Develop Reference

Find all permutations of a set of numbers and operators

I was given a code challenge to do that was related to recursion and was unable to complete it. My experience with these types of questions is very slim and this one just stumped me. Could any of you help me out just for my own education, as I've already failed the challenge?
The description:
Given a string of numbers and operators, print out all the different ways you can add parentheses to force the order of operations to be explicit, and the result for running the operations in that order.
Assume:
No weird inputs, everything is separated by one space.
Supported operators are +, *, -, = (for the = operator, if the values are the same return 1, otherwise return 0)
Print your results sorted numerically
Don't worry about the input expression size being too large
Your code should be written in javascript
Don't use eval or external libraries
Example:
node test.js "2 - 1 - 1"
((2-1)-1) = 0
(2-(1-1)) = 2
node test.js "2 * 3 - 4 * 5";
(2*(3-(4*5))) = -34
((2*3)-(4*5)) = -14
((2*(3-4))*5) = -10
(2*((3-4)*5)) = -10
(((2*3)-4)*5) = 10
node test.js "2 + 2 = 2"
((2+2)=2) = 0
(2+(2=2)) = 3
This is where I'm at so far. I'm far from getting the right output, but I feel like the logic is starting to get there. I've adapted this code from a similar, but different question.
var args = process.argv.slice(2)[0].split(" "),
numberOfOperators = 0;
args.forEach(function(val, index, array) {
if (isNaN(val)) {
++numberOfOperators;
}
});
args = args.join("");
var recurse = function(openParenCount, closeParenCount, input, pointer) {
if (openParenCount === 0 && closeParenCount === 0) {
console.log(input + "\n");
}
if (openParenCount > 0) {
input = input.slice(0, pointer) + "(" + input.slice(pointer, input.length);
recurse(openParenCount - 1, closeParenCount + 1, input, pointer+1);
}
if (closeParenCount > 0) {
input = input.slice(0, pointer+openParenCount+3) + ")" + input.slice(pointer+openParenCount+3, input.length+1);
recurse(openParenCount, closeParenCount - 1, input, pointer+3);
}
}
recurse(numberOfOperators, 0, args, 0);

a little hint:
var AP = [];
var input = process.argv.slice(2)[0];
var args = input.replace(/\s+/g, "").split(/([^\d\.]+)/g);
recurse(args, []).forEach(function(v){ console.log(v); });
function recurse(arr, into){
if(arr.length === 1){
into.push(arr[0]);
}else{
for(var i=0, j=arr.length-2; i<j; i+=2){
recurse(
AP.concat(
arr.slice(0, i),
"(" + arr.slice(i, i+3).join(" ") + ")",
arr.slice(i+3)
),
into
)
}
}
return into
}
This Implementation still has a few "bugs", and these by intent; I'm not going to do your "homework".
If you have more than 2 operators in your Equasion, the result will contain duplicates, 2nd It is not sorted, and since it is just splitting and concatenating strings, it can't compute any result.
But it shows you a way how you can implement the recursion.

Javascript function. Don't understand how function inside self-function work

would like to ask about a JavaScript function.
I don't understand below function, I thought at line 4 fib(n-1) will return 1 and the latter fib(n-2) will return 0, and then they both add together as 1.
May I know why the final result for f(10); will be 55, can't get my head around this.
Anyone can help to explain to me what happening behind the scene, please?
Thanks! ;)
var f = function fib(n) {
if (n === 0) return 0;
if (n === 1) return 1;
if (n > 1) return fib(n - 1) + fib(n - 2); // *2
};
f(10); // 55
ref: https://slides.com/concise/js/fullscreen#/35

