Cyclic Rotation
There is an array $$$a$$$ of length $$$n$$$. You may perform the following operation any number of times:  Choose two indices $$$l$$$ and $$$r$$$ where $$$1 \le l < r \le n$$$ and $$$a_l = a_r$$$. Then, set $$$a[l \ldots r] = [a_{l+1}, a_{l+2}, \ldots, a_r, a_l]$$$. You are also given another array $$$b$$$ of length $$$n$$$ which is a permutation of $$$a$$$. Determine whether it is possible to transform array $$$a$$$ into an array $$$b$$$ using the above operation some number of times.
In the first test case, we can choose $$$l=2$$$ and $$$r=5$$$ to form $$$[1, 3, 3, 2, 2]$$$.In the second test case, we can choose $$$l=2$$$ and $$$r=4$$$ to form $$$[1, 4, 2, 2, 1]$$$. Then, we can choose $$$l=1$$$ and $$$r=5$$$ to form $$$[4, 2, 2, 1, 1]$$$.In the third test case, it can be proven that it is not possible to transform array $$$a$$$ to $$$b$$$ using the operation.
Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$)  — the number of test cases. The description of the test cases follows.The first line of each test case contains an integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10 ^ 5$$$)  — the length of array $$$a$$$ and $$$b$$$.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 $$$a$$$.The third line of each test case contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$1 \le b_i \le n$$$)  — elements of the array $$$b$$$.It is guaranteed that $$$b$$$ is a permutation of $$$a$$$.It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10 ^ 5$$$
For each test case, print "YES" (without quotes) if it is possible to transform array $$$a$$$ to $$$b$$$, and "NO" (without quotes) otherwise.You can output "YES" and "NO" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response).
def solve():
    n = int(input())
    a = list(map(int, input().split()))
    b = list(map(int, input().split()))
    ok = True
    reserve = [0] * (n + 1)
    (i, j) = (n - 1, n - 1)
    while j >= 0:
        while j > 0 and b[j] == b[j - 1]:
            reserve[b[j]] += 1
            j -= 1
        if a[i] == b[j]:
            i -= 1
            j -= 1
        else:
            if not reserve[a[i]]:
                ok = False
                break
            reserve[a[i]] -= 1
            i -= 1
    return ok
t = int(input())
for _ in range(t):
    print('YES' if solve() else 'NO')
import sys
input = sys.stdin.readline

def readList():
    return list(map(int, input().split()))

def readInt():
    return int(input())

def readInts():
    return map(int, input().split())

def readStr():
    return input().strip()

def solve():
    (n, a, b) = (readInt(), readList(), readList())
    (cnt, j) = ([0] * (n + 1), 0)
    for i in range(n + 1):
        while j < n and cnt[b[j]] and (b[j - 1] == b[j]):
            cnt[b[j]] -= 1
            j += 1
        if i == n:
            return 'YES' if j == n else 'NO'
        if a[i] == b[j]:
            j += 1
        else:
            cnt[a[i]] += 1
for _ in range(int(input())):
    print(solve())
from collections import defaultdict

def solve():
    n = int(input())
    array1 = list(map(int, input().split()))
    array2 = list(map(int, input().split()))
    if array1[-1] != array2[-1]:
        return 'No'
    diff_count = defaultdict(int)
    (i, j) = (n - 2, n - 2)
    while i >= 0 and j >= 0:
        if array1[i] == array2[j]:
            i -= 1
            j -= 1
        elif array2[j] == array2[j + 1]:
            diff_count[array2[j]] += 1
            j -= 1
        elif diff_count[array1[i]] > 0:
            diff_count[array1[i]] -= 1
            i -= 1
        else:
            return 'No'
    return 'Yes'
testCases = int(input())
while testCases > 0:
    print(solve())
    testCases -= 1