Codeforces1009C——Reorder the Array

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≤n≤105) — the length of the array.
The second line contains n integers a1,a2,…,an (1≤ai≤109) — 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
7
10 1 1 1 5 5 3
Output
4
Input
5
1 1 1 1 1
Output
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.

问重排数列之后比之前大的数的位置最多有多少个
我的思路是从大到小排序，然后就双指针贪心的移动

代码：

#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int N=100020;
int a[N];
bool cmp(int a,int b){
    return a>b;
}
int main(void){
    int n;
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        scanf("%d",&a[i]);
    }
    sort(a,a+n,cmp);
    int cnt=0;
    for(int i=0,j=0;i<n && j<n;){
        if(a[j]>a[i]){
            cnt++;
            i++;
            j++;
        }
        else{
            i++;
        }
    }
    printf("%d\n",cnt);
    return 0;
}