Problem:
Score : 600 points

Problem Statement

Given are two integer sequences of N elements each: A_1,...,A_N and B_1,...,B_N.
Determine if it is possible to do the following operation at most N-2 times (possibly zero) so that, for every integer i from 1 to N, A_i \leq B_i holds:

Choose two distinct integers x and y between 1 and N (inclusive), and swap the values of A_x and A_y.

Constraints

2 \leq N \leq 10^5

1 \leq A_i,B_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N
B_1 B_2 ... B_N

Output

If the objective is achievable, print Yes; if it is not, print No.

Sample Input 1

3
1 3 2
1 2 3

Sample Output 1

Yes

We should swap the values of A_2 and A_3.

Sample Input 2

3
1 2 3
2 2 2

Sample Output 2

No

Sample Input 3

6
3 1 2 6 3 4
2 2 8 3 4 3

Sample Output 3

Yes

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

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

public class Main {
    public static void main(String[] args) {
        FastScanner scanner = new FastScanner();
        PrintWriter out = new PrintWriter(System.out, false);
        int n = scanner.nextInt();
        int[] a = new int[n], aa = new int[n], b = new int[n], bb = new int[n];
        for(int i = 0; i < n; i++) {
            a[i] = scanner.nextInt();
            aa[i] = a[i];
        }
        for(int i = 0; i < n; i++) {
            b[i] = scanner.nextInt();
            bb[i] = b[i];
        }
        Arrays.sort(aa);
        Arrays.sort(bb);
        //viability check after sorting
        for(int i = 0; i < n; i++) {
            if (aa[i] > bb[i]) {
                System.out.println("No");
                return;
            }
        }
        //whether we can do this with n-2 naive swaps
        for(int i = 1; i < n; i++) {
            if (aa[i] <= bb[i-1]) {
                System.out.println("Yes");
                return;
            }
        }
        //create a mapping of the sorted elts into 1...n
        HashMap<Integer, Integer> mapb = new HashMap<>();
        HashMap<Integer, Integer> mapa = new HashMap<>();
        for(int i = 0; i < n; i++) {
            mapa.put(aa[i], i);
            mapb.put(bb[i], i);
        }
        //get the permutation compared the what is needed for b;
        int[] perm = new int[n];
        for(int i = 0; i < n; i++) {
            perm[mapb.get(b[i])] = mapa.get(a[i]);
        }
        boolean[] vis = new boolean[n];
        int cnt = 0;
        for(int i = 0; i < n; i++) {
            if (vis[i]) continue;
            int cur = i;
            do {
                vis[cur] = true;
                cur = perm[cur];
            } while(!vis[cur]);
            cnt++;
        }
        //issue if all in one cycle, would require n-1 swaps
        if (cnt == 1) System.out.println("No");
        else System.out.println("Yes");
        out.flush();
    }
    public static class FastScanner {
        BufferedReader br;
        StringTokenizer st;
        
        public FastScanner(Reader in) {
            br = new BufferedReader(in);
        }
        
        public FastScanner() {
            this(new InputStreamReader(System.in));
        }
        
        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        
        int nextInt() {
            return Integer.parseInt(next());
        }
        
        long nextLong() {
            return Long.parseLong(next());
        }
        
        double nextDouble() {
            return Double.parseDouble(next());
        }
        
        String readNextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
}