Repetitions Decoding (Codeforces 1641B)
Tags: Others, sortings

Olya has an array of integers $$$a_1, a_2, \ldots, a_n$$$. She wants to split it into tandem repeats. Since it's rarely possible, before that she wants to perform the following operation several (possibly, zero) number of times: insert a pair of equal numbers into an arbitrary position. Help her!More formally:  A tandem repeat is a sequence $$$x$$$ of even length $$$2k$$$ such that for each $$$1 \le i \le k$$$ the condition $$$x_i = x_{i + k}$$$ is satisfied.  An array $$$a$$$ could be split into tandem repeats if you can split it into several parts, each being a subsegment of the array, such that each part is a tandem repeat.  In one operation you can choose an arbitrary letter $$$c$$$ and insert $$$[c, c]$$$ to any position in the array (at the beginning, between any two integers, or at the end).  You are to perform several operations and split the array into tandem repeats or determine that it is impossible. Please note that you do not have to minimize the number of operations. 

Input: Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \le t \le 30\,000$$$) — the number of test cases. Description of the test cases follows.The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 500$$$). The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$)  — the initial array. It is guaranteed that the sum of $$$n^2$$$ over all test cases does not exceed $$$250\,000$$$.

Output: For each test case print answer in the following format.If you cannot turn the array into a concatenation of tandem repeats, print a single integer $$$-1$$$.Otherwise print the number of operations $$$q$$$ ($$$0 \le q \le 2 \cdot n^2$$$) that you want to do. Then print the descriptions of operations.In each of the following $$$q$$$ lines print two integers $$$p$$$ and $$$c$$$ ($$$1 \le c \le 10^9$$$), which mean that you insert the integer $$$c$$$ twice after $$$p$$$ elements of the array. If the length of the array is $$$m$$$ before the operation, then the condition $$$0 \le p \le m$$$ should be satisfied.Then you should print any way to split the resulting array into tandem repeats. First, print a single integer $$$d$$$, and then print a sequence $$$t_1, t_2, \ldots, t_d$$$ of even integers of size $$$d$$$ ($$$d, t_i \ge 1$$$). These numbers are the lengths of the subsegments from left to right.Note that the size of the resulting array $$$a$$$ is $$$m = n + 2 \cdot q$$$. The following statements must hold:   $$$m = \sum\limits_{i = 1}^{d}{t_i}$$$.    For all integer $$$i$$$ such that $$$1 \le i \le d$$$, the sequence $$$a_l, a_{l+1}, \ldots, a_r$$$ is a tandem repeat, where $$$l = \sum\limits_{j = 1}^{i - 1}{t_j} + 1$$$, $$$r = l + t_i - 1$$$. It can be shown that if the array can be turned into a concatenation of tandem repeats, then there exists a solution satisfying all constraints. If there are multiple answers, you can print any.

Note: In the first test case, you cannot apply operations to the array to make it possible to split it into tandem repeats.In the second test case the array is already a tandem repeat $$$[5, 5] = \underbrace{([5] + [5])}_{t_1 = 2}$$$, thus we can do no operations at all.In the third test case, initially, we have the following array: $$$$$$[1, 3, 1, 2, 2, 3].$$$$$$ After the first insertion with $$$p = 1, c = 3$$$: $$$$$$[1, \textbf{3, 3}, 3, 1, 2, 2, 3].$$$$$$ After the second insertion with $$$p = 5, c = 3$$$: $$$$$$[1, 3, 3, 3, 1, \textbf{3, 3}, 2, 2, 3].$$$$$$ After the third insertion with $$$p = 5, c = 3$$$: $$$$$$[1, 3, 3, 3, 1, \textbf{3, 3}, 3, 3, 2, 2, 3].$$$$$$ After the fourth insertion with $$$p = 10, c = 3$$$: $$$$$$[1, 3, 3, 3, 1, 3, 3, 3, 3, 2, \textbf{3, 3}, 2, 3].$$$$$$ The resulting array can be represented as a concatenation of tandem repeats: $$$$$$\underbrace{([1, 3, 3, 3] + [1, 3, 3, 3])}_{t_1 = 8} + \underbrace{([3, 2, 3] + [3, 2, 3])}_{t_2 = 6}.$$$$$$In the fourth test case, initially, we have the following array: $$$$$$[3, 2, 1, 1, 2, 3].$$$$$$ After the first insertion with $$$p = 0, c = 3$$$: $$$$$$[\textbf{3, 3}, 3, 2, 1, 1, 2, 3].$$$$$$ After the second insertion with $$$p = 8, c = 3$$$: $$$$$$[3, 3, 3, 2, 1, 1, 2, 3, \textbf{3, 3}].$$$$$$ After the third insertion with $$$p = 5, c = 3$$$ $$$$$$[3, 3, 3, 2, 1, \textbf{3, 3}, 1, 2, 3, 3, 3].$$$$$$ After the fourth insertion with $$$p = 6, c = 2$$$: $$$$$$[3, 3, 3, 2, 1, 3, \textbf{2, 2}, 3, 1, 2, 3, 3, 3].$$$$$$ After the fifth insertion with $$$p = 7, c = 1$$$: $$$$$$[3, 3, 3, 2, 1, 3, 2, \textbf{1, 1}, 2, 3, 1, 2, 3, 3, 3].$$$$$$ The resulting array can be represented as a concatenation of tandem repeats: $$$$$$\underbrace{([3] + [3])}_{t_1 = 2} + \underbrace{([3, 2, 1] + [3, 2, 1])}_{t_2 = 6} + \underbrace{([1, 2, 3] + [1, 2, 3])}_{t_3 = 6} + \underbrace{([3] + [3])}_{t_4 = 2}.$$$$$$

