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Here's an example of what I'm trying to do...
Let's say I have the number 73,284.
The number of thousands is 73 (73,284 divided by 1,000).
The number of hundreds is 2 (284 divided by 100).
The number of tens is 8 (84 divided by 10).
The number of singles is 4 (4 is left).
What I need is a Javascript function that would take the number 73,284 and create 4 numbers from it using the criteria above.
So if the number was 73,284, I'd pass that number into the function as a parameter and the function would return an array that looks like this, [73,2,8,4].
I tried to use the Math.round() function. It seemed to work for the thousands, but not necessarily for the hundreds, tens, and singles.
Simple for loop.
const f = (n) => {
let div = 1000;
const result = [];
for (let i = 0; i < 4 && n; i++, n %= div, div /= 10) {
result.push(Math.floor(n / div));
}
return result;
}
console.log(f(73284));
You can do this by keeping track of how many thousands, hundreds and tens you have and removing these as you go. A little simple arithmetic.
function get(num){
const thousands = Math.floor(num/1000)
const hundreds = Math.floor((num-thousands*1000)/100);
const tens = Math.floor((num-thousands*1000-hundreds*100)/10);
const units = Math.floor(num-thousands*1000-hundreds*100-tens*10)
return [
thousands,
hundreds,
tens,
units
]
}
const [th,hu,te,un] = get(73284)
console.log("Thousands=",th);
console.log("Hundreds=",hu);
console.log("Tens=",te);
console.log("Units=",un);
You will need a combinaison of division with Math.floor and modulo operator (%). The Modulo operator returns the reminder of a division.
Since you will be doing this operation multiple times, this a great opportunity to use recursion. Here is what i've come up with.
First, we need to get how many of a given devider we've got in our input. To do that, you can devide it and ignore the decimal using Math.floor.
const input = 73284
const divider = 1000
const amoutOfDividerInInput = input / divider .
// this returns 73.284, we don't want the .284 so we can use Math.floor.
const amountOfDividerInInputWithoutDecimal = Math.floor(amoutOfDividerInInput)
// we print the value
console.log("there are " + amountOfDividerInInputWithoutDecimal + " 1000s in " + input).
We, then, need to check what is the reminder of that operation, this is where the modulo operator comes in.
const input = 73284
const divider = 1000
const reminderOfTheDivision = input % divider;
// this gives us 284.
We can, then, use this reminder as our new input, and divide the divider by 10, to get how many hundreds are contains in it.
// we devide the divider by 10,
const newDevider = divider / 10;
// we calculate the new amount and reminder
const amoutOfNewDividerInInput = Math.floor(reminderOfTheDivision / newDivider );
const reminderOfTheNewDivision = reminderOfTheDivision % newDivider
// we print the new values
console.log("There are " + amoutOfNewDividerInInput + " 100s in" + input)
And this goes on until we are down to the single digit.
As you can see, there are alot of repetition in this code. A good developer is trying to avoid repetition at all cost. To prevent this, we could use a loop or a recursive function. A recursive function is basically a function that call itself. I've choose to use a function, since it's cleaner.
Here is what i've come up with.
function findDividingQuantity(reminder, divider) {
// if the devider is less than 10, we simply print the reminder.
if(divider < 10) {
console.log('and ' + reminder + ' remains.');
// and we returns to stop the recursion.
return;
}
// otherwise, we get the current amount of the divider in the reminder.
const amount = Math.floor(reminder / divider);
// we then calculate how much is left after the first division,
// this will become our new reminder.
const newReminder = reminder % divider;
// knowing how many of the divider there is in the input, we can print it.
console.log(amount + " " + divider + "s.");
// we call the function again, but with the new reminder
// and the divider divided by 10
findDividingQuantity(newReminder, divider / 10)
}
const input = 73284
console.log("In " + input + " there is: ");
// we call the function with the initial divider: 1000.
findDividingQuantity(input, 1000)
Here is a simple example:
function test(num) {
const arr = num.toLocaleString().split(',');
arr[arr.length-1].toString().split('').map(x=>arr.push(x));
arr[arr.length-4]=0;
return arr.map(x=>+x).filter(x=>x);
}
test(1234); // [1,2,3,4]
test(73284); // [73,2,8,4]
This also works for the millions, and so forth.
Here, we defined a function that will convert the number to localeString and split the number. Next, we loop through the array and convert every element to numbers.
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I am trying to do the below
Number("0.00000000000122") results in 1.22e-12
But what I need to just that number to get converted from String to Number.
console.log(Number("0.00000000000122"))
You can do it like this:
console.log(Number("0.00000000000122").toFixed(20).replace(/\.?0+$/,""));
Explanation:
toFixed() will keep the number, and replace() will remove trailing zeroes
Edit:
It seems that you want the result of Number("0.00000000000122") to BOTH be a number and also keep it string as "0.00000000000122", not its scientific display.
In that case, you can save it as a number-type variable in ts file, then display it as a string in HTML
If you want to work with precise numbers you have to work with integers.
