Palindrome Number Problem Explained

A palindrome number is a number that reads the same whether you look at it from left to right or right to left. Take 121, for example. Flip it around and you still get 121. The same goes for 1221, 9009, or even a single digit like 7. Numbers like these have a kind of numerical symmetry that makes them interesting both in mathematics and in computer science.

In math, palindrome numbers show up in some genuinely fascinating places. There is even a subset called palindromic primes, where a number is both prime and a palindrome, like 11, 101, or 131. These are studied in recreational mathematics and number theory. Outside of academic math, palindrome number patterns are used in checksum algorithms, error detection in data transmission, and coding puzzles that test your understanding of digit manipulation.

In coding interviews, the palindrome number problem is a staple for a reason. It looks simple on the surface but forces you to think carefully about edge cases, number properties, and trade-offs between time and space. If you are preparing for FAANG or similar company interviews, problems involving digit manipulation and number properties are common, and this one is a great place to start.

Key Takeaways

Check If the Number Is a Palindrome: Problem Statement

Given an integer, determine whether it is a palindrome. Return true if it is, and false if it is not.

Examples

Also Read: Frequently Asked FAANG DSA Interview Questions You Should Know in 2026

Understanding Palindrome Numbers

Before jumping into code, it helps to understand what makes a number a palindrome at the digit level. When you reverse the digits of a number and get back the same number, it is a palindrome. This is the same idea behind palindrome strings like ‘racecar’ or ‘madam’, just applied to integers.

Here is a simple table showing a few numbers, their reverses, and whether they qualify:

One thing worth noting: a number ending in zero (other than zero itself) can never be a palindrome. For the reversed number to match the original, the first digit would have to be zero, which is not possible in standard integer representation. This is an important edge case that any solid solution needs to handle.

As a fun aside, some palindrome numbers are also prime numbers, making them palindromic primes. Examples include 11, 101, 131, and 313. These sit at the intersection of number theory and combinatorics and are a popular topic in math olympiad problems. For our purposes, though, we are just checking the palindrome property, not whether the number is prime.

Approach Overview

There are three solid ways to solve this problem, each with its own trade-offs. We will walk through all three, so you understand not just how to solve it but why each approach works and when you might prefer one over another.

• Solution 1: Convert the integer to a string, then check if the string is a palindrome. Simple and intuitive, but it uses extra space.
• Solution 2: Extract the second half of the number as an integer, reverse it, and compare with the first half. Uses O(1) extra space.
• Optimized Solution: Access individual digits mathematically without converting or reversing anything. Also, O(1) space, and a great demonstration of digit-level thinking.

All three approaches run in O(d) time, where d is the number of digits in the input. The main difference is in space usage and elegance. In a real interview setting, any of these would be acceptable, but examiners at top companies tend to appreciate the second or third approach since they avoid the string conversion overhead.

Solution 1: Convert to String

How It Works

The simplest approach is to convert the number into a string and then check whether that string reads the same forwards and backwards. To do this, we extract digits one at a time from the right side of the number using modulo and integer division, building the string character by character.

Once we have the string, we compare the first character with the last, the second with the second-to-last, and so on, up to the midpoint. If all pairs match, the number is a palindrome.

Step-by-Step

• If num is negative, return false immediately.
• Extract each digit from right to left using num % 10, add it to the string, then divide num by 10.
• Once all digits are extracted, iterate from index 0 to len/2 and compare str[i] with str[len – i – 1].
• If any pair mismatches, return false. Otherwise, return true.

Code

/*
 * Time: O(d)  |  Auxiliary Space: O(d)  |  Total Space: O(d)
 * d = number of digits in num
 */
bool is_palindrome(int num) {
    if (num < 0) {
        return false;
    }
    string str = "";
    int digit;
    while (num > 0) {
        digit = num % 10;
        str += ('0' + digit);
        num /= 10;
    }
    int len = str.length();
    for (int i = 0; i < len / 2; ++i) {
        if (str[i] != str[len - i - 1]) {
            return false;
        }
    }
    return true;
}

Solution 2: Extract the Second Half

How It Works

This approach avoids the string entirely. Instead of reversing the whole number, we only reverse the second half and compare it to the first half. This sidesteps any potential integer overflow issues that would occur if we reversed a very large number like 1999999999.

