E. Beautiful Palindromes

Problem Statement

We call an array $[b_1, b_2, \dots, b_m]$ of length $m$ palindromic if the following condition holds:

$b_i = b_{m-i+1}$ for all $1 \le i \le m$.

In other words, an array is palindromic if it reads the same forward and backward.

You are given an array $[a_1, a_2, \dots, a_n]$ of $n$ integers where $1 \le a_i \le n$ and an integer $k$.

You are required to perform the following operation exactly $k$ times:

1. Choose an integer $x$ such that $1 \le x \le n$.
2. Append $x$ to the end of the array $a$.

Your goal is to perform these $k$ operations in such a way that the number of palindromic subarrays in the resulting array is minimized.

Output the $k$ integers you chose for each operation, in the order they were appended.

A subarray is a contiguous part of an array.

Solution

The core idea is to greedily append integers in a way that avoids creating new palindromic subarrays, especially short ones of length 2 or 3. The strategy depends on whether the initial array a contains all integers from 1 to n.

First, we check if there is any integer in the range $[1, n]$ that is missing from the initial array a. We can do this by inserting all elements of a into a set and then checking for the existence of each integer from 1 to n.

Case 1: An integer is missing

Let’s say we find an integer mex (Minimum Excluded value) that is not in the array a. This mex is a powerful tool to prevent palindrome formation. Since mex does not appear in the original array, any new palindrome must be a subarray composed entirely of the newly appended elements.

To minimize palindromes, we can adopt a strategy where we never place two identical numbers next to each other, or separated by one other number. The algorithm is as follows:

1. For the first operation, we append mex.
2. For each subsequent operation, we choose an integer to append that is different from the last two elements of the array. This prevents the creation of new palindromes of length 2 (like [x, x]) and length 3 (like [y, x, y]). The code iterates from 1 to n to find the first valid integer that satisfies this condition.

This greedy approach effectively minimizes the creation of new, short palindromic subarrays.

Case 2: The array is a permutation of [1, ..., n]

If the initial array a contains all integers from 1 to n, it means a is a permutation. In this scenario, we cannot use a “safe” missing number. Any number we append is already present in the array.

A robust strategy to minimize new palindromes is to append numbers in a repeating, non-palindromic sequence. Since the original array a is a permutation (containing distinct numbers), it provides a good base for such a sequence.

The corrected strategy is to cyclically append the elements of the array a. For the i-th operation (where i goes from 0 to k-1), we append the element a[i % n].

This approach ensures that we are not introducing easily formed palindromes. For example, since all elements in a are distinct, this strategy avoids creating any palindromes of length 2 (like [x, x]) within the appended sequence itself, unless n=1. It also makes it much harder to form longer palindromes that span across the original and appended parts of the array.

#include <algorithm>
#include <cmath>
#include <cstdint>
#include <iostream>
#include <map>
#include <numeric>
#include <set>
#include <tuple>
#include <vector>
using namespace std;
#define int long long
// #define mod 1000000007

#define N 500005
void solve() {
    int n, k;
    cin >> n >> k;
    vector<int> a(n);
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }
    set<int> s(a.begin(), a.end());
    int not_found = -1;
    for (int i = 1; i <= n; i++) {
        if (s.find(i) == s.end()) {
            not_found = i;
            break;
        }
    }
    if (not_found == -1) {
        // If the array is a permutation, append the elements cyclically.
        for (int i = 0; i < k; i++) {
            cout << a[i % n] << " ";
        }
        cout << endl;
        return;
    }
    // if found,
    int pre = a[n-1];
    int cur = not_found;
    cout << cur << " ";
    k --;
    while(k) {
        // try to make next different than cur and pre and within 1 -> n;
        int next = pre;
        for (int i = 1; i <= n; i++) {
            if (i != cur && i != pre) {
                next = i;
                break;
            }
        }
        pre = cur;
        cur = next;
        cout << cur << " ";
        k--;
    }
    cout << endl;

}

int32_t main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
#ifdef DEBUG
    freopen("./input.txt", "r", stdin);
#endif
    int t;
    cin >> t;
    while (t--) {
#ifdef DEBUG
        static int test_num = 1;
        cout << "test case: " << test_num++ << endl;
#endif
        solve();
    }
}