Develop Reference

Finding the second lowest frequent element from an array of integers?

I'm looking to answer a coding challenge in JavaScript that I'm stuck on, here's the question:
Write a function which accepts an array of integers and returns an element of that array.
The function should determine the frequency of each element (how many times the element appears in the array) and whenever possible should return the element with the second-lowest frequency. Otherwise it should return the integer with the lowest frequency.
If there is more than one element satisfying the requirements then the second smallest one (according to value) should be returned.
Example outputs:
secondLowest( [4, 3, 1, 1, 2] ) === 1
secondLowest( [4, 3, 1, 1, 2, 2] ) === 2
secondLowest( [4, 3, 1, 2] ) === 2
This is what I've got so far although don't know how best to go about answering it after this:
function mode(array) {
if (array.length == 0) return null;
let modeMap = {};
let maxEl = array[0],
maxCount = 1;
for (let 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;
}

I was determined to give a generic, parameterized function, where no number is hardcoded.
Your example involved two hardcoded values:
The second least frequency should be selected
In cases of ties, the second least value should be selected
The following code works like this:
Get the frequency of each input value
Group together all values with the same frequency.
Sort these grouped pairs by frequency and select the nth-lowest (in your case, n=2)
If the nth-lowest frequency has multiple pairs, sort these pairs by value, and select the mth-lowest pair (in your case, m=2)
Return the value of this final pair
The m and n parameters I refer to here are called freqInd and valInd in the code. Note that in order to select the second-lowest frequency, freqInd should be 1, not 2 (since 0 would select the lowest, and therefore 1 selects the second-lowest).
let lowestFreqVal = (freqInd, valInd, values) => {
// Calculate frequencies in a map
let f = new Map();
for (let v of values) f.set(v, (f.get(v) || 0) + 1);
// Group together all val/freq pairs with the same frequency
let ff = new Map();
for (let [ val, freq ] of f) ff.set(freq, (ff.get(freq) || []).concat([ val ]));
// Sort these groups by frequency
let byFreq = [ ...ff ].sort(([ freq1 ], [ freq2 ]) => freq1 - freq2);
// Here are all the items of the `freqInd`-th lowest frequency, sorted by value
// Note that `[1]` returns an array of integers at the frequency, whereas `[0]` would return the frequency itself
let lowestItems = byFreq[ Math.min(byFreq.length - 1, freqInd) ][1]
.sort((v1, v2) => v1 - v2);
// Return the `valInd`-th lowest value
return lowestItems[ Math.min(lowestItems.length - 1, valInd) ];
};
console.log('Some random examples:');
for (let i = 0; i < 10; i++) {
// An array of random length, full of random integers
let arr = [ ...new Array(3 + Math.floor(Math.random() * 5)) ]
.map(v => Math.floor(Math.random() * 4));
// Show the result of `lowestFreqVal` on this random Array
console.log(`lowestFreqVal(1, 1, ${JSON.stringify(arr)}) = ${lowestFreqVal(1, 1, arr)}`);
}
This is not an optimal solution, since it resorts to using sort. It's known that the problem of finding some nth-maximal value in a list can be implemented to have a better runtime than sort (and a significantly better runtime when n is a small value - we can see this intuitively because if n=0, a single pass (O(n)) does the trick).

Assigning every possible combination of a range of numbers to variables

Given a range of numbers (1-25) how can I create a loop so that at each iteration I get a unique set of variables.
To give an example, if I was doing it with 4 numbers my variables would be:
on loop 1:
a = 1, b = 2, c = 3, d = 4,
on loop 2:
a = 1, b = 2, c = 4, d = 3
etc.
What I am trying to do is iterate over every possible number for each position (think sudoku)
so in a 3x3 grid: a = top left position, b = top middle, etc...
so similar to sudoku I would have a condition (each row = 65: a + b + c + d + e= 65)
so what I want to do is have a loop where I can assign all the values to variables:
for (something) {
var topLeft = (determined from loop)
var nextPosition = etc.
My solution is currently like so:
var numbers = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25];
var a,b,c,d,e,f,g,h,i,j,k,l,n,o,p,q,r,s,t,u,v,w,x,y;
var vars = [a,b,c,d,e,f,g,h,i,j,k,l,n,o,p,q,r,s,t,u,v,w,x,y];
var counter = 0;
var found = false;
while(found == false) {
for (var asdf = numbers, i = asdf.length; i--; ) {
var random = asdf.splice(Math.floor(Math.random() * (i + 1)), 1)[0];
vars[i] = random;
}
if (
{a+b+c+d+e = 65,
f+g+h+i+j = 65,
k+l+1+n+o = 65,
p+q+r+s+t = 65,
u+v+w+x+y = 65,
a+f+k+p+u = 65,
b+g+l+q+v = 65,
c+h+1+r+w = 65,
d+i+n+s+x = 65,
e+j+o+t+y = 65,
u+q+1+i+e = 65,
a+g+1+s+y = 65}
) {
console.log(a,b,c,d,e,f,g,h,i,j,k,l,n,o,p,q,r,s,t,u,v,w,x,y);
found = true;
}
counter++;
}
However the obvious problem with this is that it's just randomly selecting values. So it will take an incredible amount of time. I can't work out how to iterate over every possible combination (without having 25 for loops) so I can check which values will pass the condition.

