Codeforces Global Round 17 |
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Finished |
An array $$$[b_1, b_2, \ldots, b_m]$$$ is a palindrome, if $$$b_i = b_{m+1-i}$$$ for each $$$i$$$ from $$$1$$$ to $$$m$$$. Empty array is also a palindrome.
An array is called kalindrome, if the following condition holds:
It's possible to select some integer $$$x$$$ and delete some of the elements of the array equal to $$$x$$$, so that the remaining array (after gluing together the remaining parts) is a palindrome.
Note that you don't have to delete all elements equal to $$$x$$$, and you don't have to delete at least one element equal to $$$x$$$.
For example :
You are given an array $$$[a_1, a_2, \ldots, a_n]$$$. Determine if $$$a$$$ is kalindrome or not.
The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the length of the array.
The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le n$$$) — elements of the array.
It's guaranteed that the sum of $$$n$$$ over all test cases won't exceed $$$2 \cdot 10^5$$$.
For each test case, print YES if $$$a$$$ is kalindrome and NO otherwise. You can print each letter in any case.
4 1 1 2 1 2 3 1 2 3 5 1 4 4 1 4
YES YES NO YES
In the first test case, array $$$[1]$$$ is already a palindrome, so it's a kalindrome as well.
In the second test case, we can choose $$$x = 2$$$, delete the second element, and obtain array $$$[1]$$$, which is a palindrome.
In the third test case, it's impossible to obtain a palindrome.
In the fourth test case, you can choose $$$x = 4$$$ and delete the fifth element, obtaining $$$[1, 4, 4, 1]$$$. You also can choose $$$x = 1$$$, delete the first and the fourth elements, and obtain $$$[4, 4, 4]$$$.
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