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I'm solving Leetcode problem: 14 using Javascript

I sorted the elements and comparing the first and last string to check the common prefixes. It works for most of the cases, but not for the input ["dog","racecar","car"]. The expected output is "", but what I'm getting is "c" (The "r" in "car" and "r" in "racecar"). I can tell the code to remove the last char, but this will break the other cases such as ["car", "car", "car"]. Not sure what am I missing. Any insights would help me improve.
Thanks
var longestCommonPrefix = function(strs) {
let count=0
const sortedString = strs.sort()
const firstString = sortedString[0]
const lastString = sortedString[sortedString.length-1]
for(let i=0; i< firstString.length; i++) {
if(firstString.charAt(i) === lastString.charAt(i)) {
count++
}
}
console.log(firstString.substring(0, count))
};
longestCommonPrefix(
["dog","racecar","car"])

You need to break out of the loop as soon as a match is not found. Otherwise, for example, ra and ca match on the second index, the a - which is undesirable.
var longestCommonPrefix = function(strs) {
let count = 0
const sortedString = strs.sort()
const firstString = sortedString[0]
const lastString = sortedString[sortedString.length - 1]
for (let i = 0; i < firstString.length; i++) {
if (firstString.charAt(i) === lastString.charAt(i)) {
count++
} else {
break;
}
}
console.log(firstString.substring(0, count))
};
longestCommonPrefix(
["dog", "racecar", "car"])
or, refactored a bit
const longestCommonPrefix = (strs) => {
strs.sort();
const firstString = strs[0];
const lastString = strs[strs.length - 1];
let prefixSoFar = '';
for (let i = 0; i < firstString.length; i++) {
if (firstString[i] === lastString[i]) {
prefixSoFar += firstString[i];
} else {
return prefixSoFar;
}
}
return prefixSoFar;
};
console.log(longestCommonPrefix(["dog", "racecar", "car"]));

Delete middle-character of string array in Javascript using splice

I want to write a function that inserts the pipe symbol in the middle of the string if the string-length is even, otherwise delete the middle character and insert the pipe symbol. For instance:
isolateIt('abcd', 'efgh'); // should return ['ab|cd', 'ef|gh']
isolateIt('abcd','abcde', 'efghe'); // should return ['ab|cd', 'ab|de', 'ef|he']
My try:
function isolateIt(array) {
let newarray = [];
for (let i = 0; i < array.length; i++) {
if (array[i].length%2 !== 0) {
newarray.push(array.splice(array[i].length % 2 + 1, 1, '|'))
} else {
newarray.push(array.splice(array[i].length,'|'))
}
}
return newarray
}
When I run the function I get:
isolateIt(["1234", '12345', '1234', '123456']);
/// returns [ [], [ '1234' ], [ '|' ], [] ]
What is the correct approach?

Here's a full working example:
function spliceString(str, index, count, add) {
if (index < 0) {
index = str.length + index;
if (index < 0) {
index = 0;
}
}
return str.slice(0, index) + (add || "") + str.slice(index + count);
}
function isolate(string) {
let midpoint = string.length / 2;
if (string.length % 2 == 0) {
string = spliceString(string, midpoint, 0, "|");
} else {
string = spliceString(string, midpoint, 1, "|");
}
return string;
}
function isolateIt(...args) {
return args.map(isolate);
}
console.log(isolateIt('abcd', 'efgh', 'abcde', 'efghe'));
// logs ["ab|cd", "ef|gh", "ab|de", "ef|he"]

What you should do is to modify the string instead of splicing the whole array
function isolateIt(array) {
let newarray = [];
for (let i = 0; i < array.length; i++) {
const mid = Math.floor(array[i].length / 2)
if (array[i].length % 2 !== 0) {
newarray.push(array[i].replace(array[i][mid], "|"));
} else {
const newStr = array[i].slice(0, mid) + '|' + array[i].slice(mid)
newarray.push(newStr);
}
}
return newarray;
}
const newArray = isolateIt(["1234", "12345", "1234", "123456"]);
console.log(newArray)

from an array of objects how do I find which value comes up most often, in javascript? [duplicate]

I'm looking for an elegant way of determining which element has the highest occurrence (mode) in a JavaScript array.
For example, in
['pear', 'apple', 'orange', 'apple']
the 'apple' element is the most frequent one.

