Problem:
Score : 600 points

Problem Statement

Given are two sequences A and B, both of length N.
A and B are each sorted in the ascending order.
Check if it is possible to reorder the terms of B so that for each i (1 \leq i \leq N) A_i \neq B_i holds, and if it is possible, output any of the reorderings that achieve it.

Constraints

1\leq N \leq 2 \times 10^5

1\leq A_i,B_i \leq N

A and B are each sorted in the ascending order.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 \cdots A_N
B_1 B_2 \cdots B_N

Output

If there exist no reorderings that satisfy the condition, print No.

If there exists a reordering that satisfies the condition, print Yes on the first line.
After that, print a reordering of B on the second line, separating terms with a whitespace.

If there are multiple reorderings that satisfy the condition, you can print any of them.

Sample Input 1

6
1 1 1 2 2 3
1 1 1 2 2 3

Sample Output 1

Yes
2 2 3 1 1 1

Sample Input 2

3
1 1 2
1 1 3

Sample Output 2

No

Sample Input 3

4
1 1 2 3
1 2 3 3

Sample Output 3

Yes
3 3 1 2

Solution in Java (OpenJDK 11.0.6) (judge verdict: Wrong Answer)

import java.io.*;
import java.util.*;

public class Main {
    public static void main(String[] args) throws Exception {
        BufferedReader buffer = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(buffer.readLine());
        String [] inp = buffer.readLine().split(" ");
        Integer []arr = new Integer[n];
        int [] arr2 = new int[n];
        for (int pos = 0; pos < n; pos++) {
            arr[pos] = Integer.parseInt(inp[pos]);
        }
        inp = buffer.readLine().split(" ");
        for (int pos = 0; pos < n; pos++) {
            arr2[pos] = Integer.parseInt(inp[pos]);
        }
        TreeMap<Integer, Integer>freqForArr2 = new TreeMap<>();
        for (int i : arr2) {
            freqForArr2.put(i, freqForArr2.getOrDefault(i, 0)+1);
        }
        boolean check = true;
        for (int pos = 0; pos < n; pos++) {
            Integer higherKey = freqForArr2.higherKey(arr[pos]), lowerKey = freqForArr2.lowerKey(arr[pos]), key = -1;
            if (higherKey == null && lowerKey == null){
                check = false;
                break;
            }
            if (higherKey != null && higherKey > arr[pos])
                key = higherKey;
            else if (lowerKey < arr[pos])
                key = lowerKey;
            else {
                check = false;
                break;
            }
            freqForArr2.put(key, freqForArr2.get(key)-1);
            if (freqForArr2.get(key) == 0)
                freqForArr2.remove(key);
            arr2[pos] = key;
        }
        if (!check)
            System.out.println("No");
        else {
            System.out.println("Yes");
            for (int num : arr2) {
                System.out.print(num + " ");
            }
        }
    }
}