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Number Formatting for large integer numbers in Javascript

I know there are in-build function in javascript like toLocalString() to achieve number formatting. But this question is purely for learning and logic understanding.
I have a function in javascript that formats given number in Indian Number formatting standards (eg: 1,234 | 12,21,123 | etc)
Code
function formatter(input) {
var inputStr = input.toString(), l = inputStr.length;
var c = 1, f = 0;
console.log(l);
for (var x=l-1; x>=0; x--) {
if (x === 0) {
continue;
}
if (c === 3 && f === 0) {
inputStr = inputStr.substring(0, x) + ',' + inputStr.substring(x);
f = 1;
c = 0;
} else if (c % 2 === 0 && f === 1) {
inputStr = inputStr.substring(0, x) + ',' + inputStr.substring(x);
c = 0;
}
c++;
}
return inputStr;
}
Now this works for most part (as far as I have test, do point out bugs if you spot any). But my question is how do I handle large number in this, i.e. how do I handle values greater than 9007199254740991.

Hope this now fixes the issue:
function formatter(inputStr) {
inputStr+="";
let c = 1, f = 0;
for (let x=inputStr.length-1; x>=0; x--) {
if (x === 0) continue;
if (c === 3 && f === 0) {
inputStr = inputStr.substring(0, x) + ',' + inputStr.substring(x);
f = 1;
c = 0;
} else if (c % 2 === 0 && f === 1) {
inputStr = inputStr.substring(0, x) + ',' + inputStr.substring(x);
c = 0;
}
c++;
}
return inputStr;
}
//========= Tests =======
console.log(formatter(9007199254740991));
console.log(formatter("900719925474093454549341"));
console.log(formatter("123456678890987665443221112345667676766545434243"));

why not operator in javascript excute the first part of if statement?

i am trying to check if(a==b) then if(c==d) and after that if(a and b values) != (b and c values).
var a = 3;
var b = 3;
var c = 4;
var d = 4;
if ((a == b) != (c == d)){
document.write('Yes');
}else{
document.write('No');
}
Why not operator in javascript excute the first part of if statement?

Operators that we are using in if statements are returning boolean values ( true or false ). So your if statement is checking:
if(a==b) and returns true,
if(c==d) and retuns true,
if(true != true) returns false and it goes to else.
you have to use this like that:
var a = 3;
var b = 3;
var c = 4;
var d = 4;
if(a == b && c==d && a != c)
{
document.write('Yes');
}
else
{
document.write('No');
}
so first it will check if a==b then d==c if it's okey so checking if a != c is enough to get answer that you want.

Everything is fine , answer is NO ..
I update answer !
var a = 3;
var b = 3;
var c = 4;
var d = 4;
if ((a == b) != (c == d)){
document.write('Yes');
}else{
document.write('No');
}
/*
a == b ==> TRUE
c == d ==> TRUE
(TRUE != TRUE) ==> FALSE
Answer NO is correct
*/
I suggest this code for you :
var a = 3;
var b = 3;
var c = 5;
var d = 5;
if ( a == b == c == d ){
document.write('Yes');
}else{
document.write('NO');
}

Compare 3 values - Show the highest

Doing a survey where a user picks :
A B or C
I then need to know if the user has picked Mostly A's, B's or C's.
I'm trying a few jQuery logics' but not having much luck, Due to expression expected error.
Is there a neater / better way to show purely which variable is the highest?
My current jQuery :
var count = 0;
var count_a = 0;
var count_b = 0;
var count_c = 0;
$('.radio-select').click(function()
{
var chosen_option = $(this).val();
if(chosen_option == 'a')
{
count++;
count_a ++;
}
if(chosen_option == 'b')
{
count++;
count_b ++;
}
if(chosen_option == 'c')
{
count++;
count_c ++;
}
check_numbers(count, count_a, count_b, count_c);
})
function check_numbers(count, a, b, c)
{
parseInt(a);
parseInt(b);
parseInt(c);
if(count == '8')
{
if ((a > b ) && (a > c))
{
alert("A is Highest");
}
if ((b > a ) && (b > c))
{
alert("B is Highest");
}
if(c > b) && (c > a))
{
alert("C is highest!");
}
}
}
jsFiddle Example