This is just the mode. Here's a quick, non-optimized solution. It should be O(n).
function mode(array)
{
if(array.length == 0)
return null;
var modeMap = {};
var maxEl = array[0], maxCount = 1;
for(var i = 0; i < array.length; i++)
{
var el = array[i];
if(modeMap[el] == null)
modeMap[el] = 1;
else
modeMap[el]++;
if(modeMap[el] > maxCount)
{
maxEl = el;
maxCount = modeMap[el];
}
}
return maxEl;
}

There have been some developments in javascript since 2009 - I thought I'd add another option. I'm less concerned with efficiency until it's actually a problem so my definition of "elegant" code (as stipulated by the OP) favours readability - which is of course subjective...
function mode(arr){
return arr.sort((a,b) =>
arr.filter(v => v===a).length
- arr.filter(v => v===b).length
).pop();
}
mode(['pear', 'apple', 'orange', 'apple']); // apple
In this particular example, should two or more elements of the set have equal occurrences then the one that appears latest in the array will be returned. It's also worth pointing out that it will modify your original array - which can be prevented if you wish with an Array.slice call beforehand.
Edit: updated the example with some ES6 fat arrows because 2015 happened and I think they look pretty... If you are concerned with backwards compatibility you can find this in the revision history.

As per George Jempty's request to have the algorithm account for ties, I propose a modified version of Matthew Flaschen's algorithm.
function modeString(array) {
if (array.length == 0) return null;
var modeMap = {},
maxEl = array[0],
maxCount = 1;
for (var i = 0; i < array.length; i++) {
var el = array[i];
if (modeMap[el] == null) modeMap[el] = 1;
else modeMap[el]++;
if (modeMap[el] > maxCount) {
maxEl = el;
maxCount = modeMap[el];
} else if (modeMap[el] == maxCount) {
maxEl += "&" + el;
maxCount = modeMap[el];
}
}
return maxEl;
}
This will now return a string with the mode element(s) delimited by a & symbol. When the result is received it can be split on that & element and you have your mode(s).
Another option would be to return an array of mode element(s) like so:
function modeArray(array) {
if (array.length == 0) return null;
var modeMap = {},
maxCount = 1,
modes = [];
for (var i = 0; i < array.length; i++) {
var el = array[i];
if (modeMap[el] == null) modeMap[el] = 1;
else modeMap[el]++;
if (modeMap[el] > maxCount) {
modes = [el];
maxCount = modeMap[el];
} else if (modeMap[el] == maxCount) {
modes.push(el);
maxCount = modeMap[el];
}
}
return modes;
}
In the above example you would then be able to handle the result of the function as an array of modes.

Based on Emissary's ES6+ answer, you could use Array.prototype.reduce to do your comparison (as opposed to sorting, popping and potentially mutating your array), which I think looks quite slick.
const mode = (myArray) =>
myArray.reduce(
(a,b,i,arr)=>
(arr.filter(v=>v===a).length>=arr.filter(v=>v===b).length?a:b),
null)
I'm defaulting to null, which won't always give you a truthful response if null is a possible option you're filtering for, maybe that could be an optional second argument
The downside, as with various other solutions, is that it doesn't handle 'draw states', but this could still be achieved with a slightly more involved reduce function.

a=['pear', 'apple', 'orange', 'apple'];
b={};
max='', maxi=0;
for(let k of a) {
if(b[k]) b[k]++; else b[k]=1;
if(maxi < b[k]) { max=k; maxi=b[k] }
}

As I'm using this function as a quiz for the interviewers, I post my solution:
const highest = arr => (arr || []).reduce( ( acc, el ) => {
acc.k[el] = acc.k[el] ? acc.k[el] + 1 : 1
acc.max = acc.max ? acc.max < acc.k[el] ? el : acc.max : el
return acc
}, { k:{} }).max
const test = [0,1,2,3,4,2,3,1,0,3,2,2,2,3,3,2]
console.log(highest(test))

