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 C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>

void swap(long long int* a,long long int* b){long long int c=*a;*a=*b;*b=c;}
int compare_ascending(const void* a, const void* b){
    long long int *A = (long long int *)a;
  long long int *B = (long long int *)b;
  if (*A > *B) return 1;
  if (*A < *B) return -1;
  return 0;
}

int main(){
    int n;
    scanf("%d", &n);
    long long int a[n] , b[n];
    long long int a2[n], b2[n];
    for(int i=0; i<n; i++){
        scanf("%lld", &a[i]);
        a2[i] = a[i];
    }
    for(int i=0; i<n; i++){
        scanf("%lld", &b[i]);
        b2[i] = b[i];
    }
    qsort(a2, n, sizeof(long long int), compare_ascending);
    qsort(b2, n, sizeof(long long int), compare_ascending);
    int id[n];
    int f = 1, in = 0;
    for(int i=0; i<n; i++){
        if(a2[i] > b2[i]){
            f = 0;
            break;
        }
    }
    long long int x = 0;
    for(int i=0; i<n; i++){
        int l = 0, r = n-1;
        int p = (l+r)/2;
        while(l < r){
            if(a2[p] == a[i]){
                break;
            }else if(a2[p] < a[i]){
                l = p+1;
            }else{
                r = p-1;
            }
            p = (l+r)/2;
        }
        if(b[i] != b2[p]){
            x++;
        }
    }
    if((x+1)/2 > n-2){
        f = 0;
    }
    if(a2[n-1] < b2[0]){
        f = 1;
    }
    if(f){
        printf("Yes");
    }else{
        printf("No");
    }


    return 0;
}