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I want to write a Generator for Fibonacci numbers in Javascript;
0,1,1,2,5,7,12..... (to make the sequence you have to add the last two numbers)
But I have this problem when I assign the the output.length to a variable the code is not working, if I write it down straight instead of "newNumber" the code down is however working, but I don't understand what is wrong with the first one. Is it something wrong with the place of the variables?
function fibonacciGenerator(n) {
var output = [];
var lastNumber = output[output.length - 1];
var nPrev = output[output.length - 2];
var newNumber = lastNumber + nPrev;
if (n === 1) {
output = [0];
} else if (n === 2) {
output = [0, 1];
} else {
output = [0, 1];
for (var i = 2; i < n; i++) {
output.push(newNumber);
}
}
return output
}
console.log(fibonacciGenerator(5));
function fibonacciGen(n) {
const output = [0, 1];
// Return an empty array if n is less than 1
if (n < 1) {
return [];
}
// If n is 1 or 2, we can return the array now
if (n <2) {
return output;
}
// Loop through the remaining numbers in the sequence
for (let i = 2; i < n; i++) {
// Calculate the next number in the sequence
let lastNumber = output[output.length - 1];
let nPrev = output[output.length - 2];
let newNumber = lastNumber + nPrev;
// Add the new number to the output array
output.push(newNumber);
}
// Return the output array
return output;
}
console.log(fibonacciGen(5));
you have to declare your logic of next term in loop because first time length of output is zero
function fibonacciGenerator (n) {
var output =[];
if (n < 1) {
return [];
}
if (n===1){
output=[0];
}
else if (n===2){
output=[0,1];
}
else{
output=[0,1];
for( var i = 2; i < n; i++){
var lastNumber=output[output.length-1];
var nPrev=output[output.length-2];
var newNumber=lastNumber+nPrev;
output.push(newNumber);
}
}
return output
}
I have an array [1,2,4,5,1,7,8,9,2,3]
and i would like it to generate all subset which sum of values are less than 10
current result [[1,2,4],[5,1],[7],[8],[9],[2,3]]
expected result [[4,5,1],[9,1],[8,2],[3,7],[1,2]]
that is what i did
var a = [1,2,4,5,1,7,8,9,2,3], tempArr = []; tempSum = 0, result = [];
for (var i = 0;i< a.length; i += 1 ) {
tempSum+=a[i];
tempArr.push(a[i]);
if((tempSum+a[i+1])>10) {
result.push(tempArr);
tempSum = 0;
tempArr = [];
} else if (i == a.length-1 && tempArr.length > 0) { // if array is [1,2,3]
result.push(tempArr);
}
}
but it gives me [[1,2,4],[5,1],[7],[8],[9],[2,3]] and it has 6 subset, but i expect to get [[4,5,1],[9,1],[8,2],[3,7],[1,2]] which has 5 subset.
Below logic is in JavaScript :-
var limit = 10;
var arr = [1,2,4,5,1,7,8,9,2,3];
arr.sort();
var ans = new Array ( );
while(arr.length >0){
var ts = arr[arr.length-1];
arr.splice(arr.length-1 , 1);
var ta= new Array ( );
ta.push(ts);
var x = arr.length-1;
while(x>=0){
if(ts + arr[x] <= limit){
ts = ts + arr[x];
ta.push(arr[x]);
arr.splice(x , 1);
}
x= x-1;
}
ans.push(JSON.stringify(ta));
}
alert(ans);
It is Giving Output as required .
[9,1],[8,2],[7,3],[5,4,1],[2]
I have removed duplicates then added maxSum parameter to combine function to generate all subset which have those conditions and then sorted subsets by sum of the values and sliced them.
You could change parameters to fit it for your problem.