If you wanted a smaller way of doing it you could use inline if statements. Up to you if this is a better way, I like it though.
a = 5
b = 11
c = 6
console.log((a > b && a > c? a : (b > c ? b : c)))

Firstly you do not need to use parseInt() on a, b, c as they are already integers. And again count is an integer while you are comparing it to a string. This should work.
if(count == 8)
{
if ((a > b ) && (a > c))
{
alert("A is Highest");
}
else if ((b > a ) && (b > c))
{
alert("B is Highest");
}
else
{
alert("C is highest!");
}

You need to fetch the value returned by parseInt. Use it like: a = parseInt(a); and same for the other variables before comparing them in the if...else.

function check_numbers(count, a, b, c)
{
var x = parseInt(a),
y = parseInt(b),
z = parseInt(c);
if(count == 8)
{
var result = (x > y ? (x > z ? x : z) : (y > z ? y : z));
}
}

#StuBlackett you can consider adding the values and labels to an array then sorting Descending and returning the lable at the top.
function CompareIndexZero(a, b) {
if (a[0] < b[0]) return 1;
if (a[0] > b[0]) return -1;
return 0;
}
function myFunction() {
var count_a = 2;
var count_b = 5;
var count_c = 4;
var arrHighest = [];
arrHighest.push([count_a, "A"]);
arrHighest.push([count_b, "B"]);
arrHighest.push([count_c, "C"]);
arrHighest.sort(CompareIndexZero);
alert(arrHighest[0][1] + " is the highest");
}

Here is a modified version of check_numbers() that works as intended if I got you right. The point I want to make is the use of Math.max() to find the highest number from a selection of numbers.
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Math/max
function check_numbers(count, a, b, c) {
if(count === 8) {
var numArray = [a, b, c];
var highest = Math.max.apply(null, numArray);
console.log(highest);
if (highest === a) {
console.log('a is highest');
} else if (highest === b) {
console.log('b is highest');
} else if (highest === c) {
console.log('c is highest');
}
}
}
check_numbers(8, 1 , 2, 5);
check_numbers(8, 5, 2, 1);
check_numbers(8, 1 , 5, 2);

Have you also taken into account that multiple answers could share the highest count?
My 2 cents on that:
var count = count_a = count_b = count_c = 0;
$('.radio-select').on('click', function() {
var chosen_option = $(this).val();
if (chosen_option == 'a') {
count_a++;
}
else if (chosen_option == 'b') {
count_b++;
}
else if (chosen_option == 'c') {
count_c++;
}
if (++count == 8) {
check_numbers(count_a, count_b, count_c);
}
});
function check_numbers(a, b, c) {
var highest = ((a > b && a > c) ? a : (b > c)? b : c),
multiple = false,
alertText = '';
if (a == highest) {
alertText += 'A';
}
if (b == highest) {
if (alertText != '') {
multiple = true;
alertText += ' and ';
}
alertText += 'B';
}
if (c == highest) {
if (alertText != '') {
multiple = true;
alertText += ' and ';
}
alertText += 'C';
}
alert(alertText + ' ' + (multiple ? 'are' : 'is') + ' highest!');
}

Most effecient way to solve a formula in javascript

Say I have the following formula
a + (13 * (b/c)) + d + (12 * e) - f - 11 + (g * (h/i)) - 10 = 66;
Now the letters a-i can represent a number from 1-9. However, each number can only be used once. What would be the most efficient way to solve this? The way I have at the moment is to generate an array of random numbers e.g.
array[2,5,1,7,8,6,9,3,4]
I then fit this into the formula to check if it equals 66. If not, I generate a new array of numbers. I was thinking there should be a better way to do this though, maybe using recursion?
Just wanted to get some input in how you would tackle this?
Thanks

