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Javascript Time Complexity Analysis

Hi there I have been researching and trying to learn how to check for the time complexity of certain algorithms. I've seen this video which was very helpful.
That being said I wondered off and started trying to work out the Worsts Case and an average case of certain algorithms.
1
I believe in the following snippet it is O(n) since to ind the value for sin we have to loop the entire array.
function mySin(x, iterNum) {
var mxx = -x*x;
var sin = 1;
var n = 0;
var term = 1;
for (var i = 1; i <= 2*iterNum; i++) {
n = n + 2;
term = term * mxx / ( n*(n+1) );
sin = sin + term
}
sin = x*sin;
console.log(sin + " = my function.");
console.log(Math.sin(x) + " math.sin");
}
Thanks again
2
function calculateFibonacciSum (num) {
if(cachedNumbers[num]) {
return cachedNumbers[num];
}
if(('number' === typeof num) && num <= 0) {
throw new Error ('Fibonnci series starts with 0. Please, enter any interget greater than or equal to 0');
}
else if(('number' === typeof num) && num === 0) {
return 0;
}
else if(('number' === typeof num) && (num === 1 || num === 2)) {
return 1;
}
else {
var value = calculateFibonacciSum(num-1) + calculateFibonacciSum(num-2);
cachedNumbers[num] = value;
return value;
}
}
While for this one I think it is also O(n) since in the first if/else statement the tc is O(1) since its contestant whilst the final else statement we must loop all the numbers and if the number is not calculated then call the function again (aka recurssion).
TIA

Both of these seem correct to me. Here's a bit more explanation:
1.
This is in fact O(n), as there are n iterations of the loop, the rest constant time; and n is proportional to iterNum
2.
This one is also linear time, but only since you cache the results of previous calculations. Otherwise it would be O(2n).
It is linear time since it essentially runs a loop down to the base cases (0 and 1). In fact, you could re-write this one using a loop instead of recursion.

How can I prepend characters to a string using loops?

I have an input field that expects a 10 digit number. If the user enters and submits a number less than 10 digits, the function would simply add a "0" until the inputed value is 10 digits in length.
I haven't really used, or understand how recursive functions really work, but I'm basically looking at an efficient way of doing this. One minor issue I'm having is figuring out how to prepend the "0"s at the beginning of the string rather than appended to the end.
My thinking:
function lengthCheck(sQuery) {
for (var i = 0; i < sQuery.length; i++) {
if (sQuery.length !== 10) {
sQuery += "0";
//I'd like to add the 0s to the beggining of the sQuery string.
console.log(sQuery);
lengthCheck(sQuery);
} else return sQuery
}
}

Change:
sQuery += "0"; // added at end of string
to:
sQuery = "0" + sQuery; // added at start of string
To remove the for loop/recursion, you could slice out the desired length in one step:
function padZeros(sQuery) {
// the max amount of zeros you want to lead with
const maxLengthZeros = "0000000000";
// takes the 10 rightmost characters and outputs them in a new string
return (maxLengthZeros + sQuery).slice(-10);
}
Simple generic function using ES6 repeat:
// edge case constraints not implemented for brevity
function padZeros(sQuery = "", maxPadding = 10, outputLength = 10) {
// the max amount of zeros you want to lead with
const maxLengthZeros = "0".repeat(maxPadding);
// returns the "outputLength" rightmost characters
return (maxLengthZeros + sQuery).slice(-outputLength);
}
console.log('padZeros: ' + padZeros("1234567890"));
console.log('padZeros: ' + padZeros("123"));
console.log('padZeros: ' + padZeros(""));
Alternate version that doesn't affect strings over your set limit:
function padZerosIfShort(inputString = "", paddedOutputLength = 10) {
let inputLen = inputString.length;
// only padded if under set length, otherwise returned untouched
return (paddedOutputLength > inputLen)
? "0".repeat(paddedOutputLength - inputLen) + inputString
: inputString;
}
console.log('padZerosIfShort: ' + padZerosIfShort("1234567890", 5));
console.log('padZerosIfShort: ' + padZerosIfShort("123", 5));
console.log('padZerosIfShort: ' + padZerosIfShort("", 5));
It will ultimately depend on your needs how you want to implement this behavior.

