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How to sum a given number and it's reverse number in javaScript

const reversedNum = num =>
parseFloat(num.toString().split('').reverse().join('')) * Math.sign(num)
console.log(reversedNum(456))
Couldn't figure it out how to write code in order to sum 654 + 456
Thank You very much!

const reversedNum = num => num + +num.toString().split('').reverse().join('')

You can return sum of num and reversedNum inside a function.
const sumOfNumAndReversedNum= num => {
const reversedNum = parseFloat(num.toString().split('').reverse().join('')) * Math.sign(num)
return num + reversedNum
}
let userNumber = 456
console.log(sumOfNumAndReversedNum(userNumber))

You can write a more performant way of reversing the number than turning it into a string, flipping it, and turning it back into an integer.
One option is to go through the number backwards by popping off the last integer (e.g., 123 % 10 === 3) and adding it to your newly reversed number. You'll also need to multiply your reversed number by 10 in each iteration to move you to the next degree.
For example, given the number 123:
123 % 10 = 3;
123 /= 10 = 12;
0 * 10 + 3 = 3;
1 % 10 = 2;
12 /= 10 = 1;
3 * 10 + 2 = 32
1 % 10 = 1;
1 /= 10 = 0;
32 * 10 + 1 = 321
This method will also automatically take care of negative numbers for you, leaving you something like:
function reverse(num) {
let reversed = 0;
while (num !== 0) {
const popped = num % 10;
num = parseInt(num / 10);
if (reversed > Number.MAX_VALUE / 10 || (reversed === Number.MAX_VALUE / 10 && popped > 7)) return 0;
if (reversed < Number.MIN_VALUE / 10 || (reversed === Number.MIN_VALUE / 10 && popped < -8)) return 0;
reversed = reversed * 10 + popped;
}
return reversed;
}
Now you can simply call:
console.log(123 + reverse(123))

const reversedNum = num =>
Number(num.toString().split('').reverse().join(''))
console.log(reversedNum(456))
Do it!

How to format numbers to block of length three separated by comma, last 2 blocks or last block can be length of two but it cannot be length of one

Need to format the number into blocks of a length of three, separated with commas.
If required, the second last block can be of length two, like 987,234,65,43
The last block cannot be of length one(046,435,699,8).
The last block should be a length of two or three, it should not be one.
Examples:
"223,334,874,498"
"987,234,65,43"
"487,534,354,23"
Below is the code, but in this code, I am not able to make that last block to the length of 2 or 2 digits.
For "0464356998" ---Current Output is:046,435,699,8 ---
But Expected Output is: 046,435,69,98
function format(N) {
return N.replace(/[^0-9]/g, '')
// Replace everything which are not a number
// Spilt on every 3rd number
.match(/\d{1,3}/g)
// Join with commas
.join(',')
}
console.log(format("0464356998"));
//Current output: 046,435,699,8
//expected Output: 046,435,69,98

I'm not sure that this is best way but n is string so you can slice and you can count length of string
function format(n) {
if((n.length - 6) % 2 === 0){ // is (n.length - 6) is divisible without remainder
return `${n.slice(0,3)}, ${n.slice(3,6)}, ${n.slice(6,8)}, ${n.slice(8,10)}`
}else{
return `${n.slice(0,3)}, ${n.slice(3,6)}, ${n.slice(6,9)}, ${n.slice(9,12)}`
}
}
console.log(format("0464356998"));
console.log(format("04643569981"));
UPD For any string from 6 to ...
function formatter(n) {
let a = [];
for (let i = 0; i < n.length / 3; i++) {
a.push(n.slice(i * 3, (i + 1) * 3));
}
if (a[a.length - 1].length === 1){
let b = a[a.length - 2].substring(2,3)
let c = a[a.length - 1];
a[a.length - 2] = a[a.length - 2].slice(0, 2);
a[a.length - 1] = b.concat(c);
}
return a.join();
}
console.log(formatter("0464356998")); // 10
console.log(formatter("04643569982")); // 11
console.log(formatter("046435699823")); // 12
console.log(formatter("0464356998231")); // 13
console.log(formatter("04643569982316")); // 14
console.log(formatter("046435699823156")); // 15
console.log(formatter("0464356998231675")); // 16
console.log(formatter("04643569982315634")); // 17
console.log(formatter("046435699823167556")); // 18
console.log(formatter("046435699810464356998")); // 21
console.log(formatter("0464356998104643569981")); // 22

