Live Coding Challenge | Episode 3 | Problem Solving with Turing

The third episode of "Problem Solving with Turing", takes you through a different problem and dissects it step-by-step to help you understand the correct approach to solving coding problems that will help you ace coding challenges and crack technical interviews with ease.

Become a Turing software developer today.

[Fanaro]

Why not just turn it to a string, put it in a list and then reverse the list?

[barthsparks]

You could just convert to string and split, check if the first element is '-'. If it is, reverse the list, join, convert to number and multiply by -1 and return the product else just return without multiplying by -1

[JuanSB]

I think the part that checks for the overflow is kinda wrong cuz it relies on integers outside the 32-bit integer range

[tmstechschool]

Temesgen from Ethiopia using C#
int ReverseInt(int a)
{
int sign = 1;
if (a < 0)
sign = -1;
a = a * sign;
int result = 0;
while (a > 0)
{
result = result * 10 + a % 10;
a = a / 10;
}
return result * sign;
}

[KateNelson-jygz]

How can this be solved on JavaScript: Imagine a circular jogging track with several rest points, labeled from 0 to n-1. the distance between each rest point is known and record in an array called distance [], where distance [i] shows the distance between point a and point (i+1)% n. a person wants to find the minimum distance between two chosen rest points, start and destination, on the circular route, while being able to travel in both directions, clockwise and counterclockwise.

[BraulioRojas-tfpv]

Jose, the error you got with the ranges is because of the use of parseInt, you're going above/below the valid numbers for integers, if you use Math.floor, that will do the trick for any number.
Also, you forgot to cover the negatives.
Here's a solution.

function reverseInteger(number) {
let numberToConvert = Math.abs(number)
let result = 0
while (numberToConvert > 0) {
const rest = numberToConvert % 10
numberToConvert = Math.floor(numberToConvert / 10)
result = (result * 10) + rest
}

if (result < (2 ** 31 * -1)) return 0
if (result > (2 ** 31 - 1)) return 0

return (number < 0) ? -result: result
}

[ameerhamza]

Make you made noise just present in peaceful environment

[birukabel]

Haven't compiled it but possible C# solution
public int ReverseInteger(int num)
{
lstNums = new
int rem, coffeicinet;
coffeicinet=num/10;
rem=num%10;
int exponent=0;
int returnedNum=0;
while(coffeicinet>=10)
{
exponent++;
lstNums.Add(rem);
rem=coffeicinet%10;
coffeicinet=coffeicinet/10;
}
lstNums.Add(coffeicinet);

for(int i=0;i<lstNums.Count && exponent>=0;i++, exponent--)
{
returnedNum+=lstNums[i]*Math.Pow(10, exponent);
}

return returnedNum;
}

[hikmetqedirov]

please add Kotlin language your coding challenge.

[NeniEmSu]

I solved it like so, * (Math.abs(num) === num ? +1 : -1)