Scale Engineer

A Number After a Double Reversal

Easy
#1929Time: O(log10(num))Space: O(log10(num))
Patterns
Math

Prompt

Reversing an integer means to reverse all its digits.

  • For example, reversing 2021 gives 1202. Reversing 12300 gives 321 as the leading zeros are not retained.

Given an integer num, reverse num to get reversed1, then reverse reversed1 to get reversed2. Return true if reversed2 equals num. Otherwise return false.

 

Example 1:

Input: num = 526
Output: true
Explanation: Reverse num to get 625, then reverse 625 to get 526, which equals num.

Example 2:

Input: num = 1800
Output: false
Explanation: Reverse num to get 81, then reverse 81 to get 18, which does not equal num.

Example 3:

Input: num = 0
Output: true
Explanation: Reverse num to get 0, then reverse 0 to get 0, which equals num.

 

Constraints:

  • 0 <= num <= 106

Approaches

3 approaches with complexity analysis and trade-offs.

This approach directly simulates the process described in the problem statement. It involves converting the number to a string, reversing it, converting it back to an integer, and repeating the process. The final result is then compared with the original number.

Algorithm

  1. Define a helper function reverse(int n) that takes an integer and returns its reversed form.
  2. Inside the helper function: a. Convert the integer n to its string representation. b. Create a StringBuilder from the string. c. Use the reverse() method of StringBuilder. d. Convert the reversed StringBuilder back to a string. e. Parse the string back into an integer. This step automatically handles the dropping of leading zeros.
  3. In the main function isSameAfterReversals(int num): a. Call the helper function to get the first reversal: reversed1 = reverse(num). b. Call the helper function again on the result to get the second reversal: reversed2 = reverse(reversed1). c. Compare reversed2 with the original num and return true if they are equal, false otherwise.

Walkthrough

This method translates the problem's logic into code in the most literal way. We create a helper function that performs the reversal of a single number. This function first converts the integer to a String. Then, it uses a StringBuilder to easily reverse the sequence of characters. Finally, it parses the reversed string back into an integer. The Integer.parseInt() method conveniently ignores any leading zeros, which perfectly matches the problem's requirement (e.g., reversing 12300 gives 321, and parsing the string "00321" would also yield 321). The main function then orchestrates this process, calling the reversal function twice and comparing the final number with the original.

class Solution {    public boolean isSameAfterReversals(int num) {        int reversed1 = reverse(num);        int reversed2 = reverse(reversed1);        return reversed2 == num;    }     private int reverse(int n) {        String s = Integer.toString(n);        StringBuilder sb = new StringBuilder(s);        sb.reverse();        return Integer.parseInt(sb.toString());    }}

Complexity

Time

O(log10(num))

Space

O(log10(num))

Trade-offs

Pros

  • The logic is very straightforward and easy to follow as it directly mirrors the problem description.

  • It leverages powerful built-in Java library functions, which can make the code concise.

Cons

  • This is the least efficient approach due to the overhead of object creation (String, StringBuilder) and type conversions.

  • It consumes more memory compared to purely mathematical solutions.

Solutions

class Solution {public  boolean isSameAfterReversals(int num) { return num == 0 || num % 10 != 0; }}

Video walkthrough

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