var arr = [1,2,4,5,1,7,8,9,2,3]
MAX_SUM = 10,
MIN_SUBSET_LEN = 2,
RESULT_LEN = 5;
//remove duplicates
var uniqeSet = arr.filter(function(value, index){
return this.indexOf(value) == index
},arr);
// a function to get all subset which
// their length are greater than minLength and
// sum of values are little than maxSum
var combine = function(sourceArr, minLength, maxSum) {
var fn = function(n, src, got, all, sum) {
if(sum <= maxSum){
if (n == 0) {
if (got.length > 0) {
all.push({arr:got,sum:sum});
}
return;
}
for (var j = 0; j < src.length; j++) {
var tempSum = sum
fn(n - 1, src.slice(j + 1), got.concat([src[j]]), all, sum + src[j]);
}
}
return;
}
var all = [];
for (var i = minLength; i < sourceArr.length; i++) {
fn(i, sourceArr, [], all, 0);
}
return all;
}
var result = combine(uniqeSet, MIN_SUBSET_LEN, MAX_SUM);
var sortedSliced = result.sort(function(a1, a2){
return a2.sum - a1.sum;
}).slice(0, RESULT_LEN).map(function(m){return m.arr;});
console.log(JSON.stringify(sortedSliced));
I want to get a random digit from 0-9 and have it popped so it doesn't get repeated but I find that after the the second number is pushed it doesn't have it's number popped. Instead, some other number not yet selected is popped giving room for a repeat.
var yourNum = [],
oppNum = [],
choose = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9];
function chooseRandomNumber() {
return choose[Math.floor(Math.random() * choose.length)];
}
for (var i = 0; i < 4; i++) {
if (i === 0) {
yourNum.push(chooseRandomNumber());
if (yourNum[yourNum.length - 1] === 9) {
choose.pop();
} else {
choose.splice(yourNum[0], 1);
}
} else if (i === 1) {
yourNum.push(chooseRandomNumber());
if (yourNum[yourNum.length - 1] === 9) {
choose.pop();
} else {
choose.splice(yourNum[1], 1);
}
} else if (i === 2) {
yourNum.push(chooseRandomNumber());
if (yourNum[yourNum.length - 1] === 9) {
choose.pop();
} else {
choose.splice(yourNum[2], 1);
}
} else if (i === 3) {
yourNum.push(chooseRandomNumber());
if (yourNum[yourNum.length - 1] === 9) {
choose.pop();
} else {
choose.splice(yourNum[3], 1);
}
}
}
console.log(choose);
console.log(yourNum);
function getRand(min, max, result) {
result = result || [];
if(result.length == 4) {
return result;
}
var rand = Math.floor(Math.random()*max) + min;
if(result.indexOf(rand) === -1) {
result.push(rand);
}
return getRand(min, max, result);
}
var result = getRand(1,9);
console.log(result);
Your whole approach is way to complicated and unperformant.
A better approach:
//first we need a shuffle function
function shuffle(array){
for(var i = array.length, j, tmp; i--; ){
j = 0|(Math.random() * i);
tmp = array[j];
array[j] = array[i];
array[i] = tmp;
}
return array;
}
//now let's define a sequence of possible values
var numset = [0,1,2,3,4,5,6,7,8,9];
//shuffle the sequence and take the first 4 values
var fourRandomValues = shuffle(numset).slice(0,4);
console.log("four random values: " + fourRandomValues);
//doing this multiple times:
for(var values = []; values.length < 10;){
//shuffle again, and take the values that are now at the beginning of this sequence
values.push( shuffle(numset).slice(0,4) );
}
console.log("more random values: \n" + values.join("\n"));
Edit:
to address holi-java's approach by implementing sort of an Iterator, I'll add a way to do this with ES6 Iterators/Generators
Since Generators can be unlimited sequences we need to account for that. We do that by buffering a limited amount of values and returning them randomly; basically a shifiting frame of shuffled values.
function *shuffled(iterable, bufferSize = 256){
var buffer, numValues = 0, randomIndex;
if(Array.isArray(iterable) && iterable.length <= bufferSize){
//an optimization for (small) Arrays:
buffer = iterable.slice();
numValues = iterable.length;
}else{
buffer = Array( bufferSize )
for(var value of iterable){
//push value from the iterable to the buffer
buffer[numValues++] = value;
//buffer is full, yield a random value
if(numValues === bufferSize){
//choose a random value from the buffer
randomIndex = 0|(Math.random() * (numValues-1));
//yield it
yield buffer[randomIndex];
//overwrite the value with the last index
//that's cheaper than pop() and splice()
buffer[randomIndex] = buffer[--numValues];
}
}
}
//iterable doesn't provide any more values
//flush the buffer in a random order
while(numValues){
randomIndex = 0|(Math.random() * (numValues-1));
yield buffer[randomIndex];
buffer[randomIndex] = buffer[--numValues];
}
}
//every Array is a valid iterator
for(var v of shuffled([0,1,2,3,4,5,6,7,8,9]))
console.log(v);
That way we can shuffle a stream of values without first caching all the values in an array.