The += operator adds things to the end of strings similar to:
sQuery=sQuery+"0"
You can add characters to the front of a string like this
sQuery="0"+sQuery
I also found something interesting here. it works like this:
("00000" + sQuery).slice(-5)
You would add zeros to the front then slice off everything except the last 5. so to get 10 characters you would use:
("0000000000" + n).slice(-10)

You don't need recursion to solve this, just a simple for loop should do the trick. Try this:
function lengthCheck (sQuery) {
for (var i = sQuery.length; i<10; i++) {
sQuery = "0" + sQuery;
}
return sQuery;
}

You're looking to pad the string with zeroes. This is an example I've used before from here and will shorten your code a little bit:
function lengthCheck (sQuery) {
while (sQuery.length < 10)
sQuery = 0 + sQuery;
return sQuery;
}

I believe this has already been answered here (or similar enough to provide you the solution): How to output integers with leading zeros in JavaScript

How can I split a string into a given number of lines?

Here is my question:
Given a string, which is made up of space separated words, how can I split that into N strings of (roughly) even length, only breaking on spaces?
Here is what I've gathered from research:
I started by researching word-wrapping algorithms, because it seems to me that this is basically a word-wrapping problem. However, the majority of what I've found so far (and there is A LOT out there about word wrapping) assumes that the width of the line is a known input, and the number of lines is an output. I want the opposite.
I have found a (very) few questions, such as this that seem to be helpful. However, they are all focused on the problem as one of optimization - e.g. how can I split a sentence into a given number of lines, while minimizing the raggedness of the lines, or the wasted whitespace, or whatever, and do it in linear (or NlogN, or whatever) time. These questions seem mostly to be unanswered, as the optimization part of the problem is relatively "hard".
However, I don't care that much about optimization. As long as the lines are (in most cases) roughly even, I'm fine if the solution doesn't work in every single edge case, or can't be proven to be the least time complexity. I just need a real world solution that can take a string, and a number of lines (greater than 2), and give me back an array of strings that will usually look pretty even.
Here is what I've come up with:
I think I have a workable method for the case when N=3. I start by putting the first word on the first line, the last word on the last line, and then iteratively putting another word on the first and last lines, until my total width (measured by the length of the longest line) stops getting shorter. This usually works, but it gets tripped up if your longest words are in the middle of the line, and it doesn't seem very generalizable to more than 3 lines.
var getLongestHeaderLine = function(headerText) {
//Utility function definitions
var getLongest = function(arrayOfArrays) {
return arrayOfArrays.reduce(function(a, b) {
return a.length > b.length ? a : b;
});
};
var sumOfLengths = function(arrayOfArrays) {
return arrayOfArrays.reduce(function(a, b) {
return a + b.length + 1;
}, 0);
};
var getLongestLine = function(lines) {
return lines.reduce(function(a, b) {
return sumOfLengths(a) > sumOfLengths(b) ? a : b;
});
};
var getHeaderLength = function(lines) {
return sumOfLengths(getLongestLine(lines));
}
//first, deal with the degenerate cases
if (!headerText)
return headerText;
headerText = headerText.trim();
var headerWords = headerText.split(" ");
if (headerWords.length === 1)
return headerText;
if (headerWords.length === 2)
return getLongest(headerWords);
//If we have more than 2 words in the header,
//we need to split them into 3 lines
var firstLine = headerWords.splice(0, 1);
var lastLine = headerWords.splice(-1, 1);
var lines = [firstLine, headerWords, lastLine];
//The header length is the length of the longest
//line in the header. We will keep iterating
//until the header length stops getting shorter.
var headerLength = getHeaderLength(lines);
var lastHeaderLength = headerLength;
while (true) {
//Take the first word from the middle line,
//and add it to the first line
firstLine.push(headerWords.shift());
headerLength = getHeaderLength(lines);
if (headerLength > lastHeaderLength || headerWords.length === 0) {
//If we stopped getting shorter, undo
headerWords.unshift(firstLine.pop());
break;
}
//Take the last word from the middle line,
//and add it to the last line
lastHeaderLength = headerLength;
lastLine.unshift(headerWords.pop());
headerLength = getHeaderLength(lines);
if (headerLength > lastHeaderLength || headerWords.length === 0) {
//If we stopped getting shorter, undo
headerWords.push(lastLine.shift());
break;
}
lastHeaderLength = headerLength;
}
return getLongestLine(lines).join(" ");
};
debugger;
var header = "an apple a day keeps the doctor away";
var longestHeaderLine = getLongestHeaderLine(header);
debugger;
EDIT: I tagged javascript, because ultimately I would like a solution I can implement in that language. It's not super critical to the problem though, and I would take any solution that works.
EDIT#2: While performance is not what I'm most concerned about here, I do need to be able to perform whatever solution I come up with ~100-200 times, on strings that can be up to ~250 characters long. This would be done during a page load, so it needs to not take forever. For example, I've found that trying to offload this problem to the rendering engine by putting each string into a DIV and playing with the dimensions doesn't work, since it (seems to be) incredibly expensive to measure rendered elements.