Count how many palindrome numbers are on the segment [1, 10 ** n] (n <100)

It is necessary to calculate how many palindrome numbers are on the segment [1, 10 ** n] (n <100).
function f(n) {
let res = 10 ** (parseInt(n / 2) + n % 2);
return res - res / 10;
}
function countPalindromes(n) {
let count = 0;
for (let i = 1; i <= n; i++) {
count += f(i);
}
return count;
}
for (let i = 1; i < 100; i++) {
console.log(i, f(i), countPalindromes(i));
}
Problems:
Over time js returns the result in exponential form, and the result cannot be translated into a string.
Adding such large numbers into a column does not work
And so, how can I count the number of palindromes in the segment [1, 10 ** n] (n <100)???

Let's start with a smaller problem first. How many n digit palindromic numbers exist?
If n is even, let the number be of the form _ _ _ _ (midpoint) _ _ _ _. Now, in the first digit, you can't fill 0, so you have 9 options (1-9). For the second digit to the n/2th digit, you can fill any of the digits in 0-9. Since we're dealing with palindromes, second half of the number will have the same digits as the first half. So, using basic counting, total number of n digit palindromes when n is even is 9*[10*10*...10 (n/2)-1 times] = 9 * 10^((n/2)-1)
When n is odd, the analysis will be similar, but with two differences:
You have 10 choices (0-9) for the midpoint.
There are (n-1)/2 digits on either side of the midpoint, where the first digit can't be 0.
Again, using basic counting, number of palindromes if n is odd = 9 (for first digit) * 10 (for midpoint) * 10^((n/2)-2) (the remaining digits) = 9 * (10^(n/2)-1), similar to the even case.
Thus, number of n digit palindromes = 9 * (10^(n/2)-1) if n is greater than 1. After that, you just have to loop n from 2 to 100 to get the total count of palindromic numbers you need.

I'm not 100% sure, so correct me if I'm wrong. It works, check the code at the bottom.
There should be two kinds of palindromes: ones with odd number of digits and ones with even number of digits. You can go from one palindrome to another (of the same kind) by appending and prepending the same digit to the first one. So...
Single-digit palindromes: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
For every single-digit palindrome, we can construct 10 3-digit palindromes:
000, 101, 202, ..., 010, 111, 212, ..., 898, 999.
Similarly, for every 3-digit palindromes, we can construct 10 5-digit palindromes.
10 ** n would have up to n-digit palindromes. Let's count:
10 ** 1 single-digit, 10 ** 2 3-digit, 10 ** 3 5-digit, ..., 10 ** ((n + 1) / 2) n-digit.
The same can be done for double-digit palindromes (00, 11, 22, 33, 44, 55, 66, 77, 88, 99), to iteratively obtain palindromes with an even number of digits.
Then we need to drop the 'illegal' numbers - numbers starting with the digit 0.
It works. Yay!
// Counting without creating the numbers
function getSmartCount(n) {
let smartPalindromeCount = 0;
let startingWithZeroCount = 0;
for (let digits = 1; digits <= n; digits += 2) { // even digits
const power = (digits + 1) / 2;
smartPalindromeCount += 10 ** power;
startingWithZeroCount += 10 ** (power - 1); // the zero-x-zero numbers
}
for (let digits = 2; digits <= n; digits += 2) { // odd digits
const power = digits / 2;
smartPalindromeCount += 10 ** power;
startingWithZeroCount += 10 ** (power - 1); // the zero-x-zero numbers;
}
return smartPalindromeCount - startingWithZeroCount;
}
// Counting by creating the numbers
function getCountIteratively(n) {
const singleDigitPalindromes = '0123456789'.split('');
const doubleDigitPalindromes = singleDigitPalindromes.map(x => x + x);
const palindromes = [...singleDigitPalindromes, ...doubleDigitPalindromes];
const nestedPalindromes = [
singleDigitPalindromes,
doubleDigitPalindromes,
];
for (let digits = 3; digits <= n; digits++) {
const index = digits - 1;
const extraPalindromes = [];
const previousPalindromes = nestedPalindromes[index - 2];
for (const previousPalindrome of previousPalindromes) {
for (const character of singleDigitPalindromes) {
const newPalindrome = character + previousPalindrome + character;
extraPalindromes.push(newPalindrome);
palindromes.push(newPalindrome);
}
}
nestedPalindromes.push(extraPalindromes);
}
const legalPalindromes = palindromes.filter(palindrome => palindrome[0] !== '0');
// console.log(legalPalindromes); // uncomment to see the numbers
return legalPalindromes.length;
}
// Counting by checking every single number
function getCountByBruteForce(n) {
const bruteForcePalindromes = [];
for (let i = 1; i < 10 ** n; i++) {
const leftToRight = i.toString();
const rightToLeft = leftToRight.split('').reverse().join('');
if (leftToRight === rightToLeft) {
bruteForcePalindromes.push(i);
}
}
// console.log(bruteForcePalindromes); // uncomment to see the numbers
return bruteForcePalindromes.length;
}
for (let n = 2; n <= 5; n++) {
console.log({
n,
smart: getSmartCount(n),
byCreatingTheNumbers: getCountIteratively(n),
bruteForce: getCountByBruteForce(n),
});
}