Given a range of numbers (1-25) how can I create a loop so that at each iteration I get a unique set of variables.
It sounds like you are talking about the permutations of a set. You can find a bunch of different algorithms to do this. Here is a nice one from this StackOverflow answer:
function getArrayMutations(arr, perms = [], len = arr.length) {
if (len === 1) perms.push(arr.slice(0))
for (let i = 0; i < len; i++) {
getArrayMutations(arr, perms, len - 1)
len % 2 // parity dependent adjacent elements swap
? [arr[0], arr[len - 1]] = [arr[len - 1], arr[0]]
: [arr[i], arr[len - 1]] = [arr[len - 1], arr[i]]
}
return perms
}
getArrayMutations([1, 2, 3])
> [ [ 1, 2, 3 ],
[ 2, 1, 3 ],
[ 3, 1, 2 ],
[ 1, 3, 2 ],
[ 2, 3, 1 ],
[ 3, 2, 1 ] ]
Be careful though! Permutations are factorial which means they grow really fast.
P(n, k) =
This means that if you want to permute 25 numbers, you are looking at 1.551121e+25 possible combinations which is getting into the not-computable-in-your-lifetime territory.
What I am trying to do is iterate over every possible number for each position (think sudoku) so in a 3x3 grid: a = top left position, b = top middle, etc...
Two dimensional arrays (really just lists of lists) are a great way to store matrix data like this. It doesn't fundamentally change the math to change the representation from a single array, but it might be easier to think about. I'm not 100% sure if you want a 3x3 grid or a 5x5 grid but I'll assume 5x5 since you have 25 numbers in your example. You can easily reshape them like this:
function reshapeArray(array, n=5) {
let result = []
let row = 0
let col = 0
for (let i = 0; i < array.length; i++) {
if (col == 0) {
result[row] = []
}
result[row][col] = array[i]
col++
if (col == n) {
col = 0
row++
}
}
return result
}
reshapeArray([1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25])
> [ [ 1, 2, 3, 4, 5 ],
[ 6, 7, 8, 9, 10 ],
[ 11, 12, 13, 14, 15 ],
[ 16, 17, 18, 19, 20 ],
[ 21, 22, 23, 24, 25 ] ]
so similar to sudoku I would have a condition (each row = 65: a + b + c + d + e= 65)
Now that you have your data in an iteratable array, you can very easily check this or any other constraint. For example:
/**
* Checks if a matrix (a 2-d array like the output from reshapeArray())
* meets our criteria.
*/
function checkMatrix(matrix) {
for (let row = 0; row < matrix.length; row++) {
let rowSum = 0
for (let col = 0; col < matrix[row].length; col++) {
rowSum += matrix[row][col]
}
// The row sum does not equal 65, so this isn't the one!
if (rowSum != 65) {
return false
}
}
// All the row sums equal 65
return true
}
If you want add extra rules (like having the columns sum to 65 as well) just modify the code to check for that. You can get the value at any point in the matrix by indexing it matrix[row][col] so matrix[0][0] is the upper-left, etc.
However the obvious problem with this is that it's just randomly selecting values. So it will take an incredible amount of time. I can't work out how to iterate over every possible combination (without having 25 for loops) so I can check which values will pass the condition.
Yes, it will. Sudoku is an NP-Hard problem. If you haven't seen complexity classes before, that's just a very mathematically formal way of saying that there's no clever solution that's going to be significantly faster than just checking every possible solution. This hypothetical problem is not exactly the same, so it might be possible, but it has a very np-ish feel to it.
Currently, your pseudocode solution would look like this:
let permutations = getPermutations() // You're going to need to change this part
// because getting all the permutations
// ahead of time will take too long.
// Just picking random numbers each time is
// not actually a terrible idea. Or, look at
// generator functions (https://developer.mozilla.org/en-US/docs/Web/JavaScript/Guide/Iterators_and_Generators)
for (let permutation of permutations) {
let matrix = reshapeArray(permutation)
if (checkMatrix(matrix)) {
console.log("Found it")
console.log(matrix)
break
}
}
If there is only one possible solution that matches your criteria, you will never find it this way. If there is a relatively high density of solutions, you will probably find some. If you really want to solve this problem I would recommend first looking at it from a mathematical perspective -- can you prove that it is or isn't NP? can you make some prediction about the density of solutions?