pro: memory efficient
possible problem: if the buffer's to small the result doesn't feel random anymore since values that are generated late in the sequence simply can not be shifted entirely to the start. You see some noise but it doesn't feel random anymore.
now let's take a jump into potentially infinite sequences:
// *shuffled again, for this snippet
function *shuffled(iterable, bufferSize = 256){
var buffer, numValues = 0, randomIndex;
if(Array.isArray(iterable) && iterable.length <= bufferSize){
buffer = iterable.slice();
numValues = iterable.length;
}else{
buffer = Array( bufferSize )
for(var value of iterable){
buffer[numValues++] = value;
if(numValues === bufferSize){
randomIndex = 0|(Math.random() * (numValues-1));
yield buffer[randomIndex];
buffer[randomIndex] = buffer[--numValues];
}
}
}
while(numValues){
randomIndex = 0|(Math.random() * (numValues-1));
yield buffer[randomIndex];
buffer[randomIndex] = buffer[--numValues];
}
}
//creates an infinite sequence of numbers
function *count(){
for(var index = 0; true;)
yield index++;
}
//like limits a iterator but for iterators
function *take(n, iterator){
for(var value of iterator){
if(n-- > 0) yield value;
else break;
}
}
//create an (infinite) counter and convert it into a generator of shuffled values
//with a bufferSize of 256 entries (play a bit with that value)
var shuffledSequence = shuffled(count(), 256);
//to convert that into an Array we take the first 1000 values generated from that generator
var array = [...take(1000, shuffledSequence)];
//and log it
console.log(array.toString());
what do you mean like this below:
function next() {
function all() {
return [].concat(Array(10).fill(null).map(function (_, index) {
return index;
}));
}
function random(start, end) {
return parseInt(Math.random() * (end + 1 - start)) + start;
}
var self = this;
self.all = self.all && self.all.length && self.all || all().sort(function(){
return random(-1, 1);
});
return self.all.shift();}
var results=Array(4).fill(null).map(next);
console.log(results);
my be this is better for browser to run.
var it = {
next: function () {
var self = this;
self._all = self._all && self._all.length && self._all || self.shuffle(self.all());
return self._all.shift();
},
all: function () {
return [0,1,2,3,4,5,6,7,8,9];
},
random: function (start, end) {
if (!end && end != 0) {
end = start;
start = 0;
}
return parseInt(Math.random() * (end + 1 - start)) + start;
},
shuffle: function (array) {
var self = this;
for (var i = array.length - 1; i > 0; --i) {
self.swap(array, i, self.random(i));
}
return array;
},
swap: function (array, i, j) {
var tmp = array[i];
array[i] = array[j];
array[j] = tmp;
},
reset: function () {
this._all = null;
}
};
for (var i = 0; i < 10; i++) {
var results=[];
for(var n=0;n<4;n++) results.push(it.next());
console.log("retain:"+it._all + '>>generated:' + results);
it.reset();
}
Edit
Takes four unique numbers out of the choose array. Is that what you're after?
var yourNum = [],
oppNum = [],
choose = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9];
function chooseRandomNumber() {
return choose[Math.floor(Math.random() * choose.length)];
}
var rand;
for (var i = 0; i < 4; i++) {
rand = chooseRandomNumber();
yourNum.push(rand);
choose.splice(choose.indexOf(rand), 1);
}
console.log(choose);
console.log(yourNum);
Old approach:
var start = 0, end = 9;
function generate(count) {
var nums = [],
random;
for (var i = 0; i < count; i++) {
while (!random || nums.indexOf(random) !== -1) {
random = Math.floor(Math.random() * (end + 1)) + start;
}
nums.push(random);
}
return nums;
}
console.log(
generate(4)
);
Demo:
function getRandomInt(min, max) {
min = Math.ceil(min);
max = Math.floor(max);
return Math.floor(Math.random() * (max - min)) + min;
}
var yourNums = [];
var choose = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9];
for (var i = 0; i < 4; i++) {
yourNums = yourNums.concat(choose.splice(getRandomInt(0, choose.length), 1))
}
console.log("yourNums is:")
console.log(yourNums);
console.log("===")
console.log("choose is:")
console.log(choose);
I'm trying to script a function that takes two numbers and returns the smallest common multiple that is also divisible by all the numbers between those numbers, what I've got only works for 1,1 through 1,12, but for some reason stops working at 1,13. Other set like 12,14 work but I can't figure out why or what the pattern is.