I think there's no smarter algorithm than just trying with each permutation of the numbers in the set for equality. Which I think will be easier for you, than creating everytime an unique array, that was not already generated.
In the code bellow I have used very primitive check for equality between numbers. There might be an easier way from the standard javascript library (e.g. if number is found in the predefined set...)
var solutions = [];
for (var a = 1; a <= 9; a++) {
for (var b = 1; b <= 9; b++) {
if ( b == a ) continue;
for (var c = 1; c <= 9; c++) {
if ( c == b || c == a ) continue;
for (var d = 1; d <= 9; d++) {
if ( d == c || d == b || d == a ) continue;
for (var e = 1; e <= 9; e++) {
if ( e == d || e == c || e == b || e == a ) continue;
for (var f = 1; f <= 9; f++) {
if ( f == e || f == d || f == c || f == b || f == a ) continue;
for (var g = 1; g <= 9; g++) {
if ( g == f || g == e || g == d || g == c || g == b || g == a ) continue;
for (var h = 1; h <= 9; h++) {
if ( h == g || h == f || h == e || h == d || h == c || h == b || h == a ) continue;
for (var i = 1; i <= 9; i++) {
if ( i == h || i == g || i == f || i == e || i == d || i == c || i == b || i == a ) continue;
if ((a + (13 * (b/c)) + d + (12 * e) - f - 11 + (g * (h/i)) - 10) == 66) {
solutions.push(a + "," + b + "," + c + "," + d + "," + e + "," + f + "," + g + "," + h + "," + i);
}
}
}
}
}
}
}
}
}
}
console.log(solutions);
Which results into a lot of solutions
["1,2,6,4,7,8,3,5,9", "1,2,6,4,7,8,5,3,9", "1,3,2,4,5,8,7,9,6", "1,3,2,4,5,8,9,7,6", "1,3,2,9,5,6,4,7,8", "1,3,2,9,5,6,7,4,8", "1,3,4,7,6,5,2,9,8", "1,3,4,7,6,5,9,2,8", "1,3,6,2,7,9,4,5,8", "1,3,6,2,7,9,5,4,8", "1,3,9,4,7,8,2,5,6", "1,3,9,4,7,8,5,2,6", "1,4,8,2,7,9,3,5,6", "1,4,8,2,7,9,5,3,6", "1,5,2,3,4,8,7,9,6", "1,5,2,3,4,8,9,7,6", "1,5,2,8,4,7,3,9,6", "1,5,2,8,4,7,9,3,6", "1,5,3,9,4,2,7,8,6", "1,5,3,9,4,2,8,7,6", "1,9,6,4,5,8,3,7,2", "1,9,6,4,5,8,7,3,2", "1,9,6,7,5,2,3,4,8", "1,9,6,7,5,2,4,3,8", "2,1,4,3,7,9,5,6,8", "2,1,4,3,7,9,6,5,8", "2,3,6,1,7,9,4,5,8", "2,3,6,1,7,9,5,4,8", "2,4,8,1,7,9,3,5,6", "2,4,8,1,7,9,5,3,6", "2,8,6,9,4,1,5,7,3", "2,8,6,9,4,1,7,5,3", "2,9,6,3,5,1,4,7,8", "2,9,6,3,5,1,7,4,8", "3,1,4,2,7,9,5,6,8", "3,1,4,2,7,9,6,5,8", "3,2,1,5,4,7,8,9,6", "3,2,1,5,4,7,9,8,6", "3,2,4,8,5,1,7,9,6", "3,2,4,8,5,1,9,7,6", "3,2,8,6,5,1,7,9,4", "3,2,8,6,5,1,9,7,4", "3,5,2,1,4,8,7,9,6", "3,5,2,1,4,8,9,7,6", "3,6,4,9,5,8,1,7,2", "3,6,4,9,5,8,7,1,2", "3,9,2,8,1,5,6,7,4", "3,9,2,8,1,5,7,6,4", "3,9,6,2,5,1,4,7,8", "3,9,6,2,5,1,7,4,8", "4,2,6,1,7,8,3,5,9", "4,2,6,1,7,8,5,3,9", "4,3,2,1,5,8,7,9,6", "4,3,2,1,5,8,9,7,6", "4,3,9,1,7,8,2,5,6", "4,3,9,1,7,8,5,2,6", "4,9,6,1,5,8,3,7,2", "4,9,6,1,5,8,7,3,2", "5,1,2,9,6,7,3,4,8", "5,1,2,9,6,7,4,3,8", "5,2,1,3,4,7,8,9,6", "5,2,1,3,4,7,9,8,6", "5,3,1,7,2,6,8,9,4", "5,3,1,7,2,6,9,8,4", "5,4,1,9,2,7,3,8,6", "5,4,1,9,2,7,8,3,6", "5,4,8,9,6,7,1,3,2", "5,4,8,9,6,7,3,1,2", "5,7,2,8,3,9,1,6,4", "5,7,2,8,3,9,6,1,4", "5,9,3,6,2,1,7,8,4", "5,9,3,6,2,1,8,7,4", "6,2,8,3,5,1,7,9,4", "6,2,8,3,5,1,9,7,4", "6,3,1,9,2,5,7,8,4", "6,3,1,9,2,5,8,7,4", "6,9,3,5,2,1,7,8,4", "6,9,3,5,2,1,8,7,4", "7,1,4,9,6,5,2,3,8", "7,1,4,9,6,5,3,2,8", "7,2,8,9,6,5,1,3,4", "7,2,8,9,6,5,3,1,4", "7,3,1,5,2,6,8,9,4", "7,3,1,5,2,6,9,8,4", "7,3,2,8,5,9,1,6,4", "7,3,2,8,5,9,6,1,4", "7,3,4,1,6,5,2,9,8", "7,3,4,1,6,5,9,2,8", "7,5,2,8,4,9,1,3,6", "7,5,2,8,4,9,3,1,6", "7,6,4,8,5,9,1,3,2", "7,6,4,8,5,9,3,1,2", "7,9,6,1,5,2,3,4,8", "7,9,6,1,5,2,4,3,8", "8,2,4,3,5,1,7,9,6", "8,2,4,3,5,1,9,7,6", "8,3,2,7,5,9,1,6,4", "8,3,2,7,5,9,6,1,4", "8,5,2,1,4,7,3,9,6", "8,5,2,1,4,7,9,3,6", "8,5,2,7,4,9,1,3,6", "8,5,2,7,4,9,3,1,6", "8,6,4,7,5,9,1,3,2", "8,6,4,7,5,9,3,1,2", "8,7,2,5,3,9,1,6,4", "8,7,2,5,3,9,6,1,4", "8,9,2,3,1,5,6,7,4", "8,9,2,3,1,5,7,6,4", "9,1,2,5,6,7,3,4,8", "9,1,2,5,6,7,4,3,8", "9,1,4,7,6,5,2,3,8", "9,1,4,7,6,5,3,2,8", "9,2,8,7,6,5,1,3,4", "9,2,8,7,6,5,3,1,4", "9,3,1,6,2,5,7,8,4", "9,3,1,6,2,5,8,7,4", "9,3,2,1,5,6,4,7,8", "9,3,2,1,5,6,7,4,8", "9,4,1,5,2,7,3,8,6", "9,4,1,5,2,7,8,3,6", "9,4,8,5,6,7,1,3,2", "9,4,8,5,6,7,3,1,2", "9,5,3,1,4,2,7,8,6", "9,5,3,1,4,2,8,7,6", "9,6,4,3,5,8,1,7,2", "9,6,4,3,5,8,7,1,2", "9,8,6,2,4,1,5,7,3", "9,8,6,2,4,1,7,5,3"]
If the formula is only this, and will not change, you might try to think if there are ranges that are not included in the result set per definition. For example, if you find that some of the numbers cannot be 9, you should not iterate to 9. E.g. the multiplications in the formula might exceed a number that even added to the minimal values in the other variables to exceed 66.