Not sure what the question really is. I would store the range in an Array:
function range(start, stop = null, upperCase = false){
let b = start, e = stop, s = 'abcdefghijklmnopqrstuvwxyz', x;
const a = [], z = s.split(''), Z = s.toUpperCase().split('');
if(typeof b === 'string'){
s = z.indexOf(b.toLowerCase());
if(e === null){
x = z.length;
}
else{
x = z.indexOf(e.toLowerCase())+1;
}
if(upperCase){
return Z.slice(s, x);
}
return z.slice(s, x);
}
else if(e === null){
e = b; b = 1;
}
for(let i=b; i<=e; i++){
a.push(i);
}
return a;
}
function permuCount(array){
let c = 1;
for(let i=0,n=1,l=array.length; i<l; i++,n++){
c *= n;
}
return c;
}
function comboCount(array){
let l = array.length;
return Math.pow(l, l);
}
console.log(range(2, 23)); console.log(range(10)); console.log(range('d'));
console.log(range('g', 'p')); console.log(range('c', 'j', true));
// here is where you'll have an issue
const testArray = range(0, 9);
console.log(permuCount(testArray));
console.log(comboCount(testArray));
As you can see there are way too many combinations. Also, you should have already see the following post: Permutations in JavaScript?

For Loop that calculates sum of squred numbers in array

I am working on a challenge and in need of some help:
Write a for loop that calculates sum of squares of items in an array of numbers. Example: For array [ 1, 2, 3, 4 ] it calculates the sum of squares as 30 (i.e. 1 + 4 + 9 + 16). I have a fiddle set up if anyone wants to have a closer look. Thanks for your help!
https://jsfiddle.net/jamie_shearman/2drt56e3/23/
var aFewNumbers = [ 1, 2, 3, 7 ];
var squareOfAFewNumbers = 0;
for( var i = 0; i <= aFewNumbers; i++ ) {
squareOfAFewNumbers = squareOfAFewNumbers * aFewNumbers[i] ;
}
console.log( squareOfAFewNumbers );

Your math is wrong. As an obvious issue, the variable starts at 0 and then is multiplied by each array element; 0 times anything is 0, so it will always remain 0 no matter what the values are. Not to mention your loop isn't looking at the length of the array as a stopping condition, which it should be, since you want to iterate from the beginning to the end of the array.
You need to iterate through to the array's length, square each array element, and add that square to the variable:
for( var i = 0; i < aFewNumbers.length; i++ ) {
squareOfAFewNumbers += aFewNumbers[i] * aFewNumbers[i];
}
If you can use ES6, you can even use higher-order array functions to simplify this more:
var squareOfAFewNumbers = aFewNumbers.reduce((result, entry) => result + entry * entry, 0);

There are multiple approaches you can take for reaching the desired result, but as you've mentioned that you must write a for loop; therefore, I sorted the answers by having that in mind.
Using the for loop
let numbers = [1, 2, 3, 7],
sum = 0;
for(let i = 0; i < numbers.length; i++) {
sum += Math.pow(numbers[i], 2)
}
// Sum is now 63
Using forEach method of the array object
let numbers = [1, 2, 3, 7],
sum = 0;
numbers.forEach(number => sum += Math.pow(number, 2))
// Sum is now 63
Oneliner
let sum = [1, 2, 3, 7].reduce((a, c) => a + Math.pow(c, 2))
The reduce method uses an accumulator to store temporary results, the accumulator is passed to the callback as the first argument and the second argument is the element's value, you can read more about the reduce method here.

