Question:Given an Integer, you need to determine if it is a palindrome or not. You should not use any extra space in the process.

Input:121

Output:Palindrome

At the first go, the problems seems very easy to solve. We just need to reverse the number and check if it remains the same. Right? Don’t be deceived by this problem which seems too easy. Also note the restriction of doing it without extra space. We also need to think of a generic solution that is not language/platform specific.

First, the problem statement did not specify if negative integers qualify as palindromes. Does negative integer such as -1 qualify as a palindrome? Finding out the full requirements of a problem before coding is what every programmer must do. For the purpose of discussion here, we define negative integers as non-palindromes.

The most intuitive approach is to first represent the integer as a string, since it is more convenient to manipulate. Although this certainly does work, it violates the restriction of not using extra space. (ie, you have to allocate n characters to store the reversed integer as string, where n is the maximum number of digits). I know, this sound like an unreasonable requirement (since it uses so little space), but don’t most interview problems have such requirements?

Another approach is to first reverse the number. If the number is the same as its reversed, then it must be a palindrome. You could reverse a number by doing the following:

int reverse(int num) { assert(num >= 0); // for non-negative integers only. int rev = 0; while (num != 0) { rev = rev * 10 + num % 10; num /= 10; } return rev; }

This seemed to work too, but did you consider the possibility that the reversed number might overflow? If it overflows, the behavior is language specific (For Java the number wraps around on overflow, but in C/C++ its behavior is undefined). Yuck.

Of course, we could avoid overflow by storing and returning a type that has larger size than int (ie, long long). However, do note that this is language specific, and the larger type might not always be available on all languages.

We could construct a better and more generic solution. One pointer is that, we must start comparing the digits somewhere. And you know there could only be two ways, either expand from the middle or compare from the two ends.

It turns out that comparing from the two ends is easier. First, compare the first and last digit. If they are not the same, it must not be a palindrome. If they are the same, chop off one digit from both ends and continue until you have no digits left, which you conclude that it must be a palindrome.

Now, getting and chopping the last digit is easy. However, getting and chopping the first digit in a generic way requires some thought. The solution below takes care of it.

int isIntPalindrome(int x) { if (x < 0) return 0; int div = 1; while (x / div >= 10) { div *= 10; } while (x != 0) { int l = x / div; int r = x % 10; if (l != r) return 0; x = (x % div) / 10; div /= 100; } return 1; }

## 6 comments

Why to use two loops? Please reply, am I missing something? I read an article in code recipe that uses just one loop for O(1) space complexity.

Here is the article for your reference: https://www.code-recipe.com/post/palindrome-number

The two loops will not change affect the time complexity. Only nested loops will exponentially increase the complexity.

I think the modulo at line 17 is useless (not tested though)

Try different test cases.

10901 will fail

The code will work perfectly. Check http://ideone.com/O2F4y5

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