I have an integer containing various digits, I want to remove 4th digit from an integer. How can I achieve that ?
Example :
let number = 789012345
Here I want to remove 0
Try this :
// Input
let number = 789012345;
// Convert number into a string
let numberStr = number.toString();
// Replace the 0 with empty string
const res = numberStr.replace(numberStr[3], '');
// Convert string into a number.
console.log(Number(res));
Rohìt Jíndal's answer is excellent. I just want to point out another way you could do this with string.replace and capturing groups.
function removeDigit(input, index) {
let exp = new RegExp(`^(\\d{${index}})(\\d)(.+)$`);
return parseInt(input.toString().replace(exp, '$1$3'));
}
let output = removeDigit(789012345, 3);
console.log(output); // 78912345
In this example, I have created a new RegExp object from a template literal in order to inject the index.
The first capturing group contains all digits up to the desired index. The second contains the digit we want to remove and the third contains the remainder of the string.
We then return an integer parsed from the string combination of only the first and third capturing groups.
You can follow this procedure:
Decide if you want to remove digits by index or by value, the following demo will remove by value, which means it will remove all values that match
Convert the number into a string
Convert the string to an array with Array.from
Use Array#filter to remove target digit(s)
Use Array#join to create a string
Use + to convert to string back into a numeric value
const n = 789012345;
const m = +Array.from( n.toString() ).filter(num => +num !== 0).join("");
console.log( m );
let numberWithoutADigit = removeIthDigitFromNumber(789012345, 4);
function removeIthDigitFromNumber(n, i){
//convert the number n to string as an array of char
let o = (n + '').split('');
//remove the item at the index i (0 based) from the array
o.splice(i, 1);
//rebuilds the string from the array of char and parse the string to return a number
let number = parseInt(o.join(''));
return number;
}
let number = 789012345
let i = 3 // index 3, 4th digit in number
let arr = number.toString().split("").filter((value, index) => index!==i);
// ['7', '8', '9', '1', '2', '3', '4', '5']
let new_number = parseInt(arr.join(""))
// 78912345
console.log(new_number)
let x = 789012345
var nums = [];
let i = 0, temp = 0;
while(x > 1){
nums[i++] = (x % 10);
x = (x - (x % 10)) / 10;
}
var cnt = 0;
for(--i; i >= 0; i--){
if (cnt++ == 3) continue;
temp = temp * 10 + nums[i];
}
Related
I don't know what is the right question for this.
I want to make this number -800 become [-8,0,0] is anyone can build it
The first thing that I do, is to make the number become a string and using the map function I iterate it becomes an array like this
const number = -800;
const numberString = number.toString();
const arrayString = numberString.split``.map((x) => +x);
console.log(arrayString)
But the result is [ NaN, 8, 0, 0 ]
How to change the NaN and first index 8 become -8 without disturbing the other index. So it becomes [-8, 0, 0]
Is anybody can help me?
Thanks.
Try numberString.match(/-?\d/g) instead of split
const number = -800;
const numberString = number.toString();
const arrayString = numberString.match(/-?\d/g).map(x => +x);
console.log(arrayString)
One possible solution is to just do this operation on the absolute value and, during the map, check if the number was negative (only for the first index).
const number = -800;
const numberString = Math.abs(number).toString();
const arrayString = numberString.split``.map((x, i) => i == 0 && number < 0 ? -x : +x);
console.log(arrayString)
If you're not clear about Regex, you can use Array.from to convert a string to an array number. Then handle the first number based on the sign of original number.
console.log(convertNumberToArray(800));
console.log(convertNumberToArray(-800));
function convertNumberToArray(number){
var result = Array.from(Math.abs(number).toString(), Number);
result[0] *= number <= 0 ? -1 : 1;
return result;
}
Use String#match with regex which matches optional - with the dig it.
var number = -800;
var numberString = number.toString().match(/-?\d/g);
var numberInt = [];
for (var i = 0; i < numberString.length; i++) {
numberInt.push(parseInt(numberString[i]));
}
console.log(numberInt);
If I have a string a12c56a1b5 then out put should be a13b5c56 as character a is repeated twice so a12 becomes a13
I have tried this:
function stringCompression (str) {
var output = '';
var count = 0;
for (var i = 0; i < str.length; i++) {
count++;
if (str[i] != str[i+1]) {
output += str[i] + count;
count = 0;
}
}
console.log(output); // but it returns `a11121c15161a111b151` instead of `a13b5c56`
}
It is happening because the code is counting the occurrence of each element and appending it, even the numbers in the string.