Trying out a declarative approach here. This solution builds an object to tally up the occurrences of each word. Then filters the object down to an array by comparing the total occurrences of each word to the highest value found in the object.
const arr = ['hello', 'world', 'hello', 'again'];
const tally = (acc, x) => {
if (! acc[x]) {
acc[x] = 1;
return acc;
}
acc[x] += 1;
return acc;
};
const totals = arr.reduce(tally, {});
const keys = Object.keys(totals);
const values = keys.map(x => totals[x]);
const results = keys.filter(x => totals[x] === Math.max(...values));

This solution has O(n) complexity:
function findhighestOccurenceAndNum(a) {
let obj = {};
let maxNum, maxVal;
for (let v of a) {
obj[v] = ++obj[v] || 1;
if (maxVal === undefined || obj[v] > maxVal) {
maxNum = v;
maxVal = obj[v];
}
}
console.log(maxNum + ' has max value = ' + maxVal);
}
findhighestOccurenceAndNum(['pear', 'apple', 'orange', 'apple']);

For the sake of really easy to read, maintainable code I share this:
function getMaxOcurrences(arr = []) {
let item = arr[0];
let ocurrencesMap = {};
for (let i in arr) {
const current = arr[i];
if (ocurrencesMap[current]) ocurrencesMap[current]++;
else ocurrencesMap[current] = 1;
if (ocurrencesMap[item] < ocurrencesMap[current]) item = current;
}
return {
item: item,
ocurrences: ocurrencesMap[item]
};
}
Hope it helps someone ;)!

Here’s the modern version using built-in maps (so it works on more than things that can be converted to unique strings):
'use strict';
const histogram = iterable => {
const result = new Map();
for (const x of iterable) {
result.set(x, (result.get(x) || 0) + 1);
}
return result;
};
const mostCommon = iterable => {
let maxCount = 0;
let maxKey;
for (const [key, count] of histogram(iterable)) {
if (count > maxCount) {
maxCount = count;
maxKey = key;
}
}
return maxKey;
};
console.log(mostCommon(['pear', 'apple', 'orange', 'apple']));

Time for another solution:
function getMaxOccurrence(arr) {
var o = {}, maxCount = 0, maxValue, m;
for (var i=0, iLen=arr.length; i<iLen; i++) {
m = arr[i];
if (!o.hasOwnProperty(m)) {
o[m] = 0;
}
++o[m];
if (o[m] > maxCount) {
maxCount = o[m];
maxValue = m;
}
}
return maxValue;
}
If brevity matters (it doesn't), then:
function getMaxOccurrence(a) {
var o = {}, mC = 0, mV, m;
for (var i=0, iL=a.length; i<iL; i++) {
m = a[i];
o.hasOwnProperty(m)? ++o[m] : o[m] = 1;
if (o[m] > mC) mC = o[m], mV = m;
}
return mV;
}
If non–existent members are to be avoided (e.g. sparse array), an additional hasOwnProperty test is required:
function getMaxOccurrence(a) {
var o = {}, mC = 0, mV, m;
for (var i=0, iL=a.length; i<iL; i++) {
if (a.hasOwnProperty(i)) {
m = a[i];
o.hasOwnProperty(m)? ++o[m] : o[m] = 1;
if (o[m] > mC) mC = o[m], mV = m;
}
}
return mV;
}
getMaxOccurrence([,,,,,1,1]); // 1
Other answers here will return undefined.

Here is another ES6 way of doing it with O(n) complexity
const result = Object.entries(
['pear', 'apple', 'orange', 'apple'].reduce((previous, current) => {
if (previous[current] === undefined) previous[current] = 1;
else previous[current]++;
return previous;
}, {})).reduce((previous, current) => (current[1] >= previous[1] ? current : previous))[0];
console.log("Max value : " + result);

function mode(arr){
return arr.reduce(function(counts,key){
var curCount = (counts[key+''] || 0) + 1;
counts[key+''] = curCount;
if (curCount > counts.max) { counts.max = curCount; counts.mode = key; }
return counts;
}, {max:0, mode: null}).mode
}

Another JS solution from: https://www.w3resource.com/javascript-exercises/javascript-array-exercise-8.php
Can try this too:
let arr =['pear', 'apple', 'orange', 'apple'];
function findMostFrequent(arr) {
let mf = 1;
let m = 0;
let item;
for (let i = 0; i < arr.length; i++) {
for (let j = i; j < arr.length; j++) {
if (arr[i] == arr[j]) {
m++;
if (m > mf) {
mf = m;
item = arr[i];
}
}
}
m = 0;
}
return item;
}
findMostFrequent(arr); // apple