You can use JavaScript pow() Method to create the square and sum it to the sumSquareOfAFewNumbers.
var aFewNumbers = [ 1, 2, 3, 7 ];
var sumSquareOfAFewNumbers = 0;
aFewNumbers.forEach(function(element) {
sumSquareOfAFewNumbers += Math.pow(element, 2);
});
console.log(sumSquareOfAFewNumbers)

Finding a non-consecutive number pair in an array

Given an array with a minimum length of 3 and a maximum length of 5, which always contains uniquely occurring integers from 0 to 4 in ascending order, I need to pick out two non-consecutive numbers from it. Non-consecutive refers to their numeric value, not their position in the array.
To clarify, here are examples of valid arrays:
[ 1, 2, 3 ]
[ 0, 1, 2, 4 ]
[ 0, 3, 4 ]
For the arrays above, valid answers could be, respectively:
[ 1, 3 ]
[ 0, 2 ], [ 0, 4 ] or [ 1, 4 ]
[ 0, 3 ] or [ 0, 4 ]
Furthermore, in those cases where there is more than one valid answer, I need it to be selected at random, if at all possible (for instance I don't want to favor sequences that begin with the lowest number, which is what would occur if I always began checking from left to right and stopped checking as soon as I found one valid solution).
What would be the most efficient way of tackling this problem in Javascript?

You could use two nested iterations and build an new array for choosing as random result.
function getNonConsecutives(array) {
return array.reduce((r, a, i, aa) => r.concat(aa.slice(i + 2).map(b => [a, b])), []);
}
console.log(getNonConsecutives([ 0, 1, 2, 4 ]));
.as-console-wrapper { max-height: 100% !important; top: 0; }
According to Bee157's answer, you could use a random choice with a constraint, like length for the first index and add the needed space for the second index.
The problem is, due to the nature of choosing the first number first, the distribution of the result is not equal.
function getNonConsecutives(array) {
var i = Math.floor(Math.random() * (array.length - 2));
return [
array[i],
array[Math.floor(Math.random() * (array.length - 2 - i)) + 2 + i]
];
}
console.log(getNonConsecutives([ 0, 1, 2, 4 ]));

demoFn(array) {
var i,j, y =[];
for (i=0; i<=array.length;i++) {
for (j = i + 1; j <= array.length; j++) {
if (array[j] && array[i]) {
if (array[j] !== array[i] + 1) {
y.push([array[i], array[j]]);
}
}
}
}
}
Take a random array and check it.

You can create a function using recursion that will pick random number in each iteration and loop all other elements and if condition is met add to array.
function findN(data) {
data = data.slice();
var r = []
function repeat(data) {
if (data.length < 2) return r;
var n = parseInt(Math.random() * data.length);
data.forEach(function(e, i) {
if (i != n) {
var a = data[n];
if (Math.abs(a - e) != 1 && r.length < 2) r.push(n < i ? [a, e] : [e, a])
}
})
data.splice(n, 1);
repeat(data)
return r;
}
return repeat(data)
}
console.log(findN([1, 2, 3]))
console.log(findN([0, 1, 2, 4]))
console.log(findN([0, 3, 4]))

Something like this should do it:
const pick = nums => {
// Pick a random number
const val = nums[Math.floor(Math.random() * nums.length) + 0];
// Filter out any numbers that are numerically consecutive
const pool = nums.filter(n => Math.abs(n - val) > 1);
// Pick another random number from the remainer
const other = pool[Math.floor(Math.random() * pool.length) + 0];
// Sort + return them
return [val, other].sort();
};
console.log(pick([0, 1, 2, 4]));

