-z]/))
{
return 1;
}
}
else if(a.match(/[0-9]/) && b.match(/[a-zA-Z]/))
{
return 1;
}
else if(a > b)
{
return 1;
}
else if(a < b)
{
return -1;
}
return 0;
}
function makeBlocks(s)
{
if (s.length === 0)
{
return '';
}
let firstBlock = [...(new Set(s))].sort(lowerUpperNumber);
let firstString = firstBlock.join('');
let blocks = [];
for(let c of firstString)
{
let repeatRegex = new RegExp(c, 'g');
let repeatCount = s.match(repeatRegex).length - 1;
for(let i = 1; i <= repeatCount; i++)
{
if(blocks.length < i)
{
let newBlock = [c];
blocks.push(newBlock);
}
else
{
blocks[i - 1].push(c);
}
}
}
blocks.unshift(firstBlock);
return blocks.map(block => block.join('')).join('-');
}
console.log(makeBlocks('21AxBz'));
console.log(makeBlocks('abacad'));
console.log(makeBlocks('Any9Old4String22With7Numbers'));
console.log(makeBlocks(''));
You'll see I've not bothered generating the characters that repeat, because I can just skip the ones that don't.

I need my array to return and array back to another variable while also including its previous array members, no idea how to go about it

This is the test code that it's supposed to pass
function makeArray() {
const array = [];
const t = 10;
for (let i = 0; i < t; i++) {
array.push("I am a strange loop.");
}
return [array, t];
}
describe('loops', () => {
jsdom({
src: fs.readFileSync(path.resolve(__dirname, '..', 'loops.js'), 'utf-8'),
});
describe('forLoop(array)', () => {
it('adds `"I am ${i} strange loop${i === 0 ? \'\' : \'s\'}."` to an array 25 times', () => {
const [array, t] = makeArray();
const strangeArray = forLoop(array);
const testArray = strangeArray.slice(array.length);
const first = "I am 1 strange loop.";
const rest = "I am 24 strange loops.";
expect(strangeArray[11]).to.equal(first);
expect(strangeArray[34]).to.equal(rest);
expect(strangeArray.length).to.equal(t + 25);
});
});
});
this is my code to return the function to strangeArray what I am thinking is that 35 is the total number of members in the array and as the test pass requires me to have 'expect(strangeArray[11]).to.equal(first)' 11th value to be equal to my function return as
"I am 1 strange loop."
function forLoop(array) {
for (let i = 0; i < 35; i++) {
if (array[i] === "I am a strange loop.") {
return;
}
else {
array.push("I am ${i} strange loops.");
}
}
return [array,i];
}

Not sure what you mean exactly but I guess you just want the test to pass? The problem is that the first loop has 'loop' as singular and your indexes don't work either since they would start at 11. That's why your code doesn't work. You can just push to the original array.
function forLoop(array){
for(let i = 0; i < 25; i++){
array.push(`I am ${i} strange loop${i > 1 ? '' : 's'}.`)
}
return array
}

Can't we reassign array values inside forEach?

The problem statement is, i should replace the any digit below 5 with 0 and any digit 5 and above with 1.
I am trying to reassign values, but it is not affecting, Why?
function fakeBinary(n) {
let numbersArr = n.split('');
numbersArr.forEach(num => {
if(Number(num) < 5) {
num = '0';
} else if(Number(num) >= 5) {
num = '1';
}
});
return numbersArr.join('');
}
console.log(fakeBinary('3457'));
I except the output of 0011, but the actual output is 3457.

forEach used like that won't do anything - use map instead.
let numbersArr = n.split("").map(num => {
if (Number(num) > 5) {
num = "0";
} else if (Number(num) <= 5) {
num = "1";
}
return num;
});
return numbersArr.join("");
Note that to produce your desired output, you need to change your conditions slightly:
if (Number(num) >= 5) {
num = "1";
} else {
num = "0";
}

forEach doesn't bring the element's reference for primitive values but rather brings a copy of the value in your case. You can easily access that manually through the index, though:
function fakeBinary(n) {
let numbersArr = n.split('');
numbersArr.forEach((num, i) => {
// ^--- note that `i` is brought and used below to access the element at index [i].
if(Number(num) < 5) {
numbersArr[i] = '0';
} else if(Number(num) >= 5) {
numbersArr[i] = '1';
}
});
return numbersArr.join('');
}
console.log(fakeBinary('3457'));
Please note that you may also use other prototypes, I just tried to stay as close as possible to your solution, you may also want to use map or, even (not appropriate, though) reduce or even a regular for loop.