Javascript Find Nth Digit Number Algorithm Performance [duplicate]

I want to generate the value being searched by the position entered in the check. For example, if 20 is entered, the function should generate numbers starting from 0 and continue in ascending order until 20 digits are created, then output the value of the 20th digit in the generated number string (01234567891011121314), which is 4.
I tried this below, however it is not efficient when it comes to numbers like 1,000,000,000,
[...Array(5).keys()]; output => [0, 1, 2, 3, 4]
Edit this post to clarify I am trying to get a more efficient solution.
Here I am trying to get the answer for long numbers(1,000,000,000) in below one second.
I already have a solution but it takes more than 1 second.
[...Array(5).keys()].join("")[4]; output => 4

This is nearly identical to the Champernowne constant.
A solution from math.stackexchange is:
(Stack Overflow doesn't support MathJax, unfortunately)
The first step is to find what decade you are in. There are 9 digits from the 1 digit numbers, 2⋅90=180 digits from the 2 digit numbers for a total of 189, and generally n⋅9⋅10n−1 from the n digit numbers. Once you have found the decade, you can subtract the digits from the earlier decades. So if you want the 765th digit, the first 189 come from the first and second decades, so we want the 576th digit of the 3 digit numbers. This will come in the ⌈5763⌉=192nd number, which is 291. As 576≡3(mod3), the digit is 1
Programatically:
const getDigit = (target) => {
let i = 0;
let xDigitNumbers = 1; // eg 1 digit numbers, 2 digit numbers
let digitsSoFar = 1;
while (true) {
const digitsThisDecade = xDigitNumbers * 9 * 10 ** (xDigitNumbers - 1);
if (digitsSoFar + digitsThisDecade > target) {
// Then this is the "decade" in which the target digit is
// digitIndexThisDecade: eg, starting from '100101102', to find the last '1' in '101', digitIndexThisDecade will be 6
const digitIndexThisDecade = target - digitsSoFar;
// numIndexThisDecade: this identifies the index of the number in the decade
// eg, starting from '100101102', this could be index 2 to correspond to 101 (one-indexed)
const numIndexThisDecade = Math.floor(digitIndexThisDecade / xDigitNumbers);
// decadeStartNum: the number right before the decade starts (0, 9, 99, 999)
const decadeStartNum = 10 ** (xDigitNumbers - 1);
// num: the number in which the target index lies, eg 101
const num = decadeStartNum + numIndexThisDecade;
// digitIndexInNum: the digit index in num that the target is
// eg, for 101, targeting the last '1' will come from a digitIndexInNum of 2 (zero-indexed)
const digitIndexInNum = digitIndexThisDecade % xDigitNumbers;
return String(num)[digitIndexInNum]
}
digitsSoFar += digitsThisDecade;
xDigitNumbers++;
}
};
for (let i = 0; i < 1000; i++) {
document.write(`${i}: ${getDigit(i)}<br>`);
}

