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Minimum number of swaps to sort an array

I need to do something like this: Let's say I have an array:
[3, 4, 1, 2]
I need to swap 3 and 4, and 1 and 2, so my array looks like [4, 3, 2, 1]. Now, I can just do the sort(). Here I need to count how many iterations I need, to change the initial array to the final output. Example:
// I can sort one pair per iteration
let array = [3, 4, 1, 2, 5]
let counter = 0;
//swap 3 and 4
counter++;
// swap 1 and 2
counter++;
// 5 goes to first place
counter++
// now counter = 3 <-- what I need
EDIT: Here is what I tried. doesn't work always tho... it is from this question: Bubble sort algorithm JavaScript
let counter = 0;
let swapped;
do {
swapped = false;
for (var i = 0; i < array.length - 1; i++) {
if (array[i] < array[i + 1]) {
const temp = array[i];
array[i] = array[i + 1];
array[i + 1] = temp;
swapped = true;
counter++;
}
}
} while (swapped);
EDIT: It is not correct all the time because I can swap places from last to first, for example. Look at the example code above, it is edited now.

This is most optimal code I have tried so far, also the code is accepted as optimal
answer by hackerrank :
function minimumSwaps(arr) {
var arrLength = arr.length;
// create two new Arrays
// one record value and key separately
// second to keep visited node count (default set false to all)
var newArr = [];
var newArrVisited = [];
for (let i = 0; i < arrLength; i++) {
newArr[i]= [];
newArr[i].value = arr[i];
newArr[i].key = i;
newArrVisited[i] = false;
}
// sort new array by value
newArr.sort(function (a, b) {
return a.value - b.value;
})
var swp = 0;
for (let i = 0; i < arrLength; i++) {
// check if already visited or swapped
if (newArr[i].key == i || newArrVisited[i]) {
continue;
}
var cycle = 0;
var j = i;
while (!newArrVisited[j]) {
// mark as visited
newArrVisited[j] = true;
j = newArr[j].key; //assign next key
cycle++;
}
if (cycle > 0) {
swp += (cycle > 1) ? cycle - 1 : cycle;
}
}
return swp;
}
reference

//You are given an unordered array consisting of consecutive integers [1, 2, 3, ..., n] without any duplicates.
//still not the best
function minimumSwaps(arr) {
let count = 0;
for(let i =0; i< arr.length; i++){
if(arr[i]!=i+1){
let temp = arr[i];
arr[arr.indexOf(i+1)] =temp;
arr[i] = i+1;
count =count+1;
}
}
return count;
}

I assume there are two reasons you're wanting to measure how many iterations a sort takes. So I will supply you with some theory (if the mathematics is too dense, don't worry about it), then some practical application.
There are many sort algorithms, some of them have a predicable number of iterations based on the number of items you are sorting, some of them are luck of the draw simply based on the order of the items to be sorted and which item how you select what is called a pivot. So if optimisation is very important to you, then you'll want to select the right algorithm for the purpose of the sort algorithm. Otherwise go for a general purpose algorithm.
Here are most popular sorting algorithms for the purpose of learning, and each of them have least, worst and average running-cases. Heapsort, Radix and binary-sort are worth looking at if this is more than just an theoretical/learning exercise.
Quicksort
Worst Case: Θ(n 2)
Best case: Θ(n lg n)
Average case: Θ(n lg n)
Here is a Quicksort implementation by Charles Stover
Merge sort
Worst case: Θ(n lg n)
Best case: Θ(n lg n)
Average Case: Θ(n lg n)
(note they're all the same)
Here is a merge sort implementation by Alex Kondov
Insertion sort
Worst case: Θ(n2)
Best case: Θ(n)
Average case:Θ(n2)
(Note that its worst and average case are the same, but its best case is the best of any algorithm)
Here is an insertion sort implementation by Kyle Jensen
Selection sort
Worst case: Θ(n2)
Best case: Θ(n2)
Average case: Θ(n2)
(note they're all the same, like a merge sort).
Here is a selection sort algorithm written by #dbdavid updated by myself for ES6
You can quite easily add an iterator variable to any of these examples to count the number of swaps they make, and play around with them to see which algorithms work best in which circumstance.
If there's a very good chance the items will already be well sorted, insertion sort is your best choice. If you have absolutely no idea, of the four basic sorting algorithms quicksort is your best choice.

