# I don't feel like this is a easy question

• This question just doesn't sound like an easy one to me... ╮(╯﹏╰)╭

• Agreed. In fact when I saw the problem and saw that it was "Easy" I was thinking there must be an easy way. Eventually I gave up because I could only think to do the "assumed max palindrome" solution (which ain't easy!).

• @tankztc I think so！

• The first step for me is going to check what's palindrome...

• This question just doesn't sound like an easy one to me... ╮(╯﹏╰)╭

LOL, neither do I.

• @memo40k lol Same

• lol. How can this be easy?

• Thank you for starting this thread! I immediately felt the same way when I saw the Easy label! Self esteem restored!

• Me too chile.

• @memo40k
Palindrome happens so much in Leetcode. like a hundred of them.

• Seems that if n is an even number, you can guess out the largest palindromic number.

n=2:
101*99=99,99; take out the middle 99, which is 990/99=10, then it is (101-10)*99=91*99=9009

n=4:
10001*9999=9999,9999, take out the middle 9999, which is 999900/9999=100, then it is (10001-100)*9999=9901*9999=99000099

and so on.

• To some extent, there's an O(1) solution...................

• This is hard even compared to most LeetCode "Hard"s. One of the hardest questions on the site imo.

• Come on~here we are in leetcode,lol

• I don't know why LeetCode mark this problem as easy.

I almost tried 2 hours to solve it and failed.

• @tankztc The reason why this is easy is that there is a straight forward solution, even though that might have the worst case complexity. O(10^n)^2.

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