since you state that the array ellemnts are all unique, and that they are sorted.
It should suffice to take an random element
var index1=Math.floor(Math.random()*arr.length)
now any other element (except maybe the elemnts on position (index1 +/- 1) are not consecutive
So a new random element can be chosen excluding the first index.
var index2=Math.floor(Math.random()*arr.length);
if(index2==index1){
index2+=((index2<arr.length-1)?1:-1);
}
if(Math.abs(arr[index1]-arr[index2])<=1){
if(index2==0 && arr.length<4){
//set index2 to arr.length-1 and do check again, if not ok=> no result
if(!(arr[index1]-arr[arr.length-1]>=-1)){
return [arr[arr.length-1],arr[index1]];
}
}
else if(index2==arr.length-1 && arr.length<4){
//set index2 to 0 and do check again, if not ok=> no result
if(!(arr[index1]-arr[0]<=1)){
return [arr[0],arr[index1]];
}
}
else{
//if index2>index1 index2++
//else index2--
//always OK so no further check needed
index2+=(index2>index1?1:-1);
return [arr[index1],arr[index2]];
}
}
else{
//ok
return [arr[index1,arr[index2]];
}
return false;
if speed is not important, you can use a filter on the array to calculate a new array with all elements differing more then 1 unit of arr[index1]. and randomly select a new number from this new array.
Other attempt
function getNonConsecutive(arr){
var index1,index2,arr2;
index1=Math.floor(Math.random()*arr.length);
arr2=[].concat(arr);
arr2.splice((index1!==0?index1-1:index1),(index!==0?3:2));
if(arr2.length){
index2=Math.floor(Math.random()*arr2.length);
return [arr[index1],arr2[index2]];
}
else{
//original array has length 3 or less
arr2=[].concat(arr);
arr2.splice(index1),1);
for (var j=0,len=arr.length;j<len;j++){
if(Math.abs(arr1[index1]-arr2[j])>1){
return [arr[index1],arr2[j]];
}
}
}
return false
}

javascript array sorting / array match sorted array [duplicate]

For example, if I have these arrays:
var name = ["Bob","Tom","Larry"];
var age = ["10", "20", "30"];
And I use name.sort() the order of the "name" array becomes:
var name = ["Bob","Larry","Tom"];
But, how can I sort the "name" array and have the "age" array keep the same order? Like this:
var name = ["Bob","Larry","Tom"];
var age = ["10", "30", "20"];

You can sort the existing arrays, or reorganize the data.
Method 1:
To use the existing arrays, you can combine, sort, and separate them:
(Assuming equal length arrays)
var names = ["Bob","Tom","Larry"];
var ages = ["10", "20", "30"];
//1) combine the arrays:
var list = [];
for (var j = 0; j < names.length; j++)
list.push({'name': names[j], 'age': ages[j]});
//2) sort:
list.sort(function(a, b) {
return ((a.name < b.name) ? -1 : ((a.name == b.name) ? 0 : 1));
//Sort could be modified to, for example, sort on the age
// if the name is the same. See Bonus section below
});
//3) separate them back out:
for (var k = 0; k < list.length; k++) {
names[k] = list[k].name;
ages[k] = list[k].age;
}
This has the advantage of not relying on string parsing techniques, and could be used on any number of arrays that need to be sorted together.
Method 2: Or you can reorganize the data a bit, and just sort a collection of objects:
var list = [
{name: "Bob", age: 10},
{name: "Tom", age: 20},
{name: "Larry", age: 30}
];
list.sort(function(a, b) {
return ((a.name < b.name) ? -1 : ((a.name == b.name) ? 0 : 1));
});
for (var i = 0; i<list.length; i++) {
alert(list[i].name + ", " + list[i].age);
}
​
For the comparisons,-1 means lower index, 0 means equal, and 1 means higher index. And it is worth noting that sort() actually changes the underlying array.
Also worth noting, method 2 is more efficient as you do not have to loop through the entire list twice in addition to the sort.
http://jsfiddle.net/ghBn7/38/
Bonus Here is a generic sort method that takes one or more property names.
function sort_by_property(list, property_name_list) {
list.sort((a, b) => {
for (var p = 0; p < property_name_list.length; p++) {
prop = property_name_list[p];
if (a[prop] < b[prop]) {
return -1;
} else if (a[prop] !== a[prop]) {
return 1;
}
}
return 0;
});
}
Usage:
var list = [
{name: "Bob", age: 10},
{name: "Tom", age: 20},
{name: "Larry", age: 30},
{name: "Larry", age: 25}
];
sort_by_property(list, ["name", "age"]);
for (var i = 0; i<list.length; i++) {
console.log(list[i].name + ", " + list[i].age);
}
Output:
Bob, 10
Larry, 25
Larry, 30
Tom, 20