function smallestCommons(arr) {
arr.sort(function(a, b) {
return a-b;
});
var arr1 = [];
var arr2 = [];
for (var k = arr[0]; k<=arr[1]; k++) {
arr1.push(k);
}
function remainder(val1, val2) {
return val1%val2;
}
var b = arr1.reduce(function(a, b) {
return a*b;
});
var i = arr1[arr1.length-1]*arr1[arr1.length-2];
while (i<=b) {
for (var m = 0; m<arr1.length; m++) {
var a = remainder(i, arr1[m]);
arr2.push(a);
}
var answer = arr2.reduce(function(c, d) {
return c+d;
});
if (answer === 0) {
return i;
} else {
arr2 = [];
i++;
}
}
}
I guess you can do as follows in JavaScript; It can calculate the common LCM up to an 216 item array, such as [1,2,3,...,216] in less than 0.25 ms.
function gcd(a,b){
var t = 0;
a < b && (t = b, b = a, a = t); // swap them if a < b
t = a%b;
return t ? gcd(b,t) : b;
}
function lcm(a,b){
return a/gcd(a,b)*b;
}
var arr = [1,2,3,4,5,6,7,8,9,10,11,12,13],
brr = Array(216).fill().map((_,i) => i+1), // limit before Infinity
result = arr.reduce(lcm);
console.log(result);
console.time("limit");
result = brr.reduce(lcm);
console.timeEnd("limit");
console.log(result);
A way is to keep multiplying the largest number in your range with an increasing number and check if all the others are divisible by that. If yes, return that or continue the loop.
Here is my solution in typescript...
function findLowestCommonMultipleBetween(start: number, end: number): number {
let numbers: number[] = [];
for (let i = start; i <= end; i++) {
numbers.push(i);
}
for (let i = 1; true; i++) {
let divisor = end * i;
if (numbers.every((number) => divisor % number == 0)) {
return divisor;
}
}
}
...but for larger ranges, this is a more efficient answer :)
As far as I can tell your algorithm is giving you a correct answer.
I am far from being a professional programmer so anyone who wants please give options to improve my code or its style :)
If you want to be able to check for the answer yourself you can check this fiddle:
https://jsfiddle.net/cowCrazy/Ld8khrx7/
function multiplyDict(arr) {
arr.sort(function (a, b) {
return a - b;
});
if (arr[0] === 1) {
arr[0] = 2;
}
var currentArr = [];
for (var i = arr[0]; i <= arr[1]; i++) {
currentArr.push(i);
}
var primeDivs = allPrimes(arr[1]);
var divsDict = {};
for (var j = currentArr[0]; j <= currentArr[currentArr.length -1]; j++){
divsDict[j] = [];
if (primeDivs.indexOf(j) > -1) {
divsDict[j].push(j);
} else {
var x = j;
for (var n = 2; n <= Math.floor(j / 2); n++) {
if (x % n === 0) {
divsDict[j].push(n);
x = x / n;
n--;
continue;
}
}
}
}
return divsDict;
}
function allPrimes(num) {
var primeArr = [];
var smallestDiv = 2;
loopi:
for (var i = 2; i <= num; i++) {
loopj:
for (var j = smallestDiv; j <= largestDiv(i); j++) {
if (i % j === 0) {
continue loopi;
}
}
primeArr.push(i);
}
return primeArr;
}
function largestDiv (a) {
return Math.floor(Math.sqrt(a));
}
multiplyDict([1,13]);
it gives a dictionary of the requested array and the divisors of each element.
from there you can go on your own to check that your algorithm is doing the right job or you can check it here:
https://jsfiddle.net/cowCrazy/kr04mas7/
I hope it helps
It is true! The result of [1, 13] is 360360. and after this we have [1, 14].
14 = 2 * 7 and we now 360360 is dividable to 2 and 7 so the answer is 360360 again.
[1, 15]: 15 = 3 * 5 and result is same.
[1, 16]: result is 720720.
[1, 17]: result is: 12252240
[1, 18]: 18 = 2 * 9 and result is 12252240 same as 17
[1, 19]: for my computer this process is so heavy and can not do this. But in a strong machine it will work. I promise. But your code is not good in performance.