Brute force method:
function findSolution() {
for (var a = 1; a <= 9; a++) {
for (var b = 1; b <= 9; b++) {
for (var c = 1; c <= 9; c++) {
for (var d = 1; d <= 9; d++) {
for (var e = 1; e <= 9; e++) {
for (var f = 1; f <= 9; f++) {
for (var g = 1; g <= 9; g++) {
for (var h = 1; h <= 9; h++) {
for (var i = 1; i <= 9; i++) {
if (
[b,c,d,e,f,g,h,i].indexOf(a) >= 0 ||
[a,c,d,e,f,g,h,i].indexOf(b) >= 0 ||
[a,b,d,e,f,g,h,i].indexOf(c) >= 0 ||
[a,b,c,e,f,g,h,i].indexOf(d) >= 0 ||
[a,b,c,d,f,g,h,i].indexOf(e) >= 0 ||
[a,b,c,d,e,g,h,i].indexOf(f) >= 0 ||
[a,b,c,d,e,f,h,i].indexOf(g) >= 0 ||
[a,b,c,d,e,f,g,i].indexOf(h) >= 0 ||
[a,b,c,d,e,f,g,h].indexOf(i) >= 0
) {
continue;
}
if (a + (13 * (b/c)) + d + (12 * e) - f - 11 + (g * (h/i)) - 10 === 66) {
return [a,b,c,d,e,f,g,h,i];
}
}
}
}
}
}
}
}
}
}
}
document.getElementById('result').innerText = findSolution().join(', ')
<span id="result"></span>

