Swap Adjacent Elements CodeForces - 920C

Swap Adjacent Elements

CodeForces - 920C

You have an array a consisting of n integers. Each integer from 1 to n appears exactly once in this array.

For some indices i (1 ≤ i ≤ n - 1) it is possible to swap i-th element with (i + 1)-th, for other indices it is not possible. You may perform any number of swapping operations any order. There is no limit on the number of times you swap i-th element with (i + 1)-th (if the position is not forbidden).

Can you make this array sorted in ascending order performing some sequence of swapping operations?

Input

The first line contains one integer n (2 ≤ n ≤ 200000) — the number of elements in the array.

The second line contains n integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 200000) — the elements of the array. Each integer from 1 to n appears exactly once.

The third line contains a string of n - 1 characters, each character is either 0 or 1. If i-th character is 1, then you can swap i-th element with (i + 1)-th any number of times, otherwise it is forbidden to swap i-th element with (i + 1)-th.

Output

If it is possible to sort the array in ascending order using any sequence of swaps you are allowed to make, print YES. Otherwise, print NO.

Example

Input

6
1 2 5 3 4 6
01110

Output

YES

Input

6
1 2 5 3 4 6
01010

Output

NO

Note

In the first example you may swap a₃ and a₄, and then swap a₄ and a₅.

题意：

给你一串数字序列和一串0,1组成的字符串。如果数字序列中的元素所在下标在字符串中为‘1’，则可以和i+1交换。求是否可以得到递增的数字序列。

思路：

根据冒泡排序可知，任何一个无序序列都可以通过相邻元素交换变成有序的。那么问题就简单了。

如果当前数字不能交换，则判断它是否等于其下标+1,；

否则判断找出这一段‘1’的右边界（‘0’或 '\0'），判断这一段区间内存储的数字是否在下标区间内。如果在区间外则输出no。将j的值赋给i。

code：

#include 
#include 
#include 
using namespace std;
int n,a[200005];
char str[200005];
int main(){
    scanf("%d",&n);
    memset(str,0,sizeof(str));
    for(int i = 0; i < n; i++)
        scanf("%d",&a[i]);
    scanf("%s",str);
    int i,j;
    for(i = 0; i < n; i++){
       if(str[i] == '1'){//如果是1可交换，一直往后到0第一个不可交换的数或字符串结束为止
          int maxx = 0;//记录可交换区间内最大数
          for(j = i; ; j++){
             if(a[j] < i+1){//小于左边界说明不在区间内不符合
                printf("NO\n");
                return 0;
             }
             maxx = max(maxx,a[j]);
             if(str[j] == '0' || str[j] == 0){
                if(maxx > j+1){//如果到了右边界，且区间内的最大值超出右边界不符合
                    printf("NO\n");
                    return 0;
                }
                i = j;
                break;
             }
          }
       }
       else if(a[i] != i+1){//如果是0不可交换，那么这个数必须等于下标+1，否则不对
          printf("NO\n");
          return 0;
       }
    }
    printf("YES\n");
    return 0;
}