function minimumSwaps(arr) {
var counter = 0;
for (var i = arr.length; i > 0; i--) {
var minval = Math.min(...arr); console.log("before", arr);
var minIndex = arr.indexOf(minval);
if (minval != = arr[0]) {
var temp = arr[0];
arr[0] = arr[minIndex];
arr[minIndex] = temp; console.log("after", arr);
arr.splice(0, 1);
counter++;
}
else {
arr.splice(0, 1); console.log("in else case")
}
} return counter;
}
This is how I call my swap function:
minimumSwaps([3, 7, 6, 9, 1, 8, 4, 10, 2, 5]);
It works with Selection Sort. Logic is as follows:
Loop through the array length
Find the minimum element in the array and then swap with the First element in the array, if the 0th Index doesn't have the minimum value founded out.
Now remove the first element.
If step 2 is not present, remove the first element(which is the minimum value present already)
increase counter when we swap the values.
Return the counter value after the for Loop.
It works for all values.
However, it fails due to a timeout for values around 50,000.

The solution to this problem is not very intuitive unless you are already somewhat familiar with computer science or real math wiz, but it all comes down to the number of inversions and the resulting cycles
If you are new to computer science I recommend the following resources to supplement this solution:
GeeksforGeeks Article
Informal Proof Explanation
Graph Theory Explanation
If we define an inversion as:
arr[i]>arr[j]
where "i" is the current index and "j" is the following index --
if there are no inversions the array is already in order and requires no sorting.
For Example:
[1,2,3,4,5]
So the number of swaps is related to the number of inversions, but not directly because each inversion can lead to a series of swaps (as opposed to a singular swap EX: [3,1,2]).
So if one consider's the following array:
[4,5,2,1,3,6,10,9,7,8]
This array is composed of three cycles.
Cycle One- 4,1,3 (Two Swaps)
Cycle Two- 5,2 (One Swap)
Cycle Three- 6 (0 Swaps)
Cycle Four- 10,9,7,8 (3 Swaps)
Now here's where the CS and Math magic really kicks in: each cycle will only require one pass through to properly sort it, and this is always going to be true.
So another way to say this would be-- the minimum number of swaps to sort any cycle is the number of element in that cycle minus one, or more explicitly:
minimum swaps = (cycle length - 1)
So if we sum the minimum swaps from each cycle, that sum will equal the minimum number of swaps for the original array.
Here is my attempt to explain WHY this algorithm works:
If we consider that any sequential set of numbers is just a section of a number line, then any set starting at zero should be equal to its own index should the set be expressed as a Javascript array. This idea becomes the criteria to programmatically determined if in element is already in the correct position based on its own value.
If the current value is not equal to its own index then the program should detect a cycle start and recording its length. Once the while loop reaches the the original value in the cycle it will add the minimum number of swaps in the cycle to a counter variable.
Anyway here is my code-- it is very verbose but should work:
export const minimumSwaps = (arr) => {
//This function returns the lowest value
//from the provided array.
//If one subtracts this value the from
//any value in the array it should equal
//that value's index.
const shift = (function findLowest(arr){
let lowest=arr[0];
arr.forEach((val,i)=>{
if(val<lowest){
lowest=val;
}
})
return lowest;
})(arr);
//Declare a counter variable
//to keep track of the swaps.
let swaps = 0;
//This function returns an array equal
//in size to the original array provided.
//However, this array is composed of
//boolean values with a value of false.
const visited = (function boolArray(n){
const arr=[];
for(let i = 0; i<n;i++){
arr.push(false);
}
return arr;
})(arr.length);
//Iterate through each element of the
//of the provided array.
arr.forEach((val, i) => {
//If the current value being assessed minus
//the lowest value in the original array
//is not equal to the current loop index,
//or, if the corresponding index in
//the visited array is equal to true,
//then the value is already sorted
if (val - shift === i || visited[i]) return;
//Declare a counter variable to record
//cycle length.
let cycleLength = 0;
//Declare a variable for to use for the
//while loop below, one should start with
//the current loop index
let x = i;
//While the corresponding value in the
//corresponding index in the visited array
//is equal to false, then we
while (!visited[x]) {
//Set the value of the current
//corresponding index to true
visited[x] = true;
//Reset the x iteration variable to
//the next potential value in the cycle
x = arr[x] - shift;
//Add one to the cycle length variable
cycleLength++;
};
//Add the minimum number of swaps to
//the swaps counter variable, which
//is equal to the cycle length minus one
swaps += cycleLength - 1;
});
return swaps
}