Like this. This is a typical recursive function, with two base cases and one recursive step.
Remember that if n is 10, then fib(n - 1) is fib(9), and so on:
fib(10) = fib(9) + fib(8) = 34 + 21 = 55
fib(9) = fib(8) + fib(7) = 21 + 13 = 34
fib(8) = fib(7) + fib(6) = 13 + 8 = 21
fib(7) = fib(6) + fib(5) = 8 + 5 = 13
fib(6) = fib(5) + fib(4) = 5 + 3 = 8
fib(5) = fib(4) + fib(3) = 3 + 2 = 5
fib(4) = fib(3) + fib(2) = 2 + 1 = 3
fib(3) = fib(2) + fib(1) = 1 + 1 = 2
fib(2) = fib(1) + fib(0) = 1 + 0 = 1
fib(1) = 1
fib(0) = 0
Side note: Although your example is a good illustration of recursive functions, it's an extremely inefficient way to calculate fibonacci numbers. A much better approach is to use memoization to eliminate the bulk of the inefficiency:
var fib = (function () {
var cache = [0, 1];
return function fib(num) {
if (!(num in cache)) {
cache[num] = fib(num - 1) + fib(num - 2);
}
return cache[num];
};
})();
console.log(fib(10));
http://jsperf.com/fibonacci-memoization-2015-04-01

this is called "recursive function" which one function calls itself.
basically there is no restriction (in terms of the OS) of doing so, a function basically gets interpreted/compiled to assembly code , and this assembly code can be copied and re-run with different (or the same) arguments.

Pad a number with leading zeros in JavaScript [duplicate]

This question already has answers here:
How can I pad a value with leading zeros?
(76 answers)
Closed 3 years ago.
In JavaScript, I need to have padding.
For example, if I have the number 9, it will be "0009". If I have a number of say 10, it will be "0010". Notice how it will always contain four digits.
One way to do this would be to subtract the number minus 4 to get the number of 0s I need to put.
Is there was a slicker way of doing this?

ES2017 Update
You can use the built-in String.prototype.padStart()
n = 9;
String(n).padStart(4, '0'); // '0009'
n = 10;
String(n).padStart(4, '0'); // '0010'
Not a lot of "slick" going on so far:
function pad(n, width, z) {
z = z || '0';
n = n + '';
return n.length >= width ? n : new Array(width - n.length + 1).join(z) + n;
}
When you initialize an array with a number, it creates an array with the length set to that value so that the array appears to contain that many undefined elements. Though some Array instance methods skip array elements without values, .join() doesn't, or at least not completely; it treats them as if their value is the empty string. Thus you get a copy of the zero character (or whatever "z" is) between each of the array elements; that's why there's a + 1 in there.
Example usage:
pad(10, 4); // 0010
pad(9, 4); // 0009
pad(123, 4); // 0123
pad(10, 4, '-'); // --10

function padToFour(number) {
if (number<=9999) { number = ("000"+number).slice(-4); }
return number;
}
Something like that?
Bonus incomprehensible-but-slicker single-line ES6 version:
let padToFour = number => number <= 9999 ? `000${number}`.slice(-4) : number;
ES6isms:
let is a block-scoped variable (as opposed to var’s functional scoping)
=> is an arrow function that, among other things, replaces function and is prepended by its parameters
If an arrow function takes a single parameter, you can omit the parentheses (hence number =>)
If an arrow function body has a single line that starts with return, you can omit the braces and the return keyword and simply use the expression
To get the function body down to a single line, I cheated and used a ternary expression

Try:
String.prototype.lpad = function(padString, length) {
var str = this;
while (str.length < length)
str = padString + str;
return str;
}
Now test:
var str = "5";
alert(str.lpad("0", 4)); //result "0005"
var str = "10"; // note this is string type
alert(str.lpad("0", 4)); //result "0010"
DEMO
In ECMAScript 2017 , we have new method padStart and padEnd which has below syntax.
"string".padStart(targetLength [,padString]):
So now we can use
const str = "5";
str.padStart(4, "0"); // "0005"

Funny, I recently had to do this.
function padDigits(number, digits) {
return Array(Math.max(digits - String(number).length + 1, 0)).join(0) + number;
}
Use like:
padDigits(9, 4); // "0009"
padDigits(10, 4); // "0010"
padDigits(15000, 4); // "15000"
Not beautiful, but effective.