To find the LCM in N numbers.
It is Compatible with ES6, and consider that is there is no control for boundaries in case that we need to find for large numbers.
var a = [10, 40, 50, 7];
console.log(GetMinMultiple(a));
function GetMinMultiple(data) {
var maxOf = data.reduce((max, p) => p > max ? p : max, 0);
var incremental = maxOf;
var found = false;
do {
for (var j = 0; j < data.length; j++) {
if (maxOf % data[j] !== 0) {
maxOf += incremental;
break;
}
else {
if (j === data.length - 1) {
found = true;
break;
}
}
}
} while (!found);
return maxOf;
}
https://jsfiddle.net/djp30gfz/
Here is my solution in Typescript
function greatestCommonDivider(x: number, y: number): number {
if (y === 0) {
return x;
}
return greatestCommonDivider(y, x % y);
}
function singleLowestCommonMultiply(x: number, y: number): number {
return (x * y) / greatestCommonDivider(x, y);
}
function lowestCommonMultiply(...numbers: number[]): number {
/**
* For each number, get it's lowest common multiply with next number.
*
* Then using new number, compute new lowest common multiply
*/
return numbers.reduce((a, b) => {
return singleLowestCommonMultiply(a, b);
});
}
lowestCommonMultiply(2, 3); // Outputs 6
lowestCommonMultiply(2, 3, 5); // Outputs 30
Playground - click here
var store = ['1','2','2','3','4'];
I want to find out that 2 appear the most in the array. How do I go about doing that?
I would do something like:
var store = ['1','2','2','3','4'];
var frequency = {}; // array of frequency.
var max = 0; // holds the max frequency.
var result; // holds the max frequency element.
for(var v in store) {
frequency[store[v]]=(frequency[store[v]] || 0)+1; // increment frequency.
if(frequency[store[v]] > max) { // is this frequency > max so far ?
max = frequency[store[v]]; // update max.
result = store[v]; // update result.
}
}
Solution with emphasis to Array.prototype.forEach and the problem of getting more than one key if the max count is shared among more items.
Edit: Proposal with one loop, only.
var store = ['1', '2', '2', '3', '4', '5', '5'],
distribution = {},
max = 0,
result = [];
store.forEach(function (a) {
distribution[a] = (distribution[a] || 0) + 1;
if (distribution[a] > max) {
max = distribution[a];
result = [a];
return;
}
if (distribution[a] === max) {
result.push(a);
}
});
console.log('max: ' + max);
console.log('key/s with max count: ' + JSON.stringify(result));
console.log(distribution);
arr.sort();
var max=0,result,freq = 0;
for(var i=0; i < arr.length; i++){
if(arr[i]===arr[i+1]){
freq++;
}
else {
freq=0;
}
if(freq>max){
result = arr[i];
max = freq;
}
}
return result;
Make a histogram, find the key for the maximum number in the histogram.
var hist = [];
for (var i = 0; i < store.length; i++) {
var n = store[i];
if (hist[n] === undefined) hist[n] = 0;
else hist[n]++;
}
var best_count = hist[store[0]];
var best = store[0];
for (var i = 0; i < store.length; i++) {
if (hist[store[i]] > best_count) {
best_count = hist[store[i]];
best = store[i];
}
}
alert(best + ' occurs the most at ' + best_count + ' occurrences');
This assumes either there are no ties, or you don't care which is selected.
Another ES6 option. Works with strings or numbers.
function mode(arr) {
const store = {}
arr.forEach((num) => store[num] ? store[num] += 1 : store[num] = 1)
return Object.keys(store).sort((a, b) => store[b] - store[a])[0]
}
If the array is sorted this should work:
function popular(array) {
if (array.length == 0) return [null, 0];
var n = max = 1, maxNum = array[0], pv, cv;
for(var i = 0; i < array.length; i++, pv = array[i-1], cv = array[i]) {
if (pv == cv) {
if (++n >= max) {
max = n; maxNum = cv;
}
} else n = 1;
}
return [maxNum, max];
};
popular([1,2,2,3,4,9,9,9,9,1,1])
[9, 4]
popular([1,2,2,3,4,9,9,9,9,1,1,10,10,10,10,10])
[10, 5]
This version will quit looking when the count exceeds the number of items not yet counted.
It works without sorting the array.