In this code,
for (var i = 0; i < str.length; i++) {
count++;
if (str[i] != str[i+1]) {
output += str[i] + count;
count = 0;
}
}
in first iteration i = 0, str[i] = 'a' and str[i + 1] = '1' for the given string a12c56a1b5 which are not equal hence, it will generate the output as a1 for first iteration, then a111 for second iteration since str[i] = '1' and str[i + 1] = '2' now, and so on.
We can achieve this by first separating the characters from the count. Assuming, that there would be characters from a-z and A-Z only followed by the count. We can do something like this, str.match(/[a-zA-Z]+/g) to get the characters: ["a", "c", "a", "b"] and str.match(/[0-9]+/g) to get their counts: ["12", "56", "1", "5"], put them in an object one by one and add if it already exists.
Something like this:
function stringCompression(str) {
var characters = str.match(/[a-zA-Z]+/g);
var counts = str.match(/[0-9]+/g);
var countMap = {};
for (var i = 0; i < characters.length; i++) {
if (countMap[characters[i]]) {
countMap[characters[i]] += parseInt(counts[i]);
} else {
countMap[characters[i]] = parseInt(counts[i]);
}
}
var output = Object.keys(countMap)
.map(key => key + countMap[key])
.reduce((a, b) => a + b);
console.log(output);
}
stringCompression('a12c56a1b5')
Using regex to extract word characters and numbers. Keeps an object map res to track and sum up following numbers. sorts and converts back to a string.
As an example, the for-of loop iteration flow with str=a12c56a1b5:
c='a', n='12'
res['a'] = (+n = 12) + ( (res['a'] = undefined)||0 = 0)
or ie: res['a'] = 12 + 0
c='c', n='56'
res['c'] = 56 + 0
c='a', n='1'
res['a'] = 1 + (res['a'] = 12 from iteration 1.) = 13
c='b', n='5'
res['b'] = 5 + 0
thus res = { 'a': 13, 'c': 56, 'b': 5 } after the for-of loop finishes
function stringCompression (str) {
// build object map with sums of following numbers
const res = {}
for(const [,c,n] of str.matchAll(/(\w+)(\d+)/g))
res[c] = +n + (res[c]||0)
// convert object map back to string
output = Object.entries(res)
output.sort(([a],[b])=>a<b ? -1 : a>b ? 1 : 0)
output = output.map(([a,b])=>`${a}${b}`).join('')
console.log(output); // but it returns `a11121c15161a111b151` instead of `a13b5c56`
}
stringCompression('a12c56a1b5')
[,c,n] = [1,2,3] is equivalent to c=2, n=3. It is called destructuring.
matchAll matches on a regex. It's a relatively new shorthand for calling .exec repeatedly to execute a regular expression that collects all the results that the regular expression matches on.
(\w+)(\d+) is a regex for two capture groups,
\w+ is for one or more alpha characters, \d+ is for one or more digits.
for(const [,c,n] of str.matchAll...) is equivalent to:
for each M of str.matchAll...
const c = M[1], n = M[2]`
res[c]||0 is shorthand for:
"give me res[c] if it is truthy (not undefined, null or 0), otherwise give me 0"
+n uses the unary operator + to force an implicit conversion to a number. JavaScript specs for + unary makes it convert to number, since + unary only makes sense with numbers.
It is basically the same as using Number(n) to convert a string to an number.
Conversion back to a string:
Object.entries converts an object {"key":value} to an array in the form of [ [key1, value1], [key2, value2] ]. This allows manipulating the elements of an object like an array.
.sort sorts the array. I destructured the keys to sort on the keys, so "a" "b" "c" are kept in order.
.map takes an array, and "maps" it to another array. In this case I've mapped each [key,value] to a string key+value, and then taking the final mapped array of key+value strings and joined them together to get the final output.
In case it asks you to sort it alphabetically, I added #user120242's sorting code snippet to #saheb's entire answer (in between Object.keys(countMap) and .map(...). That worked for me. I tried using #user120242's whole answer, but it did not pass all the tests since it did not add the repeated letters for longer strings. But #user120242's answer did work. It just need to be sorted alphabetically and it passed all the test cases in HackerRank. I had this question for a coding assessment (called "Better Coding Compression").
P.S. I also removed checking the capital letters from #saheb's code since that wasn't required for my coding challenge.