You did say you had a number-
String.prototype.padZero= function(len, c){
var s= '', c= c || '0', len= (len || 2)-this.length;
while(s.length<len) s+= c;
return s+this;
}
Number.prototype.padZero= function(len, c){
return String(this).padZero(len,c);
}

You could do something like this:
function pad ( num, size ) {
return ( Math.pow( 10, size ) + ~~num ).toString().substring( 1 );
}
Edit: This was just a basic idea for a function, but to add support for larger numbers (as well as invalid input), this would probably be better:
function pad ( num, size ) {
if (num.toString().length >= size) return num;
return ( Math.pow( 10, size ) + Math.floor(num) ).toString().substring( 1 );
}
This does 2 things:
If the number is larger than the specified size, it will simply return the number.
Using Math.floor(num) in place of ~~num will support larger numbers.

This is not really 'slick' but it's faster to do integer operations than to do string concatenations for each padding 0.
function ZeroPadNumber ( nValue )
{
if ( nValue < 10 )
{
return ( '000' + nValue.toString () );
}
else if ( nValue < 100 )
{
return ( '00' + nValue.toString () );
}
else if ( nValue < 1000 )
{
return ( '0' + nValue.toString () );
}
else
{
return ( nValue );
}
}
This function is also hardcoded to your particular need (4 digit padding), so it's not generic.

For fun, instead of using a loop to create the extra zeros:
function zeroPad(n,length){
var s=n+"",needed=length-s.length;
if (needed>0) s=(Math.pow(10,needed)+"").slice(1)+s;
return s;
}

Since you mentioned it's always going to have a length of 4, I won't be doing any error checking to make this slick. ;)
function pad(input) {
var BASE = "0000";
return input ? BASE.substr(0, 4 - Math.ceil(input / 10)) + input : BASE;
}
Idea: Simply replace '0000' with number provided... Issue with that is, if input is 0, I need to hard-code it to return '0000'. LOL.
This should be slick enough.
JSFiddler: http://jsfiddle.net/Up5Cr/

broken toFixed implementation [duplicate]

This question already has answers here:
Javascript toFixed Not Rounding
(23 answers)
Closed 2 years ago.
The default implementation of javascript's "Number.toFixed" appears to be a bit broken.
console.log((8.555).toFixed(2)); // returns 8.56
console.log((8.565).toFixed(2)); // returns 8.57
console.log((8.575).toFixed(2)); // returns 8.57
console.log((8.585).toFixed(2)); // returns 8.59
I need a rounding method that is more consistent than that.
In the range between 8.500 and 8.660 the following numbers don't round up correctly.
8.575
8.635
8.645
8.655
I've tried to fix the prototype implementation as follows, but it's only half way there. Can anyone suggest any change that would make it work more consistently?
Number.prototype.toFixed = function(decimalPlaces) {
var factor = Math.pow(10, decimalPlaces || 0);
var v = (Math.round(this * factor) / factor).toString();
if (v.indexOf('.') >= 0) {
return v + factor.toString().substr(v.length - v.indexOf('.'));
}
return v + '.' + factor.toString().substr(1);
};

This is because of floating-point errors.
Compare (8.575).toFixed(20) with (8.575).toFixed(3) and imagine this proposition: 8.575 < real("8.575"), where real is an imaginary function that creates a real number with infinite precision.
That is, the original number is not as expected and the inaccuracy has already been introduced.
One quick "workabout" I can think of is: Multiply by 1000 (or as appropriate), get the toFixed(0) of that (still has a limit, but it's absurd), then shove back in the decimal form.
Happy coding.

