Challenge: https://www.codewars.com/kata/57c7930dfa9fc5f0e30009eb/train/javascript
Hi I have been trying this problem for many hours but unfortunately my code is taking too long to pass:
function closestPower(num) {
num = Math.floor(num);
if (num < 4) return 4;
// check if input is perfect power
let base = 2;
while (base < 10) {
let exponent = Math.trunc(getBaseLog(base , num));
if ( Math.pow(base, exponent) === num ) {
return num;
}
base++;
}
// check for upper and lower
base = 2;
const verifyObj = {upper:null, lower:null}; // verify
let upperPower = num + 1;
let lowerPower = num - 1;
while (!verifyObj.upper || !verifyObj.lower)
{
// no perfect power
if (lowerPower <= 2 ) verifyObj.lower = "Not found";
if (upperPower === Infinity ) verifyObj.upper = "Not found";
// up til base 9
if (base === 10) {
if (!verifyObj.upper) upperPower++;
if (!verifyObj.lower) lowerPower--;
base = 2;
}
// upper
if (!verifyObj.upper) {
let exponent = Math.trunc(getBaseLog(base , upperPower));
if ( Math.pow(base, exponent) === upperPower ) {
verifyObj.upper = upperPower;
}
}
// lower
if (!verifyObj.lower) {
let exponent = Math.trunc(getBaseLog(base , lowerPower));
if ( Math.pow(base, exponent) === lowerPower ) {
verifyObj.lower = lowerPower;
}
}
base++;
}
console.log(verifyObj) // {upper:64, lower: 49}
// nearest power
if ((upperPower - num) < (num - lowerPower)) {
return upperPower;
}
else return lowerPower;
}
closestPower(56.5); // 49
function getBaseLog(x, y) {
return Math.log(y) / Math.log(x);
}
I realized that my code is redundant as all i need to know if a “base” and “exponent” are more than 1 to determine a perfect power. Any formulas or ideas?
Some issues:
There is no reason why base should not be allowed to be 10 or more
Trying with upperPower at each increment is taking too many iterations. The distance to the next power might be rather big.
I would suggest the following algorithm:
Let the exponent to try with start at 2, and then increment by 1. Calculate which could be the corresponding base. The real base can be found by raising n to the inverse exponent (i.e. 1/exp). Then there are only 2 interesting integer bases to consider: by rounding downwards and upwards.
Here is an implementation:
function closestPower(n) {
if (n <= 6) return 4;
let result = -1;
let closest = n;
for (let factor, exp = 2; (factor = n ** (1 / exp)) > 1.9; ++exp) {
let above = Math.ceil(factor);
for (let intfactor = Math.floor(factor); intfactor <= above; intfactor++) {
let power = intfactor ** exp;
let diff = Math.abs(power - n);
if (diff == 0) return n;
if (diff < closest || diff == closest && power < n) {
closest = diff;
result = power;
}
}
}
return result;
}
// Some tests:
const tests = [
[0, 4], [9, 9], [30, 32], [34, 32], [56.5, 49],
[123321456654, 123321773584]
];
for (let [n, expected] of tests) {
let result = closestPower(n);
if (result === expected) continue;
console.log(`closestPower(${n}) returned ${result}, but expected ${expected}`);
}
console.log("all tests done");
Here's my algorithm
first i will get the exponent from base that less than of the n then I added the current base of the loop with the n then get the base log.
function closestPower(n) {
if(n < 4) return 4
let closest = []
let base = 2
while(base < n) {
const exponent = Math.floor(Math.log(n + base) / Math.log(base))
const power = Math.pow(base,exponent)
if(exponent === 1) break
if(power === n) return n
closest.push(power)
base++
}
return closest.reduce((prev, curr) => (Math.abs(curr - n) < Math.abs(prev - n) ? curr : prev))
}
console.log(closestPower(0))
console.log(closestPower(9))
console.log(closestPower(30))
console.log(closestPower(34))
console.log(closestPower(56.5))
console.log(closestPower(123321456654))
You can see I use "" to make the if statement return nothing in certain cases, but in reality it returns a blank space. I don't want that white space, what could I use instead?
if (n >= 20000 && n < 100000) {
result = `${tens[Math.floor(n / 10000) - 1]} ${
Math.floor((n % 10000) / 1000) != 0
? num[Math.floor((n % 10000) / 1000)]
: ""
} thousand ${n % 1000 != 0 ? number2words(n % 1000) : ""}`;
return result.trim();
}
You will have to restructure you code and move the include the space when the condition is truthy.
const tensText = tens[Math.floor(n / 10000) - 1];
const divisible = Math.floor((n % 10000) / 1000);
const hundredsText = divisible != 0 ? ` ${num[divisible]}` : '';
const thousandsText = n % 1000 != 0 ? ` ${number2words(n % 1000)}` : '';
result = `${tensText}${hundredsText} thousand${thousandsText}`;
To return nothing, you'd use return;
However, if one case returns a string, you should probably always return a string for consistency.
