Palindrome Number — Level 2

A developer-focused deep dive into palindrome checking in production. Learn about Integer Overflow, Memory Pressure, and why the "Half-Reversal" strategy is professional-grade.

Author

Mr. Oz

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8 mins

Level 2

Memory overflow and O(1) space complexity efficiency comparison

Author

Mr. Oz

Date

Read

8 mins

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In a production environment, Palindrome Number is rarely about the "Reverse a String" trick. Interviewers use this problem to see if you understand Integer Overflow and Memory Pressure.

  1. Why String Conversion is a "Junior" Move

    There are three main reasons why converting to a String is problematic in production:

    Heap Allocation

    Integer.toString(n) creates a new String object. Strings are immutable in Java. This lands on the Heap.

    Garbage Collection Pressure

    If you are running this check inside a loop processing millions of transactions per second (like a high-frequency trading bot or a sensor data aggregator), you are creating millions of short-lived objects. The Garbage Collector (GC) will eventually have to "Stop the World" to clean them up, causing latency spikes.

    Array Overhead

    A String is backed by a byte[] or char[]. There is header overhead for the object and the array length.

  2. The "Half-Reversal" Professional Strategy

    The most robust way is reversing only half of the integer. But why half?

    Because if you reverse a 32-bit integer completely, the reversed version might exceed 2,147,483,647 (the max value of a 32-bit signed int). By stopping at the middle, we guarantee the number stays within the bounds of a standard int.

    public boolean isPalindrome(int x) {
        // Negative numbers are not palindromes
        // Numbers ending in 0 (except 0 itself) are not palindromes
        if (x < 0 || (x % 10 == 0 && x != 0)) {
            return false;
        }
        
        int revertedNumber = 0;
        while (x > revertedNumber) {
            revertedNumber = revertedNumber * 10 + x % 10;
            x /= 10;
        }
        
        // For even-length: x == revertedNumber
        // For odd-length: x == revertedNumber / 10 (middle digit ignored)
        return x == revertedNumber || x == revertedNumber / 10;
    }
  3. Big O and the User Experience

    Time: O(log N)

    This is incredibly fast. For a 10-digit number (the max for a 32-bit int), the loop only runs about 5-10 times.

    Space: O(1)

    This is the "holy grail." It means the memory usage is constant regardless of how big the number is.

    On a mobile device with limited RAM or an embedded system (like a smart fridge), this is the difference between a smooth app and a crash.

  4. Edge Cases to Handle

    • Negative numbers — Always return false (the minus sign breaks symmetry)
    • Numbers ending in 0 — Can't be palindromes (except 0 itself), e.g., 10, 100, 1000
    • Single digit numbers — Always palindromes (0-9)
    • Odd vs even length — Handle the middle digit for odd-length numbers

Tips and Pitfalls

  • Don't reverse the entire number — you risk Integer Overflow.
  • The String approach works for interviews but fails production scrutiny.
  • Consider using long if you need to handle larger numbers.
  • Benchmark with JMH if this is on a hot path in your application.