Here's a simple approach without using arrays.
let N = 1000000000, digitsCount = 0, currentNumber = 0;
console.time('Took time: ');
const digits = (x)=>{
if(x<10)
return 1;
if(x<100)
return 2;
if(x<1000)
return 3;
if(x<10000)
return 4;
if(x<100000)
return 5;
if(x<1000000)
return 6;
if(x<10000000)
return 7;
if(x<100000000)
return 8;
if(x<1000000000)
return 9;
return 10; // Default
}
while(true){
digitsCount += digits(currentNumber);
if(digitsCount >= N)
break;
currentNumber++;
}
console.timeEnd('Took time: ');
console.log(String(currentNumber)[N-digitsCount+digits(currentNumber)-1])
Output (The execution time may differ for you but it'll be under 1 second(or 1000ms).)
Took time: : 487.860ms
9

i used .join("") to convert the array to string '01234567891011121314151617181920'
then access the Nth number by Indexing string
N=20;
console.log ( [...Array(N+1).keys()].join("")[N-1] ) //OUTPUT 4
EDIT:i think ther's a solution which is you don't need to create array at all😎
its a mathematical formula
Blockquote

In my Solution , we don't need big iterations and loops...
But This Solution is Big for simple understanding...
I made it for upto 6 digits , and its very efficient...and can be made for any number of digits... And can even be reduced to small functions , but that would get too complex to understand...
So , Total numbers for Given Digits :
For 1 Digit Numbers , They are 10 (0 to 9)....
For 2 Digit Numbers , They are 9*10 => 90 , and total Digits ==> 90*2 ==> 180...
For 3 Digit Numbers , 9*10*10 => 900 , and total Digits ==> 90*3 ==> 2700...
For 4 Digit Numbers , 9*10*10*10 => 9000 , and total Digits ==> 9000*4 ==> 36000...
A function to get Total Digits for a given specified (Number of Digits)
let totalDigits = n => {
if (n == 1) return 10;
return 9 * (10 ** (n - 1)) * n;
}
Now , we set a Range of position for different Digits ,
for 1 Digit , its between 1 and 10....
for 2 Digits , It's Between 11(1+10) and 190(180+10)...(position of 1 in 10 is 11 , and Second 9 in 99 is 190)...
for 3 Digits , It's Between 191(1+10+180) and 2890(2700+180+10)...And so on
for n Digit , Function to get Range is
// This function is used to find Range for Positions... Eg : 2 digit Numbers are upto Position 190...(Position 191 is "100" first digit => 1 )
let digitN = n => {
if (n == 1) return totalDigits(1);
return digitN(n - 1) + totalDigits(n);
}
// To Finally set Ranege for a Given Digit Number... for 1 its [1,10] , for 2 its [11,190]
let positionRange = n => {
if (n == 1) return [1, 10];
else return [digitN(n - 1), digitN(n)]
}
So Final Solution is
// This Function tells the total number of digits for the given digit... Eg : there are 10 one digit Numbers , 180 Two Digit Numbers , 2700 3 Digit Numbers
let totalDigits = n => {
if (n == 1) return 10;
return 9 * (10 ** (n - 1)) * n;
}
// This function is used to find Range for Positions... Eg : 2 digit Numbers are upto Position 190...(Position 191 is "100" first digit => 1 )
let digitN = n => {
if (n == 1) return totalDigits(1);
return digitN(n - 1) + totalDigits(n);
}
// To Finally set Ranege for a Given Digit Number... for 1 its [1,10] , for 2 its [11,190]
let positionRange = n => {
if (n == 1) return [1, 10];
else return [digitN(n - 1), digitN(n)]
}
// A simple Hack to get same value for Different Consecutive Numbers , (0.3 or 0.6 or 0.9 or 1 return 1)
let getDigit = n => {
if (dataType(n) == "float") {
n = Math.floor(n);
n++;
}
return n;
}
// To check for Float or Integer Values
function dataType(x) {
if (Math.round(x) === x) {
return 'integer';
}
return 'float';
}
function f(position) {
let result, charInd, temp;
if ((position >= positionRange(1)[0]) && (position <= positionRange(1)[1])) { // Positions 1 to 10 (1 Digit Numbers)
result = position - 1;
charInd = 0
}
if ((position > positionRange(2)[0]) && (position <= positionRange(2)[1])) { // Positions 11 to 190 (2 Digit Numbers)
temp = (position - 10) / 2;
temp = getDigit(temp);
result = temp + 9;
charInd = (position - 11) % 2
}
if ((position > positionRange(3)[0]) && (position <= positionRange(3)[1])) { // Positions 191 to 2890 (3 Digit Numbers)
temp = (position - 190) / 3;
temp = getDigit(temp);
result = temp + 99;
charInd = (position - 191) % 3
}
if ((position > positionRange(4)[0]) && (position <= positionRange(4)[1])) { // Positions 2891 to 38890 (4 Digit Numbers)
temp = (position - 2890) / 4;
temp = getDigit(temp);
result = temp + 999;
charInd = (position - 2891) % 4
}
if ((position > positionRange(5)[0]) && (position <= positionRange(5)[1])) { // Positions 38890 to 488890 (5 Digit Numbers)
temp = (position - 38890) / 5;
temp = getDigit(temp);
result = temp + 9999;
charInd = (position - 38891) % 5
}
if ((position > positionRange(6)[0]) && (position <= positionRange(6)[1])) { // Positions 488890 to 5888890 (6 Digit Numbers)
temp = (position - 488890) / 6 ;
temp = getDigit(temp);
result = temp + 99999;
charInd = (position - 488891) % 6
}
finalChar = String(result)[charInd];
console.log("Given Position => ", position, " Result Number => ", result, "Char Index ==> ", charInd, "Final Char => ", finalChar);
}
let d1 = Date.now();
f(138971); // Given Position => 138971 Result Number => 30016 Char Index ==> 0 Final Char => 3
let d2 = Date.now();
console.log(d2-d1) ; // 351