Thanks for the answer pst. My implementation almost worked, but didn't in some cases because of floating point errors.
this line in my function is the culprit:
Math.round(this * factor)
(it's on the Number.prototype, so "this" is the number);
8.575 * 100 comes out to 857.4999999999999, which in turn rounds down.
this is corrected by changing the line to read as follows:
Math.round(Math.round(this * factor * 100) / 100)
My entire workaround is now changed to:
Number.prototype.toFixed = function(decimalPlaces) {
var factor = Math.pow(10, decimalPlaces || 0);
var v = (Math.round(Math.round(this * factor * 100) / 100) / factor).toString();
if (v.indexOf('.') >= 0) {
return v + factor.toString().substr(v.length - v.indexOf('.'));
}
return v + '.' + factor.toString().substr(1);
};

A consistent solution would be to add a fixed tolerance (epsilon) to each number before rounding. It should be small, but not too small.
For example, with an eps = 1e-9, this:
console.log((8.555).toFixed(2)); // returns 8.56
console.log((8.565).toFixed(2)); // returns 8.57
console.log((8.575).toFixed(2)); // returns 8.57
console.log((8.585).toFixed(2)); // returns 8.59
Becomes this:
console.log((8.555 + eps).toFixed(2)); // returns 8.56
console.log((8.565 + eps).toFixed(2)); // returns 8.57
console.log((8.575 + eps).toFixed(2)); // returns 8.58
console.log((8.585 + eps).toFixed(2)); // returns 8.59

Maybe it will help someone, this is fixed popular formatMoney() function, but with correct roundings.
Number.prototype.formatMoney = function() {
var n = this,
decPlaces = 2,
decSeparator = ",",
thouSeparator = " ",
sign = n < 0 ? "-" : "",
i = parseInt(n = Math.abs(+n || 0)) + "",
j = (j = i.length) > 3 ? j % 3 : 0,
decimals = Number(Math.round(n +'e'+ decPlaces) +'e-'+ decPlaces).toFixed(decPlaces),
result = sign + (j ? i.substr(0, j) + thouSeparator : "") + i.substr(j).replace(/(\d{3})(?=\d)/g, "$1" + thouSeparator) + (decPlaces ? decSeparator + Math.abs(decimals-i).toFixed(decPlaces).slice(2) : "");
return result;
};
(9.245).formatMoney(); // returns 9,25
(7.5).formatMoney(); // returns 7,50
(8.575).formatMoney(); // returns 8,58

Check my answer
function toFixed( num, precision ) {
return (+(Math.round(+(num + 'e' + precision)) + 'e' + -precision)).toFixed(precision);
}

Round number up to the nearest multiple of 3

How would I go about rounded a number up the nearest multiple of 3?
i.e.
25 would return 27
1 would return 3
0 would return 3
6 would return 6

if(n > 0)
return Math.ceil(n/3.0) * 3;
else if( n < 0)
return Math.floor(n/3.0) * 3;
else
return 3;

Simply:
3.0*Math.ceil(n/3.0)
?

Here you are!
Number.prototype.roundTo = function(num) {
var resto = this%num;
if (resto <= (num/2)) {
return this-resto;
} else {
return this+num-resto;
}
}
Examples:
y = 236.32;
x = y.roundTo(10);
// results in x = 240
y = 236.32;
x = y.roundTo(5);
// results in x = 235

I'm answering this in psuedocode since I program mainly in SystemVerilog and Vera (ASIC HDL). % represents a modulus function.
round_number_up_to_nearest_divisor = number + ((divisor - (number % divisor)) % divisor)
This works in any case.
The modulus of the number calculates the remainder, subtracting that from the divisor results in the number required to get to the next divisor multiple, then the "magic" occurs. You would think that it's good enough to have the single modulus function, but in the case where the number is an exact multiple of the divisor, it calculates an extra multiple. ie, 24 would return 27. The additional modulus protects against this by making the addition 0.

