Kalindrome Array solution codeforces

Kalindrome Array solution codeforces

An array [b1,b2,,bm][b1,b2,…,bm] is a palindrome, if bi=bm+1ibi=bm+1−i for each ii from 11 to mm. Empty array is also a palindrome.

An array is called kalindrome, if the following condition holds:

  • It’s possible to select some integer xx and delete some of the elements of the array equal to xx, 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 xx, and you don’t have to delete at least one element equal to xx.

For example : Kalindrome Array solution codeforces

  • [1,2,1][1,2,1] is kalindrome because you can simply not delete a single element.
  • [3,1,2,3,1][3,1,2,3,1] is kalindrome because you can choose x=3x=3 and delete both elements equal to 33, obtaining array [1,2,1][1,2,1], which is a palindrome.
  • [1,2,3][1,2,3] is not kalindrome.

You are given an array [a1,a2,,an][a1,a2,…,an]. Determine if aa is kalindrome or not.

Input Kalindrome Array solution codeforces

The first line contains a single integer tt (1t1041≤t≤104) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer nn (1n21051≤n≤2⋅105) — the length of the array.

The second line of each test case contains nn integers a1,a2,,ana1,a2,…,an (1ain1≤ai≤n) — elements of the array.

It’s guaranteed that the sum of nn over all test cases won’t exceed 21052⋅105.


For each test case, print YES if aa is kalindrome and NO otherwise. You can print each letter in any case.

Example Kalindrome Array solution codeforces


1 2
1 2 3
1 4 4 1 4

output Kalindrome Array solution codeforces


In the first test case, array [1][1] is already a palindrome, so it’s a kalindrome as well.

In the second test case, we can choose x=2x=2, delete the second element, and obtain array [1][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=4x=4 and delete the fifth element, obtaining [1,4,4,1][1,4,4,1]. You also can choose x=1x=1, delete the first and the fourth elements, and obtain [4,4,4][4,4,4].

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