This solution is simple and fast.
function minimumSwaps(arr) {
let minSwaps = 0;
for (let i = 0; i < arr.length; i++) {
// at this position what is the right number to be here
// for example at position 0 should be 1
// add 1 to i if array starts with 1 (1->n)
const right = i+1;
// is current position does not have the right number
if (arr[i] !== right) {
// find the index of the right number in the array
// only look from the current position up passing i to indexOf
const rightIdx = arr.indexOf(right, i);
// replace the other position with this position value
arr[rightIdx] = arr[i];
// replace this position with the right number
arr[i] = right;
// increment the swap count since a swap was done
++minSwaps;
}
}
return minSwaps;
}

Here is my solution, but it timeouts 3 test cases with very large inputs. With smaller inputs, it works and does not terminate due to timeout.
function minimumSwaps(arr) {
let swaps = 0;
for (let i = 0; i < arr.length; i++) {
if (arr[i] === i + 1) continue;
arr.splice(i, 1, arr.splice(arr.indexOf(i + 1), 1, arr[i])[0]); //swap
swaps++;
}
return swaps;
}
I'm learning how to make it more performant, any help is welcome.

This is my solution to the Main Swaps 2 problem in JavaScript. It passed all the test cases. I hope someone finds it useful.
//this function calls the mainSwaps function..
function minimumSwaps(arr){
let swaps = 0;
for (var i = 0; i < arr.length; i++){
var current = arr[i];
var targetIndex = i + 1;
if (current != targetIndex){
swaps += mainSwaps(arr, i);
}
}
return swaps;
}
//this function is called by the minimumSwaps function
function mainSwaps(arr, index){
let swapCount = 0;
let currentElement = arr[index];
let targetIndex = currentElement - 1;
let targetElement = arr[currentElement - 1];
while (currentElement != targetElement){
//swap the elements
arr[index] = targetElement;
arr[currentElement - 1] = currentElement;
//increase the swapcount
swapCount++;
//store the currentElement, targetElement with their new values..
currentElement = arr[index];
targetElement = arr[currentElement - 1];
}
return swapCount;
}
var myarray = [2,3,4,1,5];
var result = console.log(minimumSwaps(myarray));

you can also do it with a map. But its O(nlogn)
const minSwaps = (arr) =>{
let arrSorted = [...arr].sort((a,b)=>a-b);
let indexMap = new Map();
// fill the indexes
for(let i=0; i<arr.length; i++){
indexMap.set(arr[i],i);
}
let count = 0;
for(let i=0; i<arrSorted.length;i++){
if(arr[i] != arrSorted[i]){
count++;
// swap the index
let newIdx = indexMap.get(arrSorted[i]);
indexMap.set(arr[i],newIdx);
indexMap.set(arrSorted[i],i);
// sawp the values
[arr[i],arr[newIdx]] =[arr[newIdx],arr[i]];
}
}
return count;
}

Trim string fetched from a Google Spreadsheet range and return the largest value

I am building a small script and, on it, I need to "read" the largest value in a range of cells. The cells look like this: "a-10", "a-3", "a-4" and so on, but their real values are 10, 3 and 4 respectively. I cannot remove the "a-" from the cells.
What I have so far is this:
Code:
function getLargestValue() {
var ss=SpreadsheetApp.getActiveSpreadsheet();
var sheets=['Sheet1','Sheet2','Sheet3'];
var values=[];
for(var i=0;i<sheets.length;i++)
{
var sht=ss.getSheetByName(sheets[i]);
var rng=sht.getDataRange();
var rngA=rng.getValues();
for(var j=1;j<rngA.length;j++)
{
rngA[j][0] = substring(2, rngA[j][0]);
values.push(rngA[j][0]);
}
}
values.sort(function(a, b) {return a - b;});
var max = values[values.length-1];
SpreadsheetApp.getUi().alert('The max is ' + max); //Just checking to be sure it works
return max;
// must show the largest value on "Max" sheet
}
Link to spreadsheet:
https://docs.google.com/spreadsheets/d/16WrsztYlOV7qDOMBqGYU8az26Bw9NoqdQ4Imk_hZM90/edit?usp=sharing
I still need to trim the strings and insert the largest value in a different sheet. How could I do that?
Thank you.

