Codeforces #497(div.2) C. Reorder the Array

C. Reorder the Array

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given an array of integers. Vasya can permute (change order) its integers. He wants to do it so that as many as possible integers will become on a place where a smaller integer used to stand. Help Vasya find the maximal number of such integers.

For instance, if we are given an array $[10,20,30,40]$, we can permute it so that it becomes $[20,40,10,30]$. Then on the first and the second positions the integers became larger ($20>10$, $40>20$) and did not on the third and the fourth, so for this permutation, the number that Vasya wants to maximize equals $2$. Read the note for the first example, there is one more demonstrative test case.

Help Vasya to permute integers in such way that the number of positions in a new array, where integers are greater than in the original one, is maximal.

Input

The first line contains a single integer $n$ ($1\le n\le {10}^5$) — the length of the array.

The second line contains $n$ integers $a_1,a_2,\dots ,a_n$ ($1\le a_i\le {10}^9$) — the elements of the array.

Output

Print a single integer — the maximal number of the array's elements which after a permutation will stand on the position where a smaller element stood in the initial array.

Examples

input

Copy

7
10 1 1 1 5 5 3

output

Copy

4

input

Copy

5
1 1 1 1 1

output

Copy

0

Note

In the first sample, one of the best permutations is $[1,5,5,3,10,1,1]$. On the positions from second to fifth the elements became larger, so the answer for this permutation is 4.

In the second sample, there is no way to increase any element with a permutation, so the answer is 0.

数组里的数字可以任意更换位置

问以最优方式替换 

最多有多少个数字在替换后比替换前大

先排序 

双指针往后指

后指针指的元素比前指针大 

答案++

后指针指到头 

不可替换 结束

代码 ↓

#include <iostream>
#include <algorithm>
using namespace std;

const int MAXN = 1e5 + 10;

int arr[MAXN];

int main()
{
	int N, ans = 0;
	cin>>N;
	
	for(int i = 0; i < N; i++)
		cin>>arr[i];
		
	sort(arr, arr + N);
	
	for(int i = 0, j = 1; i < N; i++)
	{
		while(arr[i] >= arr[j] && j <= N - 1)
			j++;
		
		if(j == N)
			break;
		
		ans ++;
		j++;
		
	}
	
	cout<<ans<<endl;
	
	return 0;
}