Sample Inputs:
4
2
5 7
2
5 5
6
1 3 1 2 2 3
6
3 2 1 1 2 3

Sample Outputs:
-1
0
1
2
4
1 3
5 3
5 3
10 3
2
8 6 
5
0 3
8 3
5 3 
6 2 
7 1
4
2 6 6 2

Solution (Python):
def cut(l):
    if not l:
        return [[], []]
    if len(l) & 1:
        return None
    d = {}
    for num in l:
        d[num] = d.get(num, 0) + 1
    for key in d:
        if d[key] & 1:
            return None
    length = len(l)
    if l[:length // 2] == l[length // 2:]:
        return [[], [length]]
    first_index = 0
    second_index = l[1:].index(l[first_index]) + 1
    cut_res = []
    part_length = []
    next_list = []
    for i in range(1, second_index):
        cut_res.append([second_index + i, l[i]])
        next_list.append(l[i])
    part_length.append(2 * second_index)
    next_list.reverse()
    next_list = next_list + l[second_index + 1:]
    (next_cut_res, next_part_length) = cut(next_list)
    for (p, c) in next_cut_res:
        cut_res.append([p + part_length[0], c])
    for next_length in next_part_length:
        part_length.append(next_length)
    return [cut_res, part_length]
t = int(input())
for _ in range(t):
    n = int(input())
    l = list(map(int, input().split()))
    res = cut(l)
    if res == None:
        print(-1)
        continue
    (cut_res, part_length) = res
    print(len(cut_res))
    for (p, c) in cut_res:
        print(p, c, sep=' ')
    print(len(part_length))
    print(' '.join(list(map(str, part_length))))

Solution (Python):
from collections import defaultdict
import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
t = int(input())
ans = []
for _ in range(t):
    n = int(input())
    a = list(map(int, input().split()))
    c = defaultdict(lambda : 0)
    for i in sorted(a):
        c[i] ^= 1
    if max(c.values()):
        ans0 = -1
        ans.append(str(ans0))
        continue
    u = n - 1
    ans0 = []
    s = []
    for _ in range(n // 2):
        au = a[u]
        v = u - 1
        while au ^ a[v]:
            v -= 1
        s.append(2 * (u - v))
        l = u - v - 1
        x = []
        for i in range(1, l + 1):
            x.append((v, a[u - i]))
            v += 1
        for (i, j) in x:
            ans0.append(' '.join(map(str, (i, j))))
            a[i] = j
        u -= 2
    q = len(ans0)
    ans.append(str(q))
    if q:
        ans.append('\n'.join(ans0))
    d = len(s)
    ans.append(str(d))
    ans.append(' '.join(map(str, reversed(s))))
sys.stdout.write('\n'.join(ans))

Solution (Python):
from sys import stdin, stdout
from collections import defaultdict
t = int(stdin.readline())
for _ in range(t):
    n = int(stdin.readline())
    a = [int(z) for z in stdin.readline().split()]
    count = defaultdict(int)
    for i in range(n):
        count[a[i]] += 1
    possible = True
    for (key, value) in count.items():
        if value % 2 == 1:
            possible = False
            break
    if not possible:
        stdout.write('-1\n')
    else:
        pointer = 0
        answer = []
        answer2 = []
        for bar in range(n // 2):
            right = 1
            to_add = []
            while a[right] != a[0]:
                to_add.append(a[right])
                right += 1
            for i in range(right - 1):
                answer.append((pointer + right + 1 + i, to_add[i]))
            pointer += 2 * right
            answer2.append(2 * right)
            a = to_add[::-1] + a[right + 1:]
        stdout.write(str(len(answer)) + '\n')
        for i in range(len(answer)):
            stdout.write('{} {}\n'.format(answer[i][0], answer[i][1]))
        stdout.write(str(len(answer2)) + '\n')
        stdout.write(' '.join(map(str, answer2)) + '\n')