This solution can return multiple elements of an array in case of a tie. For example, an array
arr = [ 3, 4, 3, 6, 4, ];
has two mode values: 3 and 6.
Here is the solution.
function find_mode(arr) {
var max = 0;
var maxarr = [];
var counter = [];
var maxarr = [];
arr.forEach(function(){
counter.push(0);
});
for(var i = 0;i<arr.length;i++){
for(var j=0;j<arr.length;j++){
if(arr[i]==arr[j])counter[i]++;
}
}
max=this.arrayMax(counter);
for(var i = 0;i<arr.length;i++){
if(counter[i]==max)maxarr.push(arr[i]);
}
var unique = maxarr.filter( this.onlyUnique );
return unique;
};
function arrayMax(arr) {
var len = arr.length, max = -Infinity;
while (len--) {
if (arr[len] > max) {
max = arr[len];
}
}
return max;
};
function onlyUnique(value, index, self) {
return self.indexOf(value) === index;
}

const frequence = (array) =>
array.reduce(
(acc, item) =>
array.filter((v) => v === acc).length >=
array.filter((v) => v === item).length
? acc
: item,
null
);
frequence([1, 1, 2])

var array = [1, 3, 6, 6, 6, 6, 7, 7, 12, 12, 17],
c = {}, // counters
s = []; // sortable array
for (var i=0; i<array.length; i++) {
c[array[i]] = c[array[i]] || 0; // initialize
c[array[i]]++;
} // count occurrences
for (var key in c) {
s.push([key, c[key]])
} // build sortable array from counters
s.sort(function(a, b) {return b[1]-a[1];});
var firstMode = s[0][0];
console.log(firstMode);

Here is my solution to this problem but with numbers and using the new 'Set' feature. Its not very performant but i definitely had a lot of fun writing this and it does support multiple maximum values.
const mode = (arr) => [...new Set(arr)]
.map((value) => [value, arr.filter((v) => v === value).length])
.sort((a,b) => a[1]-b[1])
.reverse()
.filter((value, i, a) => a.indexOf(value) === i)
.filter((v, i, a) => v[1] === a[0][1])
.map((v) => v[0])
mode([1,2,3,3]) // [3]
mode([1,1,1,1,2,2,2,2,3,3,3]) // [1,2]
By the way do not use this for production this is just an illustration of how you can solve it with ES6 and Array functions only.

const mode = (str) => {
return str
.split(' ')
.reduce((data, key) => {
let counter = data.map[key] + 1 || 1
data.map[key] = counter
if (counter > data.counter) {
data.counter = counter
data.mode = key
}
return data
}, {
counter: 0,
mode: null,
map: {}
})
.mode
}
console.log(mode('the t-rex is the greatest of them all'))

Here is my solution :-
function frequent(number){
var count = 0;
var sortedNumber = number.sort();
var start = number[0], item;
for(var i = 0 ; i < sortedNumber.length; i++){
if(start === sortedNumber[i] || sortedNumber[i] === sortedNumber[i+1]){
item = sortedNumber[i]
}
}
return item
}
console.log( frequent(['pear', 'apple', 'orange', 'apple']))

Try it too, this does not take in account browser version.
function mode(arr){
var a = [],b = 0,occurrence;
for(var i = 0; i < arr.length;i++){
if(a[arr[i]] != undefined){
a[arr[i]]++;
}else{
a[arr[i]] = 1;
}
}
for(var key in a){
if(a[key] > b){
b = a[key];
occurrence = key;
}
}
return occurrence;
}
alert(mode(['segunda','terça','terca','segunda','terça','segunda']));
Please note that this function returns latest occurence in the array
when 2 or more entries appear same number of times!

With ES6, you can chain the method like this:
function findMostFrequent(arr) {
return arr
.reduce((acc, cur, ind, arr) => {
if (arr.indexOf(cur) === ind) {
return [...acc, [cur, 1]];
} else {
acc[acc.indexOf(acc.find(e => e[0] === cur))] = [
cur,
acc[acc.indexOf(acc.find(e => e[0] === cur))][1] + 1
];
return acc;
}
}, [])
.sort((a, b) => b[1] - a[1])
.filter((cur, ind, arr) => cur[1] === arr[0][1])
.map(cur => cur[0]);
}
console.log(findMostFrequent(['pear', 'apple', 'orange', 'apple']));
console.log(findMostFrequent(['pear', 'apple', 'orange', 'apple', 'pear']));
If two elements have the same occurrence, it will return both of them. And it works with any type of element.