Finding Gapful number in javascript

A number is gapful if it is at least 3 digits long and is divisible by the number formed by stringing the first and last numbers together.
The smallest number that fits this description is 100. First digit is
1, last digit is 0, forming 10, which is a factor of 100. Therefore,
100 is gapful.
Create a function that takes a number n and returns the closest gapful
number (including itself). If there are 2 gapful numbers that are
equidistant to n, return the lower one.
Examples gapful(25) ➞ 100
gapful(100) ➞ 100
gapful(103) ➞ 105
so to solve this i wrote the code that loops from the given number to greater than that and find out if it is or not by
function getFrequency(array){
var i=array
while(i>=array){
let a=i.toString().split('')
let b=a[0]+a[a.length-1]
b= +b
if(i%b==0) return i
i++
}
}
console.log(getFrequency(103))
Thats fine but what if the gapful number is less than the number passed in the function ?
like if i pass 4780 the answer is 4773 so in my logic how do i check simultaneoulsy smaller and greater than the number passed ?
I am only looping for the numbers greater than the number provided in function

You can alternate between subtracting and adding. Start at 0, then check -1, then check +1, then check -2, then check +2, etc:
const gapful = (input) => {
let diff = 0; // difference from input; starts at 0, 1, 1, 2, 2, ...
let mult = 1; // always 1 or -1
while (true) {
const thisNum = input + (diff * mult);
const thisStr = String(thisNum);
const possibleFactor = thisStr[0] + thisStr[thisStr.length - 1];
if (thisNum % possibleFactor === 0) {
return thisNum;
}
mult *= -1;
if (mult === 1) {
diff++;
}
}
};
console.log(
gapful(100),
gapful(101),
gapful(102),
gapful(103),
gapful(104),
gapful(105),
gapful(4780),
);

You could take separate functions, one for a check if a number is a gapful number and another to get left and right values and for selecting the closest number.
function isGapful(n) {
if (n < 100) return false;
var temp = Array.from(n.toString());
return n % (temp[0] + temp[temp.length - 1]) === 0;
}
function getClosestGapful(n) {
var left = n,
right = n;
while (!isGapful(right)) right++;
if (n < 100) return right;
while (!isGapful(left)) left--;
return n - left <= right - n
? left
: right;
}
console.log(getClosestGapful(25)); // 100
console.log(getClosestGapful(100)); // 100
console.log(getClosestGapful(103)); // 105
console.log(getClosestGapful(4780)); // 4773