## Problem Statement

http://fruth.com/wp-content/2015/6/smap.html Given a 32-bit signed integer, determine its reverse. Since we have signed integer as input we can have negative values, so for Input => -123 Output => -321, also need to take care of Overflow cases which should results in Output => 0 (Invalid).

## Solution

follow link The first idea that comes to mind is to convert the number into string and reverse the string using builtin reverse function. We also have to take care of properly handling negative inputs. Overflow case is also handled by comparing resulting number with Integer min and max value.

enter site def reverse(x: Int): Int = { val isNegative = if (x < 0) 1 else 0 val y = if (isNegative == 1) (x.toLong*(-1)).toString.reverse else x.toString.reverse if (y.toLong > Int.MaxValue || y.toLong < Int.MinValue ) // Overflow case 0 else { if (isNegative == 1) y.toInt * (-1) else y.toInt } }

Now this approach requires using builtin string function reverse which might not be allowed, so lets think about another approach. Lets take this number `1234`

, if we do `1234 % 10`

, we get the last digit `4`

, to get the second to the last digit, we need to remove the last digit from `1234`

, we could do so by dividing it by 10, `1234 / 10 = 123`

. Then we can get the last digit again by doing a modulus by 10, `123 % 10 = 3`

, and if we multiply the last digit by 10 and add the second last digit, `4 * 10 + 3 = 43`

, it gives us the reverted number we want. Continuing this process would give us the reverted number.

def reverse(x: Int): Int = { var output : Long = 0; var input = x // method parameters are always VAL, hence convert it into VAR while(input != 0){ output = output * 10 + input % 10; input = input/10; if( output > Integer.MAX_VALUE || output < Integer.MIN_VALUE) // Overflow case return 0; } output.toInt }

## Test Cases

reverse(123456) reverse(-123) Int.MaxValue Int.MinValue Long.MaxValue.toString reverse(2147483641) reverse(2147483647) reverse(-2147483648)

## Complexity Analysis

- Time complexity : O(log
_{10}n). We divided the input by 10 for every iteration, so the time complexity is O(log_{10}n) - Space complexity : O(1)