Before your sort function I would try something like
// Use a regex to get any numbers from the cell value and make
// a new array
var numbers = values.map(function(el) { return /[0-9]+/.exec(el) }
and then at the end before returning
// Put the 'max' value in 'MAX' sheet in cell 'A1'
ss.getSheetByName('MAX').getRange("A1").setValue(max)
Side note: You can also get a maximum number like this:
var max = Math.max(1,2,3,4,5)
OR
var max = Math.max.apply(null,ARRAY_OF_NUMBERS)

function getLargestValue()
{
var ss=SpreadsheetApp.getActiveSpreadsheet();
var sheets=['Sheet1','Sheet2','Sheet3'];
var values=[];
for(var i=0;i<sheets.length;i++)
{
var sht=ss.getSheetByName(sheets[i]);
var rng=sht.getDataRange();
var rngA=rng.getValues();//this will give you a 2D array
for (var items in rngA) //foreach loop, easier than for loop
{
var result = rngA[items][0].substring(2);//split string at 2nd character
if(!isNaN(parseFloat(result)) && isFinite(result))// check whether it is a number (since cell has strings like "Values")
{
values.push(result);//adds all numbers to the array
}
}
}
values.sort(function(a, b) {return a - b;});
var max = values[values.length-1];
//SpreadsheetApp.getUi().alert('The max is ' + max); //Just checking to be sure it works
//return max;
// must show the largest value on "Max" sheet
ss.getSheetByName('MAX').getRange("A2").setValue(max)
}

Total of different currencies from a json file

Currently working on an application which requires to display a set of values in different currencies. This is part of an application but everything I provide here should be enough as this is the main section I am working with. I have a json file which is read in and it is stored into an array called valuesArray, this array has all the information, such as the amount, currency, etc. With the currencies being sorted with highest first to the lowest on display like this:
EUR 500.00
USD 200.00
This is the code that I have created but it seems like this wouldn't be effective the more currencies I have. I've just put an array declaration above the function but just above this function is where I do all the json stuff and adding it into the array. $scope.valuesArray has data at this point.
$scope.valuesArray =[];
$scope.total = function()
{
var eur_total = 0;
var usd_total = 0;
if (typeof $scope.valuesArray != 'undefined')
{
var length = $scope.valuesArray.length;
for (var i = 0; i<length ; i++)
{
switch($scope.valuesArray[i].currency)
{
case "USD":
usd_total += parseFloat($scope.valuesArray[i].value);
break;
default:
eur_total += parseFloat($scope.valuesArray[i].value);
break;
}
}
}
var cost_total= [usd_total,eur_total];
total.sort(function(a, b){return b-a});
return format_to_decimal(total[0]) + "\x0A" + format_to_decimal(total[1]);
}
In my for loop I go through every single data in the array and break each currency down within the switch statement and finding the total amount of each currencies.
The last bit is kind of temporary as I couldn't figure out a different way of how to do it. I sort the totals for the currencies I have from the highest at the top.
I return the function with a function call for format_numeric_with_commas which gives me the value in proper currency format and this displays the value. Will update this and add that code when I get to it. But I have used the indexes as a rough logic to show what I want to get out of it. So in this case, total[0] should be 500.00 and total[1] should be 200.00.
On top of this I want to be able to display the currency type for each. So like the example above.

You can try to save all the calculations in the array with currency index.
$scope.valuesArray = [];
$scope.total = function () {
var totalsArray = [];
if (typeof $scope.valuesArray != 'undefined') {
var length = $scope.valuesArray.length
for (var i = 0; i < length; i++) {
if (!totalsArray[$scope.valuesArray[i].currency]) {
totalsArray[$scope.valuesArray[i].currency] = 0;
}
totalsArray[$scope.valuesArray[i].currency] += $scope.valuesArray[i].value;
}
}
var cost_total = [];
for (var k in totalsArray) {
cost_total.push(currency:k,value:totalsArray[k]);
}
cost_total.sort(function (a, b) {
return b.value - a.value
});
return format_to_decimal(cost_total[0].value)+cost_total[0].currency + "\x0A" + format_to_decimal(cost_total[1].value);