I was trying to solve this problem on a leet code called Rotate digits.
This is the working solution of the problem in Javascript :
var IsGood = function(n){
let result = false;
while(n > 0){
let digit = n % 10; // 2 % 10
// ignore 0, 1, 8; by themselves, no change
if( (digit === 3) || (digit === 4) || (digit === 7) ) return false;
if( (digit === 2) || (digit === 5) || (digit === 6) || (digit === 9) ) result = true;
n /= 10;
}
return result;
}
var rotatedDigits = function(N) {
let count = 0;
for(let i = 1; i <= N; i++){
if(IsGood(i)) count++;
}
return count;
}
My question inside the IsGood function why are we taking n % 10 and then dividing n /=10. I have seen this implementation in almost all solutions to this problem. Can someone please explain the logic behind this?
Take the number 123
The remainder of that number when divided by 10 is 3 (123 % 10) = 3.
So n%10 is giving you the last digit of the number
If you want to test the next digit in the number you have to remove the 3. The way to do that is to divide by 10 and only take the integer part (12.3 lose the .3 part) which you get by 123 / 10
You can keep doing that until you get to a number less than 10 which must be the final digit! this algorithm is a way to examine each digit in a number using math rather than string manipulation
123 % 10 = 3 first digit
123 / 10 = 12
12 % 10 = 2 next digit
12 / 10 = 1
1 < 10 so final digit
How do you check whether or not an integer contains a digit?
For example:
var n = 12;
var m = 34;
n contains 1 // true
m contains 1 // false
What's the fastest (performance wise) way to do this without turning the integer into a string?
Refer to the following code (if the comments aren't good enough feel free to ask):
function contains(number, digit) {
if (number < 0) { // make sure negatives are dealt with properly, alternatively replace this if statement with number = Math.abs(number)
number *= -1;
}
if (number == digit) { // this is to deal with the number=0, digit=0 edge case
return true;
}
while (number != 0) { // stop once all digits are cut off
if (number % 10 == digit) { // check if the last digit matches
return true;
}
number = Math.floor(number / 10); // cut off the last digit
}
return false;
}
Here's a simple recursive form -
const contains = (q, p) =>
p < 10
? p === q
: p % 10 === q || contains(q, p / 10 >>> 0)
console.log(contains(1, 12)) // true
console.log(contains(1, 34)) // false
console.log(contains(9, 14293)) // true
console.log(contains(9, 1212560283)) // false
if (n.toString().includes("1")) {
/// Do something
}
try this:
let n = 1234;
let flag = false;
while (n > 0){
r = n % 10;
if(r == 1){
flag = true;
break;
}
n = (n - (n % 10)) / 10;
}
console.log("n contains 1 = "+flag);
I have a number assigned to a variable, like that:
var myVar = 1234;
Now I want to get the second digit (2 in this case) from that number without converting it to a string first. Is that possible?
So you want to get the second digit from the decimal writing of a number.
The simplest and most logical solution is to convert it to a string :
var digit = (''+myVar)[1];
or
var digit = myVar.toString()[1];
If you don't want to do it the easy way, or if you want a more efficient solution, you can do that :
var l = Math.pow(10, Math.floor(Math.log(myVar)/Math.log(10))-1);
var b = Math.floor(myVar/l);
var digit = b-Math.floor(b/10)*10;
Demonstration
For people interested in performances, I made a jsperf. For random numbers using the log as I do is by far the fastest solution.
1st digit of number from right → number % 10 = Math.floor((number / 1) % 10)
1234 % 10; // 4
Math.floor((1234 / 1) % 10); // 4
2nd digit of number from right → Math.floor((number / 10) % 10)
Math.floor((1234 / 10) % 10); // 3
3rd digit of number from right → Math.floor((number / 100) % 10)
Math.floor((1234 / 100) % 10); // 2
nth digit of number from right → Math.floor((number / 10^n-1) % 10)
function getDigit(number, n) {
return Math.floor((number / Math.pow(10, n - 1)) % 10);
}
number of digits in a number → Math.max(Math.floor(Math.log10(Math.abs(number))), 0) + 1 Credit to: https://stackoverflow.com/a/28203456/6917157
function getDigitCount(number) {
return Math.max(Math.floor(Math.log10(Math.abs(number))), 0) + 1;
}
nth digit of number from left or right
function getDigit(number, n, fromLeft) {
const location = fromLeft ? getDigitCount(number) + 1 - n : n;
return Math.floor((number / Math.pow(10, location - 1)) % 10);
}
Get rid of the trailing digits by dividing the number with 10 till the number is less than 100, in a loop. Then perform a modulo with 10 to get the second digit.
if (x > 9) {
while (x > 99) {
x = (x / 10) | 0; // Use bitwise '|' operator to force integer result.
}
secondDigit = x % 10;
}
else {
// Handle the cases where x has only one digit.
}
A "number" is one thing.
The representation of that number (e.g. the base-10 string "1234") is another thing.