// O(n)
var arr = [1, 2, 3, 2, 3, 3, 5, 6];
var duplicates = {};
max = '';
maxi = 0;
arr.forEach((el) => {
duplicates[el] = duplicates[el] + 1 || 1;
if (maxi < duplicates[el]) {
max = el;
maxi = duplicates[el];
}
});
console.log(max);

I came up with a shorter solution, but it's using lodash. Works with any data, not just strings. For objects can be used:
const mostFrequent = _.maxBy(Object.values(_.groupBy(inputArr, el => el.someUniqueProp)), arr => arr.length)[0];
This is for strings:
const mostFrequent = _.maxBy(Object.values(_.groupBy(inputArr, el => el)), arr => arr.length)[0];
Just grouping data under a certain criteria, then finding the largest group.

Here is my way to do it so just using .filter.
var arr = ['pear', 'apple', 'orange', 'apple'];
function dup(arrr) {
let max = { item: 0, count: 0 };
for (let i = 0; i < arrr.length; i++) {
let arrOccurences = arrr.filter(item => { return item === arrr[i] }).length;
if (arrOccurences > max.count) {
max = { item: arrr[i], count: arrr.filter(item => { return item === arrr[i] }).length };
}
}
return max.item;
}
console.log(dup(arr));

Easy solution !
function mostFrequentElement(arr) {
let res = [];
for (let x of arr) {
let count = 0;
for (let i of arr) {
if (i == x) {
count++;
}
}
res.push(count);
}
return arr[res.indexOf(Math.max(...res))];
}
array = [13 , 2 , 1 , 2 , 10 , 2 , 1 , 1 , 2 , 2];
let frequentElement = mostFrequentElement(array);
console.log(`The frequent element in ${array} is ${frequentElement}`);
Loop on all element and collect the Count of each element in the array that is the idea of the solution

Here is my solution :-
const arr = [
2, 1, 10, 7, 10, 3, 10, 8, 7, 3, 10, 5, 4, 6, 7, 9, 2, 2, 2, 6, 3, 7, 6, 9, 8,
9, 10, 8, 8, 8, 4, 1, 9, 3, 4, 5, 8, 1, 9, 3, 2, 8, 1, 9, 6, 3, 9, 2, 3, 5, 3,
2, 7, 2, 5, 4, 5, 5, 8, 4, 6, 3, 9, 2, 3, 3, 10, 3, 3, 1, 4, 5, 4, 1, 5, 9, 6,
2, 3, 10, 9, 4, 3, 4, 5, 7, 2, 7, 2, 9, 8, 1, 8, 3, 3, 3, 3, 1, 1, 3,
];
function max(arr) {
let newObj = {};
arr.forEach((d, i) => {
if (newObj[d] != undefined) {
++newObj[d];
} else {
newObj[d] = 0;
}
});
let nwres = {};
for (let maxItem in newObj) {
if (newObj[maxItem] == Math.max(...Object.values(newObj))) {
nwres[maxItem] = newObj[maxItem];
}
}
return nwres;
}
console.log(max(arr));

I guess you have two approaches. Both of which have advantages.
Sort then Count or Loop through and use a hash table to do the counting for you.
The hashtable is nice because once you are done processing you also have all the distinct elements. If you had millions of items though, the hash table could end up using a lot of memory if the duplication rate is low. The sort, then count approach would have a much more controllable memory footprint.

var mode = 0;
var c = 0;
var num = new Array();
var value = 0;
var greatest = 0;
var ct = 0;
Note: ct is the length of the array.
function getMode()
{
for (var i = 0; i < ct; i++)
{
value = num[i];
if (i != ct)
{
while (value == num[i + 1])
{
c = c + 1;
i = i + 1;
}
}
if (c > greatest)
{
greatest = c;
mode = value;
}
c = 0;
}
}