Try this. For any reasonable N, it should do the job:
function format(srcString, lines) {
var target = "";
var arr = srcString.split(" ");
var c = 0;
var MAX = Math.ceil(srcString.length / lines);
for (var i = 0, len = arr.length; i < len; i++) {
var cur = arr[i];
if(c + cur.length > MAX) {
target += '\n' + cur;
c = cur.length;
}
else {
if(target.length > 0)
target += " ";
target += cur;
c += cur.length;
}
}
return target;
}
alert(format("this is a very very very very " +
"long and convoluted way of creating " +
"a very very very long string",7));

You may want to give this solution a try, using canvas. It will need optimization and is only a quick shot, but I think canvas might be a good idea as you can calculate real widths. You can also adjust the font to the really used one, and so on. Important to note: This won't be the most performant way of doing things. It will create a lot of canvases.
DEMO
var t = `However, I don't care that much about optimization. As long as the lines are (in most cases) roughly even, I'm fine if the solution doesn't work in every single edge case, or can't be proven to be the least time complexity. I just need a real world solution that can take a string, and a number of lines (greater than 2), and give me back an array of strings that will usually look pretty even.`;
function getTextTotalWidth(text) {
var canvas = document.createElement("canvas");
var ctx = canvas.getContext("2d");
ctx.font = "12px Arial";
ctx.fillText(text,0,12);
return ctx.measureText(text).width;
}
function getLineWidth(lines, totalWidth) {
return totalWidth / lines ;
}
function getAverageLetterSize(text) {
var t = text.replace(/\s/g, "").split("");
var sum = t.map(function(d) {
return getTextTotalWidth(d);
}).reduce(function(a, b) { return a + b; });
return sum / t.length;
}
function getLines(text, numberOfLines) {
var lineWidth = getLineWidth(numberOfLines, getTextTotalWidth(text));
var letterWidth = getAverageLetterSize(text);
var t = text.split("");
return createLines(t, letterWidth, lineWidth);
}
function createLines(t, letterWidth, lineWidth) {
var i = 0;
var res = t.map(function(d) {
if (i < lineWidth || d != " ") {
i+=letterWidth;
return d;
}
i = 0;
return "<br />";
})
return res.join("");
}
var div = document.createElement("div");
div.innerHTML = getLines(t, 7);
document.body.appendChild(div);

I'm sorry this is C#. I had created my project already when you updated your post with the Javascript tag.
Since you said all you care about is roughly the same line length... I came up with this. Sorry for the simplistic approach.
private void DoIt() {
List<string> listofwords = txtbx_Input.Text.Split(' ').ToList();
int totalcharcount = 0;
int neededLineCount = int.Parse(txtbx_LineCount.Text);
foreach (string word in listofwords)
{
totalcharcount = totalcharcount + word.Count(char.IsLetter);
}
int averagecharcountneededperline = totalcharcount / neededLineCount;
List<string> output = new List<string>();
int positionsneeded = 0;
while (output.Count < neededLineCount)
{
string tempstr = string.Empty;
while (positionsneeded < listofwords.Count)
{
tempstr += " " + listofwords[positionsneeded];
if ((positionsneeded != listofwords.Count - 1) && (tempstr.Count(char.IsLetter) + listofwords[positionsneeded + 1].Count(char.IsLetter) > averagecharcountneededperline))//if (this is not the last word) and (we are going to bust the average)
{
if (output.Count + 1 == neededLineCount)//if we are writting the last line
{
//who cares about exceeding.
}
else
{
//we're going to exceed the allowed average, gotta force this loop to stop
positionsneeded++;//dont forget!
break;
}
}
positionsneeded++;//increment the needed position by one
}
output.Add(tempstr);//store the string in our list of string to output
}
//display the line on the screen
foreach (string lineoftext in output)
{
txtbx_Output.AppendText(lineoftext + Environment.NewLine);
}
}

