Decided to head over to LeetCode and pick the first problem that caught my eye. Here it goes…

Problem Statement

Reverse digits of an integer. The input is assumed to be a 32-bit signed integer. Your function should return 0 when the reversed integer overflows.

Initial Thoughts

I know that I can retrieve the singles digit of any number by doing a modulo operation with that number and 10. I can remove the singles digit from the number by dividing that number by 10 (since the remainder will be trunctated). I should be able to continually do these operations in a loop until I have reversed the number. This should work the same exact way for positive and negative numbers.

The part that stumped me was to return a 0 if an integer overflow occurred. I needed some sort of method that would throw an exception if an overflow occurred. I could not think of a method to use off the top of my head, so I did a quick Google search and was able to find the Math.multiplyExact and Math.addExact methods. These methods will do the arithmetic that I need, and will throw an unchecked Arithmetic Exception if an Integer overflow occurs.

  public int reverse(int x) {
      int reverseNum = 0;
      while(x != 0) {
          try {
              reverseNum = Math.multiplyExact(reverseNum, 10);
              reverseNum = Math.addExact(reverseNum, (x%10));
          } catch(ArithmeticException e) {
              return 0;
          }
          x = x/10;
      }
      return reverseNum;
  }

Complexity

Time: O(log n)  –> Dividing the input by 10 on every loop.

Space: O(1)      –> Constant space. I only declare one variable in this method (reverseNum).