Here's how mine looked like:
function stringCompression(str) {
var characters = str.match(/[a-zA-Z]+/g);
var counts = str.match(/[0-9]+/g);
var countMap = {};
for (var i = 0; i < characters.length; i++) {
if (countMap[characters[i]]) {
countMap[characters[i]] += parseInt(counts[i]);
} else {
countMap[characters[i]] = parseInt(counts[i]);
}
}
var output = Object.keys(countMap)
.sort(([a],[b])=>a<b ? -1 : a>b ? 1 : 0)
.map(key => key + countMap[key])
.reduce((a, b) => a + b);
console.log(output);
}
stringCompression('a12c56a1b5')
I have an integer and want to create an array from each of its numbers.
let interget = 345;
//I want create the array [3,4,5]
Is there any easy way to do this using Array.from() or would I need to convert the number to a string first?
easy way by converting to string
(inputNumber + "").split("").map(char => +char)
basically we split string and convert each character back to number
doing it manually
function getDigits(n) {
const ans = [];
while(n > 0){
let digit = n % 10;
ans.push(digit);
n -= digit;
n /= 10;
}
return ans.reverse();
}
Or you can do it like this:
var result = Array.from('345', Number)
console.log(result);
I want to convert a number to its corresponding alphabet letter. For example:
1 = A
2 = B
3 = C
Can this be done in javascript without manually creating the array?
In php there is a range() function that creates the array automatically. Anything similar in javascript?
Yes, with Number#toString(36) and an adjustment.
var value = 10;
document.write((value + 9).toString(36).toUpperCase());
You can simply do this without arrays using String.fromCharCode(code) function as letters have consecutive codes. For example: String.fromCharCode(1+64) gives you 'A', String.fromCharCode(2+64) gives you 'B', and so on.
Snippet below turns the characters in the alphabet to work like numerical system
1 = A
2 = B
...
26 = Z
27 = AA
28 = AB
...
78 = BZ
79 = CA
80 = CB
var alphabet = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
var result = ""
function printToLetter(number){
var charIndex = number % alphabet.length
var quotient = number/alphabet.length
if(charIndex-1 == -1){
charIndex = alphabet.length
quotient--;
}
result = alphabet.charAt(charIndex-1) + result;
if(quotient>=1){
printToLetter(parseInt(quotient));
}else{
console.log(result)
result = ""
}
}
I created this function to save characters when printing but had to scrap it since I don't want to handle improper words that may eventually form
Just increment letterIndex from 0 (A) to 25 (Z)
const letterIndex = 0
const letter = String.fromCharCode(letterIndex + 'A'.charCodeAt(0))
console.log(letter)
UPDATE (5/2/22): After I needed this code in a second project, I decided to enhance the below answer and turn it into a ready to use NPM library called alphanumeric-encoder. If you don't want to build your own solution to this problem, go check out the library!
I built the following solution as an enhancement to #esantos's answer.
The first function defines a valid lookup encoding dictionary. Here, I used all 26 letters of the English alphabet, but the following will work just as well: "ABCDEFG", "ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789", "GFEDCBA". Using one of these dictionaries will result in converting your base 10 number into a base dictionary.length number with appropriately encoded digits. The only restriction is that each of the characters in the dictionary must be unique.
function getDictionary() {
return validateDictionary("ABCDEFGHIJKLMNOPQRSTUVWXYZ")
function validateDictionary(dictionary) {
for (let i = 0; i < dictionary.length; i++) {
if(dictionary.indexOf(dictionary[i]) !== dictionary.lastIndexOf(dictionary[i])) {
console.log('Error: The dictionary in use has at least one repeating symbol:', dictionary[i])
return undefined
}
}
return dictionary
}
}
We can now use this dictionary to encode our base 10 number.
function numberToEncodedLetter(number) {
//Takes any number and converts it into a base (dictionary length) letter combo. 0 corresponds to an empty string.
//It converts any numerical entry into a positive integer.
if (isNaN(number)) {return undefined}
number = Math.abs(Math.floor(number))
const dictionary = getDictionary()
let index = number % dictionary.length
let quotient = number / dictionary.length
let result
if (number <= dictionary.length) {return numToLetter(number)} //Number is within single digit bounds of our encoding letter alphabet
if (quotient >= 1) {
//This number was bigger than our dictionary, recursively perform this function until we're done
if (index === 0) {quotient--} //Accounts for the edge case of the last letter in the dictionary string
result = numberToEncodedLetter(quotient)
}
if (index === 0) {index = dictionary.length} //Accounts for the edge case of the final letter; avoids getting an empty string
return result + numToLetter(index)
function numToLetter(number) {
//Takes a letter between 0 and max letter length and returns the corresponding letter
if (number > dictionary.length || number < 0) {return undefined}
if (number === 0) {
return ''
} else {
return dictionary.slice(number - 1, number)
}
}
}
An encoded set of letters is great, but it's kind of useless to computers if I can't convert it back to a base 10 number.