(Adapted from here, How to partition an array of integers in a way that minimizes the maximum of the sum of each partition?)
If we consider the word lengths as a list of numbers, we can binary search the partition.
Our max length ranges from 0 to sum (word-length list) + (num words - 1), meaning the spaces. mid = (range / 2). We check if mid can be achieved by partitioning into N sets in O(m) time: traverse the list, adding (word_length + 1) to the current part while the current sum is less than or equal to mid. When the sum passes mid, start a new part. If the result includes N or less parts, mid is achievable.
If mid can be achieved, try a lower range; otherwise, a higher range. The time complexity is O(m log num_chars). (You'll also have to consider how deleting a space per part, meaning where the line break would go, features into the calculation.)
JavaScript code (adapted from http://articles.leetcode.com/the-painters-partition-problem-part-ii):
function getK(arr,maxLength) {
var total = 0,
k = 1;
for (var i=0; i<arr.length; i++) {
total += arr[i] + 1;
if (total > maxLength) {
total = arr[i];
k++;
}
}
return k;
}
function partition(arr,n) {
var lo = Math.max(...arr),
hi = arr.reduce((a,b) => a + b);
while (lo < hi) {
var mid = lo + ((hi - lo) >> 1);
var k = getK(arr,mid);
if (k <= n){
hi = mid;
} else{
lo = mid + 1;
}
}
return lo;
}
var s = "this is a very very very very "
+ "long and convoluted way of creating "
+ "a very very very long string",
n = 7;
var words = s.split(/\s+/),
maxLength = partition(words.map(x => x.length),7);
console.log('max sentence length: ' + maxLength);
console.log(words.length + ' words');
console.log(n + ' lines')
console.log('')
var i = 0;
for (var j=0; j<n; j++){
var str = '';
while (true){
if (!words[i] || str.length + words[i].length > maxLength){
break
}
str += words[i++] + ' ';
}
console.log(str);
}

Using the Java String Split() Method to split a string we will discover How and Where to Apply This String Manipulation Technique:
We'll examine the Java Split() method's explanation and discover how to apply it. The principles are explained simply and with enough programming examples, either as a separate explanation or in the comment part of the programs.
The Java String Split() method is used to divide or split the calling Java String into pieces and return the Array, as the name implies. The delimiters("", " ", ) or regular expressions that we have supplied separately for each component or item of an array.
Syntax
String[ ] split(String regExp)
First Case: It involves initializing a Java String variable with a variety of words separated by spaces, using the Java String Split() method, and evaluating the results. We can effectively print each word without the space using the Java Split() function.
Second Case: In this case, we initialize a Java String variable and attempt to split or deconstruct the main String variable to use the String Split() method utilizing a substring of the initialized String variable.
Third Case: In this case, we will attempt to split a String using its character by taking a String variable (a single word).
You can check out other approaches to this problem on YouTube and even coding websites on google such as Coding Ninjas