Accessing indexes of a string inside an array and taking the sum

Ok so I am trying to access each individual number in the strings inside of this array.
var array = ['818-625-9945','999-992-1313','888-222-2222','999-123-1245'];
var str = "";
for (i=0; i<array.length; i++) {
str = array[i];
}
The problem is that this is the output: '999-992-1313'
and not the first element array[0]: '818-625-9945'
When I try doing a nested for loop to go through each element inside the string I am having trouble stating those elements.
var array = ['818-625-9945','999-992-1313','888-222-2222','999-123-1245'];
for (i=0; i<array.length; i++) {
for (j=0; j<array[i].length; j++) {
console.log(array[i][j]);
}
}
I do not know how to access each individual number inside of the string array[i]. I would like to find a way to make a counter such that if I encounter the number '8' I add 8 to the total score, so I can take the sum of each individual string element and see which number has the highest sum.
var array = ['818-625-9945','999-992-1313','888-222-2222','999-123-1245'];
for (i=0; i<array.length; i++) {
for (j=0; j<array[i].length; j++) {
if (array[i](j).indexOf('8') !== -1) {
// add one to total score
// then find a way to increase the index to the next index (might need help here also please)
}
}
}

Mabe this works for you. It utilized Array.prototype.reduce(), Array.prototype.map() and String.prototype.split().
This proposal literates through the given array and splits every string and then filter the gotten array with a check for '8'. The returned array is taken as count and added to the return value from the former iteration of reduce - and returned.
var array = ['818-625-9945', '999-992-1313', '888-222-2222', '999-123-1245'],
score = array.reduce(function (r, a) {
return r + a.split('').filter(function (b) { return b === '8'; }).length;
}, 0);
document.write('Score: ' + score);
A suggested approach with counting all '8' on every string:
var array = ['818-625-9945', '999-992-1313', '888-222-2222', '999-123-1245'],
score = array.map(function (a) {
return a.split('').filter(function (b) { return b === '8'; }).length;
});
document.write('Score: ' + score);

Actually rereading your question gave me a better idea of what you want. You simply want to count and retrieve the number of 8's per string and which index in your array conforms with this maximum 8 value. This function retrieves the index where the value was found in the array, how many times 8 was found and what is the string value for this result. (or returns an empty object in case you give in an empty array)
This you could easily do with:
'use strict';
var array = ['818-625-9945', '999-992-1313', '888-222-2222', '999-123-1245'];
function getHighestEightCountFromArray(arr) {
var max = 0,
result = {};
if (arr && arr.forEach) {
arr.forEach(function(value, idx) {
var cnt = value.split('8').length;
if (max < cnt) {
// found more nr 8 in this section (nl: cnt - 1)
max = cnt;
// store the value that gave this max
result = {
count: cnt - 1,
value: value,
index: idx
};
}
});
}
return result;
}
console.log(getHighestEightCountFromArray(array));
The only thing here is that when an equal amount of counts is found, it will still use the first one found, here you could decide which "maximum"
should be preferred(first one in the array, or the newest / latest one in the array)
OLD
I'm not sure which sums you are missing, but you could do it in the following way.
There I first loop over all the items in the array, then I use the String.prototype.split function to split the single array items into an array which would then contain ['818', '625', '9945']. Then for each value you can repeat the same style, nl: Split the value you are receiving and then loop over all single values. Those then get convert to a number by using Number.parseInt an then all the values are counted together.
There are definitelly shorter ways, but this is a way how you could do it
'use strict';
var array = ['818-625-9945','999-992-1313','888-222-2222','999-123-1245'],
sumPerIndex = [],
totalSum = 0;
array.forEach(function(item, idx) {
var values = item.split('-'), subArray = [], itemSum = 0;
values.forEach(function(value) {
var singleItems = value.split(''),
charSum = 0;
singleItems.forEach(function(char) {
charSum += parseInt(char);
});
itemSum += charSum;
subArray.push(charSum);
console.log('Sum for chars of ' + value + ' = ' + charSum);
});
sumPerIndex.push(subArray);
totalSum += itemSum;
console.log('Sum for single values of ' + item + ' = ' + itemSum);
});
console.log('Total sum of all elements: ' + totalSum);
console.log('All invidual sums', sumPerIndex);