function encodedLetterToNumber(encoded) {
//Takes any number encoded with the provided encode dictionary
const dictionary = getDictionary()
let result = 0
let index = 0
for (let i = 1; i <= encoded.length; i++) {
index = dictionary.search(encoded.slice(i - 1, i)) + 1
if (index === 0) {return undefined} //Attempted to find a letter that wasn't encoded in the dictionary
result = result + index * Math.pow(dictionary.length, (encoded.length - i))
}
return result
}
Now to test it out:
console.log(numberToEncodedLetter(4)) //D
console.log(numberToEncodedLetter(52)) //AZ
console.log(encodedLetterToNumber("BZ")) //78
console.log(encodedLetterToNumber("AAC")) //705
UPDATE
You can also use this function to take that short name format you have and return it to an index-based format.
function shortNameToIndex(shortName) {
//Takes the short name (e.g. F6, AA47) and converts to base indecies ({6, 6}, {27, 47})
if (shortName.length < 2) {return undefined} //Must be at least one letter and one number
if (!isNaN(shortName.slice(0, 1))) {return undefined} //If first character isn't a letter, it's incorrectly formatted
let letterPart = ''
let numberPart= ''
let splitComplete = false
let index = 1
do {
const character = shortName.slice(index - 1, index)
if (!isNaN(character)) {splitComplete = true}
if (splitComplete && isNaN(character)) {
//More letters existed after the numbers. Invalid formatting.
return undefined
} else if (splitComplete && !isNaN(character)) {
//Number part
numberPart = numberPart.concat(character)
} else {
//Letter part
letterPart = letterPart.concat(character)
}
index++
} while (index <= shortName.length)
numberPart = parseInt(numberPart)
letterPart = encodedLetterToNumber(letterPart)
return {xIndex: numberPart, yIndex: letterPart}
}
this can help you
static readonly string[] Columns_Lettre = new[] { "A", "B", "C"};
public static string IndexToColumn(int index)
{
if (index <= 0)
throw new IndexOutOfRangeException("index must be a positive number");
if (index < 4)
return Columns_Lettre[index - 1];
else
return index.ToString();
}
I want to Split a number into its digit (for example 4563 to 4 , 5 , 6 , 3 ) then addiction this digits. (for example: 4+5+6+3=18)
I can write code for 3 digit or 2 digit and ... numbers seperately but I cant write a global code for each number.
so this is my code for 2 digit numbers:
var a = 23
var b = Math.floor(a/10); // 2
var c = a-b*10; // 3
var total = b+c; // 2+3
console.log(total); // 5
and this is my code for 3 digit numbers:
var a = 456
var b = Math.floor(a/100); // 4
var c = a-b*100; // 56
var d = Math.floor(c/10); // 5
var e = c-d*10; // 6
var total = b+d+e; // 4+5+6
console.log(total); // 15
but I cant write a code to work with each number.How can I write a global code for each number?
In modern browsers you can do an array operation like
var num = 4563;
var sum = ('' + num).split('').reduce(function (sum, val) {
return sum + +val
}, 0)
Demo: Fiddle
where you first create an array digits then use reduce to sum up the values in the array
var num = 4563;
var sum = 0;
while(num > 0) {
sum += num % 10;
num = Math.floor(num / 10);
}
console.log(sum);
Do number%10(modulus) and then number/10(divide) till the number is not 0
I hope the following example is useful to you:
var text="12345";
var total=0;
for (i=0;i<text.length;i++)
{
total+= parseInt(text[i]);
}
alert(total);
This solution converts the number to string, splits it into characters and process them in the callback function (prev is the result from the previous call, current is the current element):
var a = 456;
var sum = a.toString().split("").reduce(function(prev, current){
return parseInt(prev) + parseInt(current)
})
Here is how I would approach the problem. The trick I used was to split on the empty string to convert the string to an array and then use reduce on the array.
function digitSum(n) {
// changes the number to a string and splits it into an array
return n.toString().split('').reduce(function(result, b){
return result + parseInt(b);
}, 0);
}
As mentioned by several other posters (hat tip to my commenter), there are several other good answers to this question as well.
Here is my solution using ES6 arrow functions as call back.
- Convert the number into a string.
- Split the string into an array.
- Call the map method on that array.
- Callback function parse each digit to an array.
let number = 65535;
//USING MAP TO RETURN AN ARRAY TO DIGITS
let digits = number.toString()
.split("")
.map(num => parseInt(num));
//OUTPUT TO DOM
digits.forEach(
digit =>
document.querySelector("#out").innerHTML += digit + "<br>"
);
<p id="out"></p>
1) You can cast input number to string, using .toString() method and expand it into array with spread (...) operator
const splitNumber = n => [ ...n.toString() ]
2) Another short way - using recursion-based solution like:
const splitNumber = n => n ? [ ...splitNumber(n/10|0), n%10 ] : []