You are a coach at your local university. There are
n
n
students under your supervision, the programming skill of the
i
i
-th student is
a
i
ai
.
You have to create a team for a new programming competition. As you know, the more students some team has the more probable its victory is! So you have to create a team with the maximum number of students. But you also know that a team should be balanced. It means that the programming skill of each pair of students in a created team should differ by no more than
5
5
.
Your task is to report the maximum possible number of students in a balanced team.
Input
The first line of the input contains one integer
n
n
(
1≤n≤2⋅
10
5
1≤n≤2⋅105
) — the number of students.
The second line of the input contains
n
n
integers
a
1
,
a
2
,…,
a
n
a1,a2,…,an
(
1≤
a
i
≤
10
9
1≤ai≤109
), where
a
i
ai
is a programming skill of the
i
i
-th student.
Output
Print one integer — the maximum possible number of students in a balanced team.
Examples
Input
Copy
6
1 10 17 12 15 2
Output
Copy
3
Input
Copy
10
1337 1337 1337 1337 1337 1337 1337 1337 1337 1337
Output
Copy
10
Input
Copy
6
1 1000 10000 10 100 1000000000
Output
Copy
1
题解:题意给n个数,求子集的最小值和最大值之差不超过5,求最大的子集的元素数.
我的思路:先对数组排序,这里如果直接暴力的话,O(n^2)的复杂度,肯定会tle,
优化一下,因为是排过序的,所以子集的左端和右端都是极值,如果右值与左值之差大于5,左值的位置右移,
再次优化,如果前一值与后一值之差大于5,I 移动的后一值的位置.
#include <bits/stdc++.h>
const int N=2e5+5;
using namespace std;
int a[N],b[N];
int ans=0,cnt=0,flag=0,maxn=0;
void check(int m){
if(m>maxn) maxn=m;
}
int cal(int i,int j){
return a[j]-a[i];
}
int main(){
int n;
scanf("%d",&n);
for(int i=1;i<=n;i++){
scanf("%d",&a[i]);
}
sort(a+1,a+1+n);
int i=1,j=i+1;
while(i<=n-1&&j<=n){
if(cal(j-1,j)>5){
check(j-1-i);
i=j;
j++;
}
else if(cal(i,j)>5){
check(j-1-i);
i++;
if(i==j) j++;
//cout<<"jjg"<<endl;
}
else{
j++;
//cout<<"jj"<<endl;
}
//if(!flag) break;
}
check(j-1-i);
printf("%d\n",maxn+1);
return 0;
}