We repeatedly extract the rightmost digit from num and build a new integer called sec_half. We stop when sec_half is greater than or equal to num. At that point, num holds the first half and sec_half holds the reversed second half.

For even-digit numbers, num should equal sec_half. For odd-digit numbers, we can discard the middle digit by checking num == sec_half / 10.

Step-by-Step

• If num is 0, return true.
• If num is negative or ends in zero (and is not zero itself), return false.
• Repeatedly extract the last digit of num and append it to sec_half, while removing that digit from num.
• Stop when num <= sec_half.
• Check num == sec_half (even digits) or num == sec_half / 10 (odd digits).

Code

/*
 * Time: O(d)  |  Auxiliary Space: O(1)  |  Total Space: O(1)
 */
bool is_palindrome(int num) {
    if (num == 0) {
        return true;
    }
    if (num < 0 || num % 10 == 0) {
        return false;
    }
    int sec_half = 0;
    int digit;
    while (num > sec_half) {
        digit    = num % 10;
        sec_half = sec_half * 10 + digit;
        num      /= 10;
    }
    return (num == sec_half || num == sec_half / 10);
}

A tricky edge case here is a number like 1100. If we follow the algorithm naively, we eventually get num = 1 and sec_half = 1 (since trailing zeros collapse). The early check for num % 10 == 0 prevents this from incorrectly returning true.

Optimized Solution: Check Without Reversing

How It Works

This is the most mathematically elegant approach. Instead of reversing anything, we directly access digits at any position using arithmetic. For a number with d digits, the i-th digit from the left is: (num / 10^(d – i)) % 10.

Using this formula, we compare the first digit with the last, the second with the second-to-last, and so on until we reach the middle. No string, no reversal, no extra space.

Step-by-Step

• Handle the edge case for negative numbers and zero.
• Calculate d = log10(num) + 1 to find the number of digits.
• Set div1 = 10^(d-1) to extract leftmost digits and div2 = 1 to extract rightmost digits.
• For each pair from outside in, check (num / div1) % 10 == (num / div2) % 10.
• Move div1 inward by dividing by 10 and div2 inward by multiplying by 10.
• If all pairs match, return true.

Code

/*
 * Time: O(d)  |  Auxiliary Space: O(1)  |  Total Space: O(1)
 */
bool is_palindrome(int num) {
    if (num < 0) {
        return false;
    }
    if (num == 0) {
        return true;
    }
    int num_of_digits = log10(num) + 1;
    int div1          = pow(10, num_of_digits - 1);
    int div2          = 1;
    for (int i = 0; i < num_of_digits / 2; i++) {
        if ((num / div1) % 10 != (num / div2) % 10) {
            return false;
        }
        div1 /= 10;
        div2 *= 10;
    }
    return true;
}

Edge Cases

Edge cases are where many candidates slip up in interviews, even if they nail the general logic. Here are the key ones to watch for with the palindrome number problem:

• Negative numbers: -121 reversed is ‘121-‘, which is not the same. All negative numbers return false. This is the first check in every solution.
• Numbers ending in zero: A number like 100 reversed would start with 0, which is not a valid integer. So 100 is not a palindrome. Any positive number whose last digit is zero (except zero itself) is automatically not a palindrome.
• Single-digit numbers: Any number from 0 to 9 is a palindrome by definition since there is nothing to reverse and compare.
• Zero: Zero is a palindrome. It reads the same both ways.
• Large numbers: Numbers near the 32-bit integer limit can cause overflow if you try to reverse the entire number as an integer. Solution 2 avoids this by only reversing half the digits.