A Number is not an integer type, it's a float so numbers are represented in memory as a float. The toString method of the number just prints out the value in memory. The integer bit and the dot location.
const scale = 18n;
const digits = 10n ** scale;
const POUND = (number) => BigInt(number) * digits;
const toPound = (number, fraction = false) => {
if (!fraction) {
return String(number / digits);
}
const str = String(number);
if (scale < 1n) {
return str;
}
return `${str.substring(0, Number(BigInt(str.length) - scale))}.${str.substr(-Number(scale))}`;
};
const a = POUND(11000);
const b = POUND(20000);
console.log('scale', String(scale));
console.log('a', toPound(a));
console.log('b', toPound(b));
console.log('mul', toPound((a * b) / digits));
console.log('add', toPound(a + b));
console.log('div', toPound(b / (a / digits), true));
console.log('sub', toPound(b - a));
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I am solving one problem .but getting correct output for small array but my solution fail when array size is large
solution
/**
* #param {number[]} digits
* #return {number[]}
*/
var plusOne = function(digits) {
let str = parseInt(digits.join(''))+1+''
return str.split('')
};
Question
The digits are stored such that the most significant digit is at the head of the list, and each element in the array contain a single digit.
You may assume the integer does not contain any leading zero, except the number 0 itself.
Example 1:
Input: [1,2,3]
Output: [1,2,4]
Explanation: The array represents the integer 123.
Example 2:
Input: [4,3,2,1]
Output: [4,3,2,2]
Explanation: The array represents the integer 4321.
above cases are passed
failed cases
Input
[6,1,4,5,3,9,0,1,9,5,1,8,6,7,0,5,5,4,3]
Output
[6,1,4,5,3,9,0,1,9,5,1,8,6,7,0,5,0,0,0]
Expected
[6,1,4,5,3,9,0,1,9,5,1,8,6,7,0,5,5,4,4]
let str = parseInt(digits.join(''))+1+''
With too many array elements, you are simply creating a number that is outside of the integer range here.
An implicit conversion to a float has to happen, and with that you get the inherent loss of precision - which then leads to several zeros in those places at the end, when you reverse the process.
Access the last array element specifically, and add 1 to the value. But if that last element had the value 9 already, you will of course have to repeat that same process for the previous one … basic application of “carry the one”.
var input = [6,1,4,5,3,9,0,1,9,5,1,8,6,7,0,5,5,4,9];
var output = [];
// loop over input array in reverse order
for(var i=input.length-1, toadd = 1; i>-1; --i) {
// add value `toadd` to current digit
var incremented = input[i] + toadd;
if(incremented < 10) {
output.unshift(incremented); // add to front of output array
toadd = 0; // reset toadd for all following digits, if we had no overflow
}
else {
output.unshift(0); // overflow occurred, so we add 0 to front of array instead
}
}
console.log(output)
This does take an “overflow” for 9 digits at the end into account; it does not handle the case when all digits are 9 though, if you need to handle that edge case as well, please implement that yourself.
This one recursively checks the last digits:
const test1 = [6,1,4,5,3,9,0,1,9,5,1,8,6,7,0,5,5,4,3]
const test2 = [6,1,4,5,3,9,0,1,9,5,1,8,6,7,0,5,5,4,9]
const plusOne = function(digits) {
if(digits[digits.length - 1] === 9){
digits = [...plusOne(digits.slice(0, -1)), 0]
}else{
digits[digits.length - 1]++
}
return digits;
};
console.log(plusOne(test1));
console.log(plusOne(test2));
You could do something like this :
var arr = [6,1,4,5,3,9,0,1,9,5,1,8,6,7,0,5,5,4,3,9,9]
function plusOne(index){
if( index < 0 ) return ; // Base case
if(arr[index] === 9){
arr[index] = 0;
plusOne(index-1); // For carry ahead
}else{
arr[index] += 1
}
}
plusOne(arr.length-1) // start with last digit
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I want to make an algorithm, for a NodeJS app, that converta any given string to a 1 to 3 digit number (better if the number is between 1-500).
e.g
ExampleString -> 214
Can anyone help me find a good solution?
EDIT:
I want to get a crime coefficient number from a username (string).
Ok, you can use JS function to get charCode of letter
let str = "some string example";
let sum = 0;
for (let i=0; i<str.length; i++) {
sum += parseInt(str[i].charCodeAt(0), 10); // Sum all codes
}
// Now we have some value as Number in sum, lets convert it to 0..1 value to scale to needed value
let rangedSum = parseFloat('0.' + String(sum)); // Looks dirty but works
let resultValue = Math.round(rangedSum * 500) + 1; // Same alogorythm as using Math.random(Math.round() * (max-min)) + min;
I hope it helps.
So as you are using nodejs, you can use crypto library to get md5 hash of string and then get it as HEX.
const crypto = require('crypto');
let valueHex = crypto.createHash('md5').update('YOUR STRING HERE').digest('hex');
// then get it as decimal based value
let valueDec = parseInt(valueHex, 16);
// and apply the same algorythm as above to scale it between 1-500
function coeficient() {
return Math.floor(Math.random() * 500) + 1;
}
console.log(coeficient());
console.log(coeficient());
console.log(coeficient());
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https://jsbin.com/wujusajowa/1/edit?html,js,output
I can sum the numbers of options. Like (5+5+5=15)
But I don't know a way to multiply the input with the sum of selects.
For example, What should I do to do 6 x (5+5+5) and get 90 ?
Use <input type="number">, define a global variable to store value of <input>; attach change event to <input> element to update global variable; use variable at change event of <select> element if variable is defined, else use 1 as multiplier
var input = 0;
$('select').change(function(){
var sum = 0;
$('select :selected').each(function() {
sum += Number($(this).val());
});
$("#toplam").html(sum * (input || 1));
}).change();
$("#miktar").on("change", function() {
input = this.valueAsNumber;
});
jsbin https://jsbin.com/cujohisahi/1/edit?html,js,output
Here's a simple example:
var mySum = 6 * ( 5 + 5 + 5 );
document.getElementById('result').innerHTML = mySum;
<div id="result"></div>