You can try this:
// using splice()
// get the element with the highest occurence in an array
function mc(a) {
var us = [], l;
// find all the unique elements in the array
a.forEach(function (v) {
if (us.indexOf(v) === -1) {
us.push(v);
}
});
l = us.length;
while (true) {
for (var i = 0; i < l; i ++) {
if (a.indexOf(us[i]) === -1) {
continue;
} else if (a.indexOf(us[i]) != -1 && a.length > 1) {
// just delete it once at a time
a.splice(a.indexOf(us[i]), 1);
} else {
// default to last one
return a[0];
}
}
}
}
// using string.match method
function su(a) {
var s = a.join(),
uelms = [],
r = {},
l,
i,
m;
a.forEach(function (v) {
if (uelms.indexOf(v) === -1) {
uelms.push(v);
}
});
l = uelms.length;
// use match to calculate occurance times
for (i = 0; i < l; i ++) {
r[uelms[i]] = s.match(new RegExp(uelms[i], 'g')).length;
}
m = uelms[0];
for (var p in r) {
if (r[p] > r[m]) {
m = p;
} else {
continue;
}
}
return m;
}

How to find the most duplicate "values" in javascript array?

my question is actually similar to: Extracting the most duplicate value from an array in JavaScript (with jQuery)
I Found this but it always return one value only which is 200.
var arr = [100,100,200,200,200,300,300,300,400,400,400];
var counts = {}, max = 0, res;
for (var v in arr) {
counts[arr[v]] = (counts[arr[v]] || 0) + 1;
if (counts[arr[v]] > max) {
max = counts[arr[v]];
res = arr[v];
}
}
console.log(res + " occurs " + counts[res] + " times");
pls help me to return values not just one...
The result is should like this:
200,300,400
.
pls help thank you!

You have to iterate your counts to find the max occurred result.
var arr = [100,100,200,200,200,300,300,300,400,400,400];
var counts = {}, max = 0, res;
for (var v in arr) {
counts[arr[v]] = (counts[arr[v]] || 0) + 1;
if (counts[arr[v]] > max) {
max = counts[arr[v]];
res = arr[v];
}
}
var results = [];
for (var k in counts){
if (counts[k] == max){
//console.log(k + " occurs " + counts[k] + " times");
results.push(k);
}
}
console.log(results);

Create a Object iterating the arry containing the indexes of most repeated values, like below
var arr = [100,100,200,200,200,300,300,300,400,400,400];
valObj = {}, max_length = 0, rep_arr = [];
arr.forEach(function(el,i){
if(valObj.hasOwnProperty(el)){
valObj[el] += 1;
max_length = (valObj[el] > max_length) ? valObj[el] : max_length
}
else{
valObj[el] = 1;
}
});
Object.keys(valObj).forEach(function(val){
(valObj[val] >= max_length) && (rep_arr.push(val))
});
console.log(rep_arr);
After the object is created with key as array value and value as array indexes of that value, you can play/parse that. Hope this helps.

Iterating an array using for..in is not a good idea. Check this link for more information.
Hopefully below snippet will be useful
var arr = [100, 100, 200, 200, 200, 300, 300, 300, 400, 400, 400];
//Use a reduce fuction to create an object where 100,200,300
// will be keys and its value will the number of times it has
//repeated
var m = arr.reduce(function(i, v) {
if (i[v] === undefined) {
i[v] = 1
} else {
i[v] = i[v] + 1;
}
return i;
}, {});
// Now get the maximum value from that object,
//getMaxRepeated will be 3 in this case
var getMaxRepeated = Math.max(...Object.values(m));
//An array to hold elements which are repeated 'getMaxRepeated' times
var duplicateItems = [];
// now iterate that object and push the keys which are repeated
//getMaxRepeated times
for (var keys in m) {
if (m[keys] === getMaxRepeated) {
duplicateItems.push(keys)
}
}
console.log(duplicateItems)