This old question was revived by a recent answer, and I think I have a simpler technique than the answers so far:
const evenSplit = (text = '', lines = 1) => {
if (lines < 2) {return [text]}
const baseIndex = Math .round (text .length / lines)
const before = text .slice (0, baseIndex) .lastIndexOf (' ')
const after = text .slice (baseIndex) .indexOf (' ') + baseIndex
const index = after - baseIndex < baseIndex - before ? after : before
return [
text .slice (0, index),
... evenSplit (text .slice (index + (before > -1 ? 1 : 0)), lines - 1)
]
}
const text = `However, I don't care that much about optimization. As long as the lines are (in most cases) roughly even, I'm fine if the solution doesn't work in every single edge case, or can't be proven to be the least time complexity. I just need a real world solution that can take a string, and a number of lines (greater than 2), and give me back an array of strings that will usually look pretty even.`
const display = (lines) => console .log (lines .join ('\n'))
display (evenSplit (text, 7))
display (evenSplit (text, 5))
display (evenSplit (text, 12))
display (evenSplit (`this should be three lines, but it has a loooooooooooooooooooooooooooooooong word`, 3))
.as-console-wrapper {max-height: 100% !important; top: 0}
It works by finding the first line then recurring on the remaining text with one fewer lines. The recursion bottoms out when we have a single line. To calculate the first line, we take an initial target index which is just an equal share of the string based on its length and the number of lines. We then check to find the closest space to that index, and split the string there.
It does no optimization, and could certainly be occasionally misled by long words, but mostly it just seems to work.

JavaScript - Convert 24 digit hexadecimal number to decimal, add 1 and then convert back?

For an ObjectId in MongoDB, I work with a 24 digit hexadecimal number. Because I need to keep track of a second collection, I need to add 1 to this hexadecimal number.
In my case, here's my value
var value = "55a98f19b27585d81922ba0b"
What I'm looking for is
var newValue = "55a98f19b25785d81922ba0c"
I tried to create a function for this
function hexPlusOne(hex) {
var num = (("0x" + hex) / 1) + 1;
return num.toString(16);
}
This works with smaller hex numbers
hexPlusOne("eeefab")
=> "eeefac"
but it fails miserably for my hash
hexPlusOne(value)
=> "55a98f19b275840000000000"
Is there a better way to solve this?

This version will return a string as long as the input string, so the overflow is ignored in case the input is something like "ffffffff".
function hexIncrement(str) {
var hex = str.match(/[0-9a-f]/gi);
var digit = hex.length;
var carry = 1;
while (digit-- && carry) {
var dec = parseInt(hex[digit], 16) + carry;
carry = Math.floor(dec / 16);
dec %= 16;
hex[digit] = dec.toString(16);
}
return(hex.join(""));
}
document.write(hexIncrement("55a98f19b27585d81922ba0b") + "<BR>");
document.write(hexIncrement("ffffffffffffffffffffffff"));
This version may return a string which is 1 character longer than the input string, because input like "ffffffff" carries over to become "100000000".
function hexIncrement(str) {
var hex = str.match(/[0-9a-f]/gi);
var digit = hex.length;
var carry = 1;
while (digit-- && carry) {
var dec = parseInt(hex[digit], 16) + carry;
carry = Math.floor(dec / 16);
dec %= 16;
hex[digit] = dec.toString(16);
}
if (carry) hex.unshift("1");
return(hex.join(""));
}
document.write(hexIncrement("55a98f19b27585d81922ba0b") + "<BR>");
document.write(hexIncrement("ffffffffffffffffffffffff"));
I was curious te see whether user2864740's suggestion of working with 12-digit chunks would offer any advantage. To my surprise, even though the code looks more complicated, it's actually around twice as fast. But the first version runs 500,000 times per second too, so it's not like you're going to notice in the real world.
function hexIncrement(str) {
var result = "";
var carry = 1;
while (str.length && carry) {
var hex = str.slice(-12);
if (/^f*$/i.test(hex)) {
result = hex.replace(/f/gi, "0") + result;
carry = 1;
} else {
result = ("00000000000" + (parseInt(hex, 16) + carry).toString(16)).slice(-hex.length) + result;
carry = 0;
}
str = str.slice(0,-12);
}
return(str.toLowerCase() + (carry ? "1" : "") + result);
}
document.write(hexIncrement("55a98f19b27585d81922ba0b") + "<BR>");
document.write(hexIncrement("000000000000ffffffffffff") + "<BR>");
document.write(hexIncrement("0123456789abcdef000000000000ffffffffffff"));