Array.prototype.most= function(){
var L= this.length, freq= [], unique= [],
tem, max= 1, index, count;
while(L>= max){
tem= this[--L];
if(unique.indexOf(tem)== -1){
unique.push(tem);
index= -1, count= 0;
while((index= this.indexOf(tem, index+1))!= -1){
++count;
}
if(count> max){
freq= [tem];
max= count;
}
else if(count== max) freq.push(tem);
}
}
return [freq, max];
}
//test
var A= ["apples","oranges","oranges","oranges","bananas",
"bananas","oranges","bananas"];
alert(A.most()) // [oranges,4]
A.push("bananas");
alert(A.most()) // [bananas,oranges,4]
I solved it this way for finding the most common integer
function mostCommon(arr) {
// finds the first most common integer, doesn't account for 2 equally common integers (a tie)
freq = [];
// set all frequency counts to 0
for(i = 0; i < arr[arr.length-1]; i++) {
freq[i] = 0;
}
// use index in freq to represent the number, and the value at the index represent the frequency count
for(i = 0; i < arr.length; i++) {
freq[arr[i]]++;
}
// find biggest number's index, that's the most frequent integer
mostCommon = freq[0];
for(i = 0; i < freq.length; i++) {
if(freq[i] > mostCommon) {
mostCommon = i;
}
}
return mostCommon;
}
This is my solution.
var max_frequent_elements = function(arr){
var a = [], b = [], prev;
arr.sort();
for ( var i = 0; i < arr.length; i++ ) {
if ( arr[i] !== prev ) {
a.push(arr[i]);
b.push(1);
} else {
b[b.length-1]++;
}
prev = arr[i];
}
var max = b[0]
for(var p=1;p<b.length;p++){
if(b[p]>max)max=b[p]
}
var indices = []
for(var q=0;q<a.length;q++){
if(b[q]==max){indices.push(a[q])}
}
return indices;
};
All the solutions above are iterative.
Here's a ES6 functional mutation-less version:
Array.prototype.mostRepresented = function() {
const indexedElements = this.reduce((result, element) => {
return result.map(el => {
return {
value: el.value,
count: el.count + (el.value === element ? 1 : 0),
};
}).concat(result.some(el => el.value === element) ? [] : {value: element, count: 1});
}, []);
return (indexedElements.slice(1).reduce(
(result, indexedElement) => (indexedElement.count > result.count ? indexedElement : result),
indexedElements[0]) || {}).value;
};
It could be optimized in specific situations where performance is the bottleneck, but it has a great advantage of working with any kind of array elements.
The last line could be replaced with:
return (indexedElements.maxBy(el => el.count) || {}).value;
With:
Array.prototype.maxBy = function(fn) {
return this.slice(1).reduce((result, element) => (fn(element) > fn(result) ? element : result), this[0]);
};
for clarity
If the array contains strings try this solution
function GetMaxFrequency (array) {
var store = array;
var frequency = []; // array of frequency.
var result; // holds the max frequency element.
for(var v in store) {
var target = store[v];
var numOccurences = $.grep(store, function (elem) {
return elem === target;
}).length;
frequency.push(numOccurences);
}
maxValue = Math.max.apply(this, frequency);
result = store[$.inArray(maxValue,frequency)];
return result;
}
var store = ['ff','cc','cc','ff','ff','ff','ff','ff','ff','yahya','yahya','cc','yahya'];
alert(GetMaxFrequency(store));
A fairly short solution.
function mostCommon(list) {
var keyCounts = {};
var topCount = 0;
var topKey = {};
list.forEach(function(item, val) {
keyCounts[item] = keyCounts[item] + 1 || 1;
if (keyCounts[item] > topCount) {
topKey = item;
topCount = keyCounts[item];
}
});
return topKey;
}
document.write(mostCommon(['AA', 'AA', 'AB', 'AC']))
This solution returns an array of the most appearing numbers in an array, in case multiple numbers appear at the "max" times.
function mode(numbers) {
var counterObj = {};
var max = 0;
var result = [];
for(let num in numbers) {
counterObj[numbers[num]] = (counterObj[numbers[num]] || 0) + 1;
if(counterObj[numbers[num]] >= max) {
max = counterObj[numbers[num]];
}
}
for (let num in counterObj) {
if(counterObj[num] == max) {
result.push(parseInt(num));
}
}
return result;
}