The following would do the trick assuming that all items in arr are numbers:
//added some numbers assuming numbers are not sorted
var arr = [300,400,200,100,100,200,200,200,300,300,300,400,400,400];
var obj = arr.reduce(//reduce arr to object of: {"100":2,"200":4,"300":4,"400":4}
(o,key)=>{//key is 100,200, ... o is {"100":numberOfOccurrences,"200":numberOf...}
o[key] = (o[key])?o[key]+1:1;
return o;
},
{}
);
// obj is now: {"100":2,"200":4,"300":4,"400":4}
//create an array of [{key:100,occurs:2},{key:200,occurs:4}...
var sorted = Object.keys(obj).map(
key=>({key:parseInt(key),occurs:obj[key]})
)//sort the [{key:100,occurs:2},... by highest occurrences then lowest key
.sort(
(a,b)=>
(b.occurs-a.occurs===0)
? a.key - b.key
: b.occurs - a.occurs
);
console.log(
sorted.filter(//only the highest occurrences
item=>item.occurs===sorted[0].occurs
).map(//only the number; not the occurrences
item=>item.key
)
);

Try as following ==>
function getDuplicate( arr ){
let obj = {}, dup = [];
for(let i = 0, l = arr.length; i < l; i++){
let val = arr[i];
if( obj[val] /**[hasOwnProperty]*/ ) {
/**[is exists]*/
if(dup.find(a => a == val) ) continue;
/**[put Unique One]*/
dup.push(val);
continue;
};
/**[hold for further use]*/
obj[val] = true;
}
return dup;
};
Use ==>
getDuplicate([100,100,200,200,200,300,300,300,400,400,400]);

Try the following:
var candles = [100,100,200,200,200,300,300,300,400,400,400];
let tempArray = {}
for (let index = 0; index <= (candles.length - 1); index++) {
let valueToCompare = candles[index];
if (tempArray[valueToCompare]) {
tempArray[valueToCompare] = tempArray[valueToCompare] + 1;
} else {
tempArray[valueToCompare] = 1;
}
}
let highestValue;
Object.values(tempArray).forEach(item => {
if (highestValue === undefined) highestValue = item;
if (highestValue < item) highestValue = item;
});
console.log(highestValue);

How to sort the words based on count value: javascript

I Need to print the input String in Ascending order based on the number of words count. I tried this piece of code!
My question is, How to sort the elements based on count value ? i.e. greater count value word should be displayed first (descending order of count value)?
Javascript file, app.js:
var x = stringCount("this is my file teting the things apple teting this is my the is is is is ", ' ');
console.log(x);
function stringCount(haystack, needle) {
if (!needle || !haystack) {
return false;
}
else {
var words = haystack.split(needle),
count = [];
for (var i = 0, len = words.length; i < len; i++) {
if (count.hasOwnProperty(words[i])) {
count[words[i]] = parseInt(count[words[i]], 10) + 1;
}
else {
count[words[i]] = 1;
}
}
count.sort();
return count;
}
}
This is the output what I'm getting.
[ this: 2,
is: 6,
my: 2,
file: 1,
teting: 2,
the: 2,
things: 1,
apple: 1,
'': 1 ]

Change:
return count;
to:
return Object.keys(count).map(k => [k, count[k]])
.sort((a, b) => a[1] < b[1]).map(e => e[0]);

You could sort the keys by the values.
function stringCount(haystack, needle) {
if (!needle || !haystack) {
return false;
}
var words = haystack.split(needle),
count = {};
words.forEach(function (a) {
count[a] = (count[a] || 0) + 1;
});
return count;
}
var x = stringCount("this is my file teting the things apple teting this is my the is is is is ", ' ');
console.log(x);
console.log(Object.keys(x).sort(function (a, b) { return x[b] - x[a]; }));
.as-console-wrapper { max-height: 100% !important; top: 0; }

Your count variable is an object. Although you defined it as an array, it has not real array elements, but word-properties. So you are actually sorting an empty array.
Here is how you could adapt your code to make it work:
var x = stringCount("this is my file teting the things apple teting this is my the is is is is ", ' ');
console.log(x);
function stringCount(haystack, needle) {
if (!needle || !haystack) {
return false;
}
else {
var words = haystack.split(needle).filter(x => x), // filter out empty words
count = {}; // not an array
for (var i = 0, len = words.length; i < len; i++) {
if (count.hasOwnProperty(words[i])) {
count[words[i]] = parseInt(count[words[i]], 10) + 1;
}
else {
count[words[i]] = 1;
}
}
return Object.keys(count) // array of words
.sort( (a,b) => count[b] - count[a] ); // sort by count
}
}
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