Calculating standard deviation not executing loop

I've just started learning coding on code academy and I'm really new to this.
I'm trying to make this program ask the user for values which it adds to an array from which it calculates the sample standard deviation.
// This array stores the values needed
var figures;
getStandardDeviation = function() {
// I need at least two figures for a standard deviation
figures[0] = prompt("Enter a number:");
figures[1] = prompt("Enter a number:");
// Checks whether user wishes to add more values to the array
var confirm = prompt("Would you like to add another? (Y or N)").toUpperCase();
// I can't figure out why the following if statement is not executed
// It checks whether the user wishes to add more values and adds them to the array
// If not it breaks the for loop
if (confirm === "Y"){
for ( i = 0; i === 100; i++){
figures[i + 2] = prompt("Enter a number:");
confirm = prompt("Would you like to add another figure? (Y or N)").toUpperCase();
if (confirm === "N"){
break;
}
}
}
// The rest of the code works fine from here onwards
var sumx = 0;
var n = figures.length;
for(var i = 0 ; i < n ; i++) {
sumx += figures[i];
}
console.log("Sum = " + sumx);
var sumXsq = 0;
for( i = 0 ; i < n ; i++) {
sumXsq += (figures[i] * figures[i]);
}
console.log("Sum x squared = " + sumXsq);
var sxx = (sumXsq - (sumx * sumx)/n);
console.log("Sxx = " + sxx);
var v = sxx/(n - 1);
console.log("Variance = " + v);
var standardDev = Math.sqrt(v);
console.log("Standard Deviation = " + standardDev);
};
getStandardDeviation();
The program is supposed to ask me if I want to add more values to the array, then when I confirm, it gives me a prompt to add more values.
Currently, when I execute the program I input the numbers 56 and 67. The code then asks me if I wish to add more values, I then confirm this. Instead of letting me add more values it ignores this and calculates the standard deviation with the first two values (56 and 67).
The output is:
Sum = 05667
Sum x squared = 7625
Sxx = -16049819.5
Variance = -16049819.5
Standard Deviation = NaN

for ( i = 0; i === 100; i++){[...]} means
Set i to 0
If it's not true that i === 100 (that is: if i is not 100), end the loop
Do whatever I put inside the {} braces, once
Do i++
Back to 2
As the initial value for i is 0 and not 100, the code inside the loop is never executed. If you want it to go from 0 to 99, it should be for ( i = 0; i < 100; i++).
You don't actually need a for loop, though. A while loop would be better. A loop like while (true){[...]} would run until it hit a break statement. As you wouldn't have the i in that case, you could use figures.push(parseFloat(prompt("Enter a number:"))) instead (you should use parseFloat, as per what Vincent Hogendoorn said) . push adds a new value at the end of an array, so it's exactly what you need. Something like:
if (confirm === "Y"){
while (true){
figures.push(parseFloat(prompt("Enter a number:")));
confirm = prompt("Would you like to add another figure? (Y or N)").toUpperCase();
if (confirm === "N"){
break;
}
}
}
You could also change it so it doesn't ask if you want to stop if you don't have at least two values. That way you would be able to leave out that first part:
figures[0] = prompt("Enter a number:");
figures[1] = prompt("Enter a number:");

indeed your figures variable isn't defined as an array, like #James Donnely says.
Keep in mind you also fill in strings, so if you want to add up values you have to convert them to values.
you can use something like parseFloat for this.
if you don't use it, you sum up strings. 3+4 will be 34 instead of 7.

Your figures variable isn't defined as an array. Because of this figure[1] = prompt(...) never gets hit and a TypeError is thrown on var n = figures.length;.
Change:
var figures;
To:
var figures = [];
JSFiddle demo.
You can then replace the for loop you're using after if (confirm === "Y") with a recursive function:
// Push a user input number into the figures array
figures.push(prompt("Enter a number:"));
// Function to add a new number and ask if we want to add more
function addNewNumber() {
// Push a new user input number into the figures array
figures.push(prompt("Enter a number:"));
// Ask if the user wants to add another number
if (confirm("Do you want to add another number?"))
// If they do, call this function again
addNewNumber();
}
// Trigger the function for the first time
addNewNumber();
JSFiddle demo with recursion.

function StandardDeviation(numbersArr) {
//--CALCULATE AVAREGE--
var total = 0;
for(var key in numbersArr)
total += numbersArr[key];
var meanVal = total / numbersArr.length;
//--CALCULATE AVAREGE--
//--CALCULATE STANDARD DEVIATION--
var SDprep = 0;
for(var key in numbersArr)
SDprep += Math.pow((parseFloat(numbersArr[key]) - meanVal),2);
var SDresult = Math.sqrt(SDprep/numbersArr.length);
//--CALCULATE STANDARD DEVIATION--
alert(SDresult);
}
var numbersArr = [10, 11, 12, 13, 14];
StandardDeviation(numbersArr);