As mentioned in a comment to the accepted answer, you can just use this:
Math.ceil(x/3)*3
(Even though it does not return 3 when x is 0, because that was likely a mistake by the OP.)
Out of the nine answers posted before this one (that have not been deleted or that do not have such a low score that they are not visible to all users), only the ones by Dean Nicholson (excepting the issue with loss of significance) and beauburrier are correct. The accepted answer gives the wrong result for negative numbers and it adds an exception for 0 to account for what was likely a mistake by the OP. Two other answers round a number to the nearest multiple instead of always rounding up, one more gives the wrong result for negative numbers, and three more even give the wrong result for positive numbers.

This function will round up to the nearest multiple of whatever factor you provide.
It will not round up 0 or numbers which are already multiples.
round_up = function(x,factor){ return x - (x%factor) + (x%factor>0 && factor);}
round_up(25,3)
27
round up(1,3)
3
round_up(0,3)
0
round_up(6,3)
6
The behavior for 0 is not what you asked for, but seems more consistent and useful this way. If you did want to round up 0 though, the following function would do that:
round_up = function(x,factor){ return x - (x%factor) + ( (x%factor>0 || x==0) && factor);}
round_up(25,3)
27
round up(1,3)
3
round_up(0,3)
3
round_up(6,3)
6

Building on #Makram's approach, and incorporating #Adam's subsequent comments, I've modified the original Math.prototype example such that it accurately rounds negative numbers in both zero-centric and unbiased systems:
Number.prototype.mround = function(_mult, _zero) {
var bias = _zero || false;
var base = Math.abs(this);
var mult = Math.abs(_mult);
if (bias == true) {
base = Math.round(base / mult) * _mult;
base = (this<0)?-base:base ;
} else {
base = Math.round(this / _mult) * _mult;
}
return parseFloat(base.toFixed(_mult.precision()));
}
Number.prototype.precision = function() {
if (!isFinite(this)) return 0;
var a = this, e = 1, p = 0;
while (Math.round(a * e) / e !== a) { a *= 10; p++; }
return p;
}
Examples:
(-2).mround(3) returns -3;
(0).mround(3) returns 0;
(2).mround(3) returns 3;
(25.4).mround(3) returns 24;
(15.12).mround(.1) returns 15.1

(n - n mod 3)+3

$(document).ready(function() {
var modulus = 3;
for (i=0; i < 21; i++) {
$("#results").append("<li>" + roundUp(i, modulus) + "</li>")
}
});
function roundUp(number, modulus) {
var remainder = number % modulus;
if (remainder == 0) {
return number;
} else {
return number + modulus - remainder;
}
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
Round up to nearest multiple of 3:
<ul id="results">
</ul>

A more general answer that might help somebody with a more general problem: if you want to round numbers to multiples of a fraction, consider using a library. This is a valid use case in GUI where decimals are typed into input and for instance you want to coerce them to multiples of 0.25, 0.2, 0.5 etc. Then the naive approach won't get you far:
function roundToStep(value, step) {
return Math.round(value / step) * step;
}
console.log(roundToStep(1.005, 0.01)); // 1, and should be 1.01
After hours of trying to write up my own function and looking up npm packages, I decided that Decimal.js gets the job done right away. It even has a toNearest method that does exactly that, and you can choose whether to round up, down, or to closer value (default).
const Decimal = require("decimal.js")
function roundToStep (value, step) {
return new Decimal(value).toNearest(step).toNumber();
}
console.log(roundToStep(1.005, 0.01)); // 1.01
RunKit example

Using remainder operator (modulus):
(n - 1 - (n - 1) % 3) + 3

By the code given below use can change any numbers and you can find any multiple of any number
let numbers = [8,11,15];
let multiple = 3
let result = numbers.map(myFunction);
function myFunction(n){
let answer = Math.round(n/multiple) * multiple ;
if (answer <= 0)
return multiple
else
return answer
}
console.log("Closest Multiple of " + multiple + " is " + result);

if(x%3==0)
return x
else
return ((x/3|0)+1)*3