encodeURIComponent algorithm source code

I am developing an application in titanium using Javascript. I need an open source implementation of encodeURIComponent in Javascript.
Can anybody guide me or show me some implementation?

The specification for this function is in 15.1.3.4.
Modern versions (2018) of V8 implement it in C++. See src/uri.h:
// ES6 section 18.2.6.5 encodeURIComponenet (uriComponent)
static MaybeHandle<String> EncodeUriComponent(Isolate* isolate,
Handle<String> component) {
which calls into Encode defined in uri.cc.
Older versions of V8 implemented it in JavaScript and distributed under the BSD license. See line 359 of src/uri.js.
// ECMA-262 - 15.1.3.4
function URIEncodeComponent(component) {
var unescapePredicate = function(cc) {
if (isAlphaNumeric(cc)) return true;
// !
if (cc == 33) return true;
// '()*
if (39 <= cc && cc <= 42) return true;
// -.
if (45 <= cc && cc <= 46) return true;
// _
if (cc == 95) return true;
// ~
if (cc == 126) return true;
return false;
};
var string = ToString(component);
return Encode(string, unescapePredicate);
}
It's not called encodeURIComponent there, but this code in the same file, esablishes the mapping:
InstallFunctions(global, DONT_ENUM, $Array(
"escape", URIEscape,
"unescape", URIUnescape,
"decodeURI", URIDecode,
"decodeURIComponent", URIDecodeComponent,
"encodeURI", URIEncode,
"encodeURIComponent", URIEncodeComponent
));

Here is my implementation:
var encodeURIComponent = function( str ) {
var hexDigits = '0123456789ABCDEF';
var ret = '';
for( var i=0; i<str.length; i++ ) {
var c = str.charCodeAt(i);
if( (c >= 48/*0*/ && c <= 57/*9*/) ||
(c >= 97/*a*/ && c <= 122/*z*/) ||
(c >= 65/*A*/ && c <= 90/*Z*/) ||
c == 45/*-*/ || c == 95/*_*/ || c == 46/*.*/ || c == 33/*!*/ || c == 126/*~*/ ||
c == 42/***/ || c == 92/*\\*/ || c == 40/*(*/ || c == 41/*)*/ ) {
ret += str[i];
}
else {
ret += '%';
ret += hexDigits[ (c & 0xF0) >> 4 ];
ret += hexDigits[ (c & 0x0F) ];
}
}
return ret;
};

What for do you need encodeuricomponent? It is already present in JS.
Anyway, here's an example of implementation:
http://phpjs.org/functions/rawurlencode:501#comment_93984