The error comes from attempting to covert the entire 24-digit hex value to a number first because it won't fit in the range of integers JavaScript can represent distinctly2. In doing such a conversion to a JavaScript number some accuracy is lost.
However, it can be processed as multiple (eg. two) parts: do the math on the right part and then the left part, if needed due to overflow1. (It could also be processed one digit at a time with the entire addition done manually.)
Each chunk can be 12 hex digits in size, which makes it an easy split-in-half.
1 That is, if the final num for the right part is larger than 0xffffffffffff, simply carry over (adding) one to the left part. If there is no overflow then the left part remains untouched.
2 See What is JavaScript's highest integer value that a Number can go to without losing precision?
The range is 2^53, but the incoming value is 16^24 ~ (2^4)^24 ~ 2^(4*24) ~ 2^96; still a valid number, but outside the range of integers that can be distinctly represented.
Also, use parseInt(str, 16) instead of using "0x" + str in a numeric context to force the conversion, as it makes the intent arguably more clear.

Rounding to Significant Figures - Missing Zeros

I'm currently producing a JavaScript driven mathematics package, which focuses on rounding to various significant figures (S.F.) but I've run into a problem that I'm struggling to solve.
More on this problem later, but first some background for you.
The program is designed to select a completely random number within a given range and then automatically work out that number's relevant significant figures; for example:
Random Number: 0.097027 S.Fs: 9, 7, 0, 2, 7
Here is a screenshot of what I have produced to give you a visual representation:
As you can see, once the user has selected their number, they are then given the opportunity to click on four separate 'SF' buttons to view their random number presented to 1, 2, 3 and 4 S.Fs respectively.
For each S.F (1-4) the random number is rounded down, rounded up and rounded off to X SF and a scale below gives the user a more visual presentation to show why the SF value has been chosen by the program.
I've already written the vast majority of the code for this and tested it and so far the numbers are coming out how I'm expecting them to. Well nearly...
In the example I've given (0.097027); as you can see on the image I've included, the data for 4 S.F is absolutely correct and outputted accurately.
When I click on to the 3 SF button, I'd expect to see the following:
Random Number: 0.097027 3 S.F Rounded Up/Down/Off: 0.0970
However, what I'm actually getting is:
Random Number: 0.097027 3 S.F Rounded Up/Down/Off: 0.097
The program hasn't displayed the additional zero. This is a perfect example of a number in my program ending in a zero and in this case the zero is really significant and must be displayed.
The data is usually correct but there appears to be an issue with outputting significant zeros at the right time. I've researched the toFixed(x) method and if I assign toFixed(4) I get the correct required output, but because my numbers are generated randomly each time, they can range from a length of 5 figures, e.g. 89.404 up to > 10, e.g. `0.000020615.
So it looks like the toFixed method needs to be flexible/dynamic, e.g. toFixed(n) with a function run beforehand to determine exactly how many trailing zeros are needed?
Here are some key excerpts from my current solution for your consideration:
function generateNum() {
do {
genNumber = Math.random() * Math.pow (10, randomRange(-5, 5));
//Round
genNumber = roundToNSF(genNumber, 5, 0);
// This number must contain >1 digit which is 1 to 9 inclusive otherwise we may have e.g. 100. Rounding 100
}
while (!countNonZero(genNumber) || genNumber < 1E-05 || genNumber == 0);
//Round
genNumber = roundToNSF(genNumber, 5, 0);
genNumber = String(genNumber);
genNumber = Number(genNumber);
}
//----------------------------------------------------------------------------
function randomRange(min, max) {
/**
* Returns a random integer between min (inclusive) and max (inclusive)