Time and Space Complexity

All three approaches run in O(d) time, where d is the number of digits. But they differ in how much memory they use:

For most practical purposes and interview settings, Solutions 2 and 3 are preferred because they use constant auxiliary space.

Visual Explanation

Here is a step-by-step illustration of how Solution 2 works on the number 1221:

Now let’s trace the same approach on 12321 (odd number of digits):

For the optimized solution (Solution 3), think of it like two pointers moving inward. We start from both ends of the number and march toward the center, comparing one digit at a time:

Number:  1  2  3  2  1
         ^           ^   Compare 1 and 1 → Match
            ^      ^     Compare 2 and 2 → Match
               ^          Middle digit (odd length), skip

Result: Palindrome!

Common Mistakes

Here are some of the most common errors people make when implementing this problem:

• Forgetting to handle negative numbers: Not checking for negatives upfront will cause wrong answers on inputs like -121.
• Not handling numbers ending in zero: Forgetting the num % 10 == 0 check (for non-zero numbers) leads to incorrect results on inputs like 1100.
• Integer overflow when reversing the full number: Reversing a number like 1999999999 produces a value that does not fit in a 32-bit integer. Only reversing half the number (Solution 2) avoids this.
• Off-by-one in the loop: When using the two-pointer digit comparison approach, make sure you iterate to num_of_digits / 2 and not beyond, or you will double-check the middle digit.
• Treating zero incorrectly: Zero is a valid palindrome. Make sure your early return conditions do not exclude it.
• Confusing the output type: Some implementations return 0/1 instead of true/false. Know what your function signature expects.

Related Problems

Once you are comfortable with the palindrome number problem, here are some closely related problems worth practicing:

• Palindrome String: Check whether a given string is a palindrome. A natural extension of this problem to character arrays.
• Palindrome Linked List: Determine if a singly linked list is a palindrome. This one requires you to reverse part of the list and adds pointer manipulation to the mix.
• Valid Palindrome: Given a string, consider only alphanumeric characters and ignore cases, then check if it is a palindrome.
• Palindrome Partitioning: Partition a string such that every substring of the partition is a palindrome.

Working through these in sequence builds a strong intuition for symmetry-based problems, which show up regularly in technical interviews.

Also Read: Palindromic Decomposition Of A String Problem

Practice More Coding Problems

The palindrome number problem is a great starting point, but real interview preparation means consistent practice across a range of problem types. Some areas worth focusing on alongside palindromes include:

• Two-pointer problems (they share the same inside-out comparison logic)
• Digit manipulation problems (divisors, modulo, digit extraction)
• String reversal and comparison problems
• Dynamic programming on strings and arrays

Conclusion

The palindrome number problem is one of those problems that rewards careful thinking over raw coding speed. The core idea is simple: a number is a palindrome if it reads the same forward and backwards. But getting there efficiently, handling all the edge cases, and being able to explain your trade-offs is what makes the difference in an interview.

Of the three approaches covered here, the half-reversal method (Solution 2) is the most commonly preferred in interviews. It runs in O(d) time with O(1) space and neatly avoids overflow issues that come with reversing the full number. The no-reversal approach (Solution 3) is equally good and shows a strong grasp of digit mathematics, which is always a plus.

Make sure you can explain why negative numbers and numbers ending in zero are automatically excluded, and practice tracing through the algorithm on a few examples by hand before your next interview. That combination of conceptual clarity and practiced execution is what gets offers at top companies.

FAQs: Palindrome Number

Q1. Is 0 a palindrome number?

Explains why zero is a palindrome and why most implementations need a special case for it.

Q2. Are negative numbers ever palindromes?

Clarifies why the minus sign breaks symmetry and why the answer is always false.

Q3. What is the difference between a palindrome string and a palindrome number?

Highlights the key technical difference in how you access elements in each.

Q4. Can you solve the palindrome number problem without converting to a string?

Walks through both non-string approaches and why they are preferred in interviews.

References

1. Palindromic Number
2. Palindrome Number

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