To do it with forEach, you would need to use the additional arguments to reference the array and the index.
function fakeBinary(n) {
let numbersArr = n.split('');
numbersArr.forEach((num, index, arr) => {
if(Number(num) < 5) {
arr[index] = '0';
} else if(Number(num) >= 5) {
arr[index] = '1';
}
});
return numbersArr.join('');
}
console.log(fakeBinary('3457'));
But forEach is not the correct way to return a new array. You want to use map()
function fakeBinary(n) {
return n
.split('')
.map(num => (Number(num) < 5) ? '0' : '1')
.join('');
}
console.log(fakeBinary('3457'));

A shorter version using Array.from()
const fakeBinary = (str) => Array.from(str, n => +(+n >= 5)).join('');
console.log(fakeBinary('3457'));

check that a word is an isogram with pure javascript

How do i check that a given word is an isogram with pure javascript, using a function. the function must return true or false.
An isogram is a word with a repeated character.
I know this code works, but i need a better solution.
function isIsogram(word){
x = false; y = false;
for(i = 0; i < word.length; i++){
wordl = word.substring(0,i)
wordr = word.substring(i)
x = wordl.includes(word.charAt(i))
y = wordr.includes(word.charAt(i))
//console.log(x,wordl,wordr)
}
return x&&y
}
isIsogram("thomas");//False
isIsogram("moses"); //True

Remove the duplicate letter from string then check both length. if same its an isogram.
function isIsogram(str){
return str.split('').filter((item, pos, arr)=> arr.indexOf(item) == pos).length == str.length;
}
console.log(isIsogram('thomas'));
console.log(isIsogram('moses'));

One way of doing this!
function isIsogram(str){
return !str.match(/([a-z]).*\1/i);
}

Here is a simple approach using .split() and .every():
let isIsogram = (str) => str.split("").every((c, i) => str.indexOf(c) == i);
console.log(isIsogram("thomas")); /* no repeating letter */
console.log(isIsogram("moses")); /* s repeat 2 times */
console.log(isIsogram("hello")); /* l repeat 2 times */
console.log(isIsogram("world")); /* no repeating letter */
console.log(isIsogram("a b c")); /* space character repeat 2 times */
Docs:
String.prototype.split()
String.prototype.indexOf()
Array.prototype.every()

Building on kishea's answer:
function isIsogram(sWord)
{
for (iCharIndex = 0; iCharIndex < sWord.length; iCharIndex++)
if (sWord.substring(iCharIndex + 1).includes(sWord.charAt(iCharIndex)))
return false;
return true;
}
If the character at the current position (charAt) is found (includes) to the right of the current position (substring), false is returned. Otherwise the loop runs to the end and true is returned.

const isIsogram = (word) => {
return new Set(word.toLowerCase()).size === word.length
}
console.log(isIsogram('Thor'));//true
console.log(isIsogram('Loki'));//true
console.log(isIsogram('America'));//false

function isIsogram(str) {
return new Set(str.toUpperCase()).size == str.length
}

What about :
> function isIsogram(word) {
... var a = word.split('')
... var b = Array.from(new Set(a))
... return a.length === b.length;
... }
undefined
> isIsogram("mesos")
false
> isIsogram("thomas")
true
Or better (checking each char only once) :
> function isIsogram2(word) {
... for(var i=0,len=word.length-1;i<len;++i) {
..... var c = word[i]
..... if(word.indexOf(c,i+1) !== -1) return false;
..... }
... return true;
... }
undefined
> isIsogram2("mesos")
false
> isIsogram2("thomas")
true

function isIsogram(word){
return !/(.).*\1|\d/i.test(word)
}

var str=prompt('Enter a string');
var strlen=str.length;
for(i=0;i<strlen;i++){
var stc=str.charAt(i);
var flag=0;
for(j=0;j<strlen;j++){
var stk=str.charAt(j);
if(stc==stk){
flag=flag+1;
}
}
if(flag!=1){
break;
}
}
if(flag!=1){
alert('It is not an isogram');
}
else{
alert('It is an isogram');
}

While given a word, this function if splits the word into two,
That is wordl and wordr respectively.
Both splittings are checked to include a character in the original word. If wordl and wordr both contain any character in the original word. Then surely this is an isogram
function isIsogram(word){
x = false; y = false;
for(i = 0; i < word.length; i++){
wordl = word.substring(0,i)
wordr = word.substring(i)
x = wordl.includes(word.charAt(i))
y = wordr.includes(word.charAt(i))
//console.log(x,wordl,wordr)
}
return !x&&y
}
isIsogram("thomas");//True
isIsogram("moses"); //False

const isIsogram = (string) => {
const lowerCased = string.toLowerCase()
const result = lowerCased.split('').every((v,i)=>lowerCased.indexOf(v)===i)
return result
}
console.log(isIsogram('ambidExtRously')) // true
console.log(isIsogram('patteRN')) // false

function isIsogram(word) {
let uniqueCharacters = new Set(word.split(''));
uniqueCharacters = Array.from(uniqueCharacters); //get all the unique char
let charFreq = {}; //e.g {a:2, b:3}
for (element of uniqueCharacters) {
charFreq[element] = 0;
} //set the frequency of each char to zero
function updateFrequency(element) {
charFreq[element] += 1;
}//callback used directly below
word.split('').forEach(updateFrequency); //for each char encountered, update frequency
let frequencyOfCharacter = [];
for (keys in charFreq) {
frequencyOfCharacter.push(charFreq[keys]);
}
function equal(item) {
return item === frequencyOfCharacter[0];
}
//check if all the frequencies are the same, an isogram means all characters occur at the same frequency
return frequencyOfCharacter.every(equal);
}
console.log(isIsogram('try'), isIsogram('baba'), isIsogram('tests'));