* Using Math.round() will give you a non-uniform distribution!
*/
return Math.floor(Math.random() * (max - min + 1)) + min;
}
//---------------------------------------------------------------------------
//Click SF3 Button to reveal the data
function showSF3() {
//Remove any CSS properties on the buttons from previous use
removeButtonCSS();
document.getElementById('SFRounded').style.display = "block";
document.getElementById('scale').style.display = "block";
document.getElementById("SF3").className = document.getElementById("SF3").className + "buttonClick"; // this removes the blue border class
//Clear text
deleteRounded();
deleteScale();
//Run calculation
calculateAnswer();
//alert(genNumber.toFixed(4));
for (i = 3; i < 4; i++)
{
//Add The new data
sfRoundedTextBlock = document.getElementById('SFRounded');
//alert(downArray[i].toFixed(4));
//Data output to HTML.
sfRoundedTextBlock.innerHTML = sfRoundedTextBlock.innerHTML + '<p><strong>Number: </strong></br>' + String(genNumber) +
'</br>' + '<strong>Rounded down to ' + i + ' SF:</br></strong>' + downArray[i] + '</br>' +
'<strong>Rounded up to ' + i + ' SF:</br></strong>' + upArray[i] + '</br><strong>Rounded off to ' + i + ' SF:</br></strong>'
+ roundedArray[i] + '</br>' + '(See the scale below for why we choose <strong>' + roundedArray[i] + '</strong> as the rounded off value.)</p>';
}
}
//----------------------------------------------------------------------
var roundedArray = [];
var upArray = [];
var downArray = [];
var temp;
function calculateAnswer() {
//Clear Arrays
roundedArray = [];
upArray = [];
downArray = [];
// Work out the answer:
for (i = 0; i < 4; i++) {
var nSF = i + 1;
// Round OFF ...
temp = roundToNSF(genNumber, nSF, 0);
// We actually have to do this twice ...
roundedArray[nSF] = roundToNSF(temp, nSF, 0);
// Round UP ...
upArray[nSF] = roundToNSF(genNumber, nSF, 1);
// Round DOWN ...
downArray[nSF] = roundToNSF(genNumber, nSF, -1);
// e.g. x = 0.0098 rounded to 1SF is 0.010 initially (take the log of 0.0098 and try it!).
};
};
//-------------------------------------------------------------------------
//Globals
var aNumber;
var digits;
var way;
function roundToNSF(aNumber, digits, way){
// Round a number to n significant figures (can use roundToNDP provided we know how many decimal places):
if (way == undefined) { way = 0; }; // default is round off
if (aNumber !=0) {
if (aNumber > 0)
{
z = log10(aNumber);
}
else
{
z = log10(-aNumber);
};
z = Math.floor(z);
var nDP = digits - z - 1; // Rounding to nDP decimal places is equivalent to rounding to digits significant figures ...
var roundedNumber = roundToNDP(aNumber, nDP, way);
}
else {
roundedNumber = aNumber; // Number is zero ...
};
return Number(roundedNumber);
};
//---------------------------------------------------------------------------------
Update:
I'm still continuing to try and find a solution for this problem and an approach I have recently taken is to convert my randomly generated number into a searchable string variable and then use the indexOf(".") command to find the position of the decimal point (dp).
Then I've searched through my number, starting from the position of the dp to find the first instance of a significant, non-zero number [1-9].
var genNumber = 0.097027;
var rString = String(genNumber);
var positionofDP = rString.indexOf(".");
var regexp = /[1-9]/;
var positionofNonZero = Number(rString.search(regexp, positionofDP)); // Output would be '5'
I have then been able to target my search further, to determine whether my first significant number has any 'problematic' zeros in the immediate digits after it.
If there are any, then I set a Boolean variable to 'true' and then in a separate function create further text strings of my rounded off/down/up numbers, so I can then physically choose to add a '0' on to the end of the existing numerical characters.
This approach does work for me in isolated cases, but with my random number length ranging from 5-12 digits long, it still isn't dealing with all scenarios.
Maybe I need to create a dynamic toFixed(i) function? Any ideas would be greatly welcomed.