You could get the indices of name array using Array.from(name.keys()) or [...name.keys()]. Sort the indices based on their value. Then use map to get the value for the corresponding indices in any number of related arrays
const indices = Array.from(name.keys())
indices.sort( (a,b) => name[a].localeCompare(name[b]) )
const sortedName = indices.map(i => name[i]),
const sortedAge = indices.map(i => age[i])
Here's a snippet:
const name = ["Bob","Tom","Larry"],
age = ["10", "20", "30"],
indices = Array.from(name.keys())
.sort( (a,b) => name[a].localeCompare(name[b]) ),
sortedName = indices.map(i => name[i]),
sortedAge = indices.map(i => age[i])
console.log(indices)
console.log(sortedName)
console.log(sortedAge)

This solution (my work) sorts multiple arrays, without transforming the data to an intermediary structure, and works on large arrays efficiently. It allows passing arrays as a list, or object, and supports a custom compareFunction.
Usage:
let people = ["john", "benny", "sally", "george"];
let peopleIds = [10, 20, 30, 40];
sortArrays([people, peopleIds]);
[["benny", "george", "john", "sally"], [20, 40, 10, 30]] // output
sortArrays({people, peopleIds});
{"people": ["benny", "george", "john", "sally"], "peopleIds": [20, 40, 10, 30]} // output
Algorithm:
Create a list of indexes of the main array (sortableArray)
Sort the indexes with a custom compareFunction that compares the values, looked up with the index
For each input array, map each index, in order, to its value
Implementation:
/**
* Sorts all arrays together with the first. Pass either a list of arrays, or a map. Any key is accepted.
* Array|Object arrays [sortableArray, ...otherArrays]; {sortableArray: [], secondaryArray: [], ...}
* Function comparator(?,?) -> int optional compareFunction, compatible with Array.sort(compareFunction)
*/
function sortArrays(arrays, comparator = (a, b) => (a < b) ? -1 : (a > b) ? 1 : 0) {
let arrayKeys = Object.keys(arrays);
let sortableArray = Object.values(arrays)[0];
let indexes = Object.keys(sortableArray);
let sortedIndexes = indexes.sort((a, b) => comparator(sortableArray[a], sortableArray[b]));
let sortByIndexes = (array, sortedIndexes) => sortedIndexes.map(sortedIndex => array[sortedIndex]);
if (Array.isArray(arrays)) {
return arrayKeys.map(arrayIndex => sortByIndexes(arrays[arrayIndex], sortedIndexes));
} else {
let sortedArrays = {};
arrayKeys.forEach((arrayKey) => {
sortedArrays[arrayKey] = sortByIndexes(arrays[arrayKey], sortedIndexes);
});
return sortedArrays;
}
}
See also https://gist.github.com/boukeversteegh/3219ffb912ac6ef7282b1f5ce7a379ad

If performance matters, there is sort-ids package for that purpose:
var sortIds = require('sort-ids')
var reorder = require('array-rearrange')
var name = ["Bob","Larry","Tom"];
var age = [30, 20, 10];
var ids = sortIds(age)
reorder(age, ids)
reorder(name, ids)
That is ~5 times faster than the comparator function.

It is very similar to jwatts1980's answer (Update 2).
Consider reading Sorting with map.
name.map(function (v, i) {
return {
value1 : v,
value2 : age[i]
};
}).sort(function (a, b) {
return ((a.value1 < b.value1) ? -1 : ((a.value1 == b.value1) ? 0 : 1));
}).forEach(function (v, i) {
name[i] = v.value1;
age[i] = v.value2;
});

You are trying to sort 2 independet arrays by only calling sort() on one of them.
One way of achieving this would be writing your own sorting methd which would take care of this, meaning when it swaps 2 elements in-place in the "original" array, it should swap 2 elements in-place in the "attribute" array.
Here is a pseudocode on how you might try it.
function mySort(originals, attributes) {
// Start of your sorting code here
swap(originals, i, j);
swap(attributes, i, j);
// Rest of your sorting code here
}