If you want a particular digit in a decimal string ... then your best bet is to get it from a string :)
Q: You're aware that there are pitfalls with integer arithmetic in Javascript, correct?
Q: Why is it so important to not use a string? Is this a homework assignment? An interview question?
You know, I get that the question asks for how to do it without a number, but the title "JavaScript: Get the second digit from a number?" means a lot of people will find this answer when looking for a way to get a specific digit, period.
I'm not bashing the original question asker, I'm sure he/she had their reasons, but from a search practicality standpoint I think it's worth adding an answer here that does convert the number to a string and back because, if nothing else, it's a much more terse and easy to understand way of going about it.
let digit = Number((n).toString().split('').slice(1,1))
// e.g.
let digit = Number((1234).toString().split('').slice(1,1)) // outputs 2
Getting the digit without the string conversion is great, but when you're trying to write clear and concise code that other people and future you can look at really quick and fully understand, I think a quick string conversion one liner is a better way of doing it.
function getNthDigit(val, n){
//Remove all digits larger than nth
var modVal = val % Math.pow(10,n);
//Remove all digits less than nth
return Math.floor(modVal / Math.pow(10,n-1));
}
// tests
[
0,
1,
123,
123456789,
0.1,
0.001
].map(v =>
console.log([
getNthDigit(v, 1),
getNthDigit(v, 2),
getNthDigit(v, 3)
]
)
);
This is how I would do with recursion
function getDigits(n, arr=[]) {
arr.push(n % 10)
if (n < 10) {
return arr.reverse()
}
return getDigits(Math.floor(n/10),arr)
}
const arr = getDigits(myVar)
console.log(arr[2])
I don’t know why you need this logic, but following logic will get you the second number
<script type="text/javascript">
var myVal = 58445456;
var var1 = new Number(myVal.toPrecision(1));
var var2 = new Number(myVal.toPrecision(2));
var rem;
rem = var1 - var2;
var multi = 0.1;
var oldvalue;
while (rem > 10) {
oldvalue = rem;
rem = rem * multi;
rem = rem.toFixed();
}
alert(10-rem);
</script>
function getDigit(number, indexFromRight) {
var maxNumber = 9
for (var i = 0; i < indexFromRight - 2; i++) {
maxNumber = maxNumber * 10 + 9
}
if (number > maxNumber) {
number = number / Math.pow(10, indexFromRight - 1) | 0
return number % 10
} else
return 0
}
Just a simple idea to get back any charter from a number as a string or int:
const myVar = 1234;
String(myVar).charAt(1)
//"2"
parseInt(String(myVar).charAt(1))
//2
you can use this function
index = 0 will give you the first digit from the right (the ones)
index = 1 will give you the second digit from the right (the tens)
and so on
const getDigit = (num, index) => {
if(index === 0) {
return num % 10;
}
let result = undefined;
for(let i = 1; i <= index; i++) {
num -= num % 10;
num /= 10;
result = num % 10;
}
return result;
}
for Example:
getDigit(125, 0) // returns 5
gitDigit(125, 1) // returns 2
gitDigit(125, 2) // returns 1
gitDigit(125, 3) // returns 0
function left(num) {
let newarr = [];
let numstring = num.split('[a-z]').join();
//return numstring;
const regex = /[0-9]/g;
const found = numstring.match(regex);
// return found;
for(i=0; i<found.length; i++){
return found[i];
}
}
//}
console.log(left("TrAdE2W1n95!"))
function getNthDigit(n, number){
return ((number % Math.pow(10,n)) - (number % Math.pow(10,n-1))) / Math.pow(10,n-1);
}
Explanation (Number: 987654321, n: 5):
a = (number % Math.pow(10,n)) - Remove digits above => 54321
b = (number % Math.pow(10,n-1)) - Extract digits below => 4321
a - b => 50000
(a - b) / 10^(5-1) = (a - b) / 10000 => 5
var newVar = myVar;
while (newVar > 100) {
newVar /= 10;
}
if (newVar > 0 && newVar < 10) {
newVar = newVar;
}
else if (newVar >= 10 && newVar < 20) {
newVar -= 10;
}
else if (newVar >= 20 && newVar < 30) {
newVar -= 20;
}
else if (newVar >= 30 && newVar < 40) {
newVar -= 30;
}
else if (newVar >= 40 && newVar < 50) {
newVar -= 40;
}
else if (newVar >= 50 && newVar < 60) {
newVar -= 50;
}
else if (newVar >= 60 && newVar < 70) {
newVar -= 60;
}
else if (newVar >= 70 && newVar < 80) {
newVar -= 70;
}
else if (newVar >= 80 && newVar < 90) {
newVar -= 80;
}
else if (newVar >= 90 && newVar < 100) {
newVar -= 90;
}
else {
newVar = 0;
}
var secondDigit = Math.floor(newVar);
That's how I'd do it :)
And here's a JSFiddle showing it works :) http://jsfiddle.net/Cuytd/
This is also assuming that your original number is always greater than 9... If it's not always greater than 9 then I guess you wouldn't be asking this question ;)