Instead of playing with the fixed points on an Int, you could manage the string directly.
Here's a link to a little fiddle: http://jsfiddle.net/5rw5G/4/
This not intended to completely/accurately solve your problem, but might help you see another solution.
function getRoundedSFs(num, SFCount) {
// Match every "leading zeros" before and after the .
var matches = num.toString().match(/^-?(0+)\.(0*)/);
// starting with "0."
if (matches) {
var firstIndex = matches[0].length;
var prefix = matches[0];
sf = Number(num.toString().substring(firstIndex, firstIndex + SFCount + 1));
sf = Math.round(sf / 10);
sf = prefix + sf.toString();
return Number(sf).toFixed(matches[2].length+SFCount);
}
// starting with something else like -5.574487436097115
else {
matches = num.toString().match(/^(-?(\d+))\.(\d+)/);
var decimalShift = SFCount - matches[2].length;
var rounded = Math.round(num * Math.pow(10, decimalShift));
rounded /= Math.pow(10, decimalShift);
return rounded.toFixed(decimalShift);
}
}

I've gone away again and I think I have now finally managed solve my initial problem.
There was a degree of confusion on my part surrounding when to use toFixed and toPrecision. I had previously attempted to convert my rounded up, down and off numbers into strings and then subsequently search through each of these to find the decimal point (".") and then work out the amount of trailing numbers, in order to then generate the correct toFixed point.
However, this was very hit and miss, given that my random number could be up to 12 digits, so what I've now done is to properly utilise toPrecision instead. For each 'SF button' (1-4) I have used the corresponding toPrecision point, e.g for SF1:
sfRoundedTextBlock.innerHTML = sfRoundedTextBlock.innerHTML + '<p><strong>Number: </strong></br>' + String(genNumber) +
'</br>' + '<strong>Rounded down to ' + i + ' SF:</br></strong>' + downArray[i].toPrecision(1) + '</br>' +
'<strong>Rounded up to ' + i + ' SF:</br></strong>' + upArray[i].toPrecision(1) + '</br><strong>Rounded off to ' + i + ' SF:</br></strong>'
+ roundedArray[i].toPrecision(1) + '</br>' + '(See the scale below for why we choose <strong>' + roundedArray[i].toPrecision(1) + '</strong> as the rounded off value.)</p>';
//Add The new scale data (Rounded Down)
downTextBlock = document.getElementById('down');
document.getElementById("down").innerHTML = String(downArray[i].toPrecision(1));
//Add The new scale data (Rounded Up)
upTextBlock = document.getElementById('up');
document.getElementById("up").innerHTML = String(upArray[i].toPrecision(1));
This was now giving me accurate results on every occasion, but there was still one hurdle left to jump. Occasionally I would reach a random scenario where scientific notation would have to be included in my outputted answer, e.g. 21819 rounded down to 1 SF, would read out at 2e+4 instead of 20000.
To combat this I setup my up, down and rounded figures into searchable strings, and then looked through these to find any illegal/scientific characters [a-z]. If I found any, I executed a slightly different version of my output which made use of parseFloat, which stripped out the scientific notation and displayed the correct figures:
//Convert Up, Down and Rounded into Strings based on their precision
var upString = String(upArray[i].toPrecision(1));
var downString = String(downArray[i].toPrecision(1));
var roundedString = String(roundedArray[i].toPrecision(1));
//Set up a regexp to search for characters [a-z], i.e. non-numeric
var regexp = /[a-z]/g;
//Search the up, down and rounded strings for non-numeric characters
var upResult = upString.match(regexp);
var downResult = downString.match(regexp);
var roundedResult = roundedString.match(regexp);
//If any of these strings contain a letter (non-numeric) we need to add in parseFloat to strip away the scientific notation included.
var containsChar = false;
if (upResult != null || downResult != null || roundedResult != null)
{
containsChar = true;
//alert("There is SN included here");
}
//Add The new data
sfRoundedTextBlock = document.getElementById('SFRounded');
if (containsChar == true)
{
sfRoundedTextBlock.innerHTML = sfRoundedTextBlock.innerHTML + '<p><strong>Number: </strong></br>' + String(genNumber) +
'</br>' + '<strong>Rounded down to ' + i + ' SF:</br></strong>' + parseFloat(downArray[i].toPrecision(1)) + '</br>' +
'<strong>Rounded up to ' + i + ' SF:</br></strong>' + parseFloat(upArray[i].toPrecision(1)) + '</br><strong>Rounded off to ' + i + ' SF:</br></strong>'
+ parseFloat(roundedArray[i].toPrecision(1)) + '</br>' + '(See the scale below for why we choose <strong>' + parseFloat(roundedArray[i].toPrecision(1)) + '</strong> as the rounded off value.)</p>';
//Add The new scale data (Rounded Down)
downTextBlock = document.getElementById('down');
document.getElementById("down").innerHTML = String(parseFloat(downArray[i].toPrecision(1)));
//Add The new scale data (Rounded Up)
upTextBlock = document.getElementById('up');
document.getElementById("up").innerHTML = String(parseFloat(upArray[i].toPrecision(1)));
}
Having tested this extensively it seems to be working as hoped.