inspired from #jwatts1980's answer, and #Alexander's answer here I merged both answer's into a quick and dirty solution;
The main array is the one to be sorted, the rest just follows its indexes
NOTE: Not very efficient for very very large arrays
/* #sort argument is the array that has the values to sort
#followers argument is an array of arrays which are all same length of 'sort'
all will be sorted accordingly
example:
sortMutipleArrays(
[0, 6, 7, 8, 3, 4, 9],
[ ["zr", "sx", "sv", "et", "th", "fr", "nn"],
["zero", "six", "seven", "eight", "three", "four", "nine"]
]
);
// Will return
{
sorted: [0, 3, 4, 6, 7, 8, 9],
followed: [
["zr", th, "fr", "sx", "sv", "et", "nn"],
["zero", "three", "four", "six", "seven", "eight", "nine"]
]
}
*/
You probably want to change the method signature/return structure, but that should be easy though. I did it this way because I needed it
var sortMultipleArrays = function (sort, followers) {
var index = this.getSortedIndex(sort)
, followed = [];
followers.unshift(sort);
followers.forEach(function(arr){
var _arr = [];
for(var i = 0; i < arr.length; i++)
_arr[i] = arr[index[i]];
followed.push(_arr);
});
var result = {sorted: followed[0]};
followed.shift();
result.followed = followed;
return result;
};
var getSortedIndex = function (arr) {
var index = [];
for (var i = 0; i < arr.length; i++) {
index.push(i);
}
index = index.sort((function(arr){
/* this will sort ints in descending order, change it based on your needs */
return function (a, b) {return ((arr[a] > arr[b]) ? -1 : ((arr[a] < arr[b]) ? 1 : 0));
};
})(arr));
return index;
};

I was looking for something more generic and functional than the current answers.
Here's what I came up with: an es6 implementation (with no mutations!) that lets you sort as many arrays as you want given a "source" array
/**
* Given multiple arrays of the same length, sort one (the "source" array), and
* sort all other arrays to reorder the same way the source array does.
*
* Usage:
*
* sortMultipleArrays( objectWithArrays, sortFunctionToApplyToSource )
*
* sortMultipleArrays(
* {
* source: [...],
* other1: [...],
* other2: [...]
* },
* (a, b) => { return a - b })
* )
*
* Returns:
* {
* source: [..sorted source array]
* other1: [...other1 sorted in same order as source],
* other2: [...other2 sorted in same order as source]
* }
*/
export function sortMultipleArrays( namedArrays, sortFn ) {
const { source } = namedArrays;
if( !source ) {
throw new Error('You must pass in an object containing a key named "source" pointing to an array');
}
const arrayNames = Object.keys( namedArrays );
// First build an array combining all arrays into one, eg
// [{ source: 'source1', other: 'other1' }, { source: 'source2', other: 'other2' } ...]
return source.map(( value, index ) =>
arrayNames.reduce((memo, name) => ({
...memo,
[ name ]: namedArrays[ name ][ index ]
}), {})
)
// Then have user defined sort function sort the single array, but only
// pass in the source value
.sort(( a, b ) => sortFn( a.source, b.source ))
// Then turn the source array back into an object with the values being the
// sorted arrays, eg
// { source: [ 'source1', 'source2' ], other: [ 'other1', 'other2' ] ... }
.reduce(( memo, group ) =>
arrayNames.reduce((ongoingMemo, arrayName) => ({
...ongoingMemo,
[ arrayName ]: [
...( ongoingMemo[ arrayName ] || [] ),
group[ arrayName ]
]
}), memo), {});
}

You could append the original index of each member to the value, sort the array, then remove the index and use it to re-order the other array. It will only work where the contents are strings or can be converted to and from strings successfuly.
Another solution is keep a copy of the original array, then after sorting, find where each member is now and adjust the other array appropriately.

I was having the same issue and came up with this incredibly simple solution. First combine the associated ellements into strings in a seperate array then use parseInt in your sort comparison function like this:
<html>
<body>
<div id="outPut"></div>
<script>
var theNums = [13,12,14];
var theStrs = ["a","b","c"];
var theCombine = [];
for (var x in theNums)
{
theCombine[x] = theNums[x] + "," + theStrs;
}
var theSorted = theAr.sort(function(a,b)
{
var c = parseInt(a,10);
var d = parseInt(b,10);
return c-d;
});
document.getElementById("outPut").innerHTML = theS;
</script>
</body>
</html>

How about:
var names = ["Bob","Tom","Larry"];
var ages = ["10", "20", "30"];
var n = names.slice(0).sort()
var a = [];
for (x in n)
{
i = names.indexOf(n[x]);
a.push(ages[i]);
names[i] = null;
}
names = n
ages = a

Simplest explantion is the best, merge the arrays, and then extract after sorting:
create an array
name_age=["bob#10","Tom#20","Larry#30"];
sort the array as before, then extract the name and the age, you can use # to reconise where
name ends and age begins. Maybe not a method for the purist, but I have the same issue and this my approach.