Bessie the cow, always a fan of shiny objects, has taken up a hobby of mining diamonds in her spare
time! She has collected N diamonds (N≤50,000) of varying sizes, and she wants to arrange some of them
in a pair of display cases in the barn.Since Bessie wants the diamonds in each of the two cases to
be relatively similar in size, she decides that she will not include two diamonds in the same case
if their sizes differ by more than K (two diamonds can be displayed together in the same case if their
sizes differ by exactly K). Given K, please help Bessie determine the maximum number of diamonds
she can display in both cases together.
The first line of the input file contains N and K (0≤K≤1,000,000,000). The next NN lines each cont
ain an integer giving the size of one of the diamonds. All sizes will be positive and will not excee
d 1,000,000,000
Output a single positive integer, telling the maximum number of diamonds that Bessie can showcase in
total in both the cases.
#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;
int n,m,i,r,a[50005],f[50005],g[50005],ans;
int main()
{
scanf("%d%d",&n,&m);
for(i=1;i<=n;i++)
scanf("%d",&a[i]);
sort(a+1,a+n+1);
r=1;
for(i=1;i<=n;i++)
{
while(a[r+1]-a[i]<=m&&r<n) r++;
f[i]=r-i+1;
}
for(i=n;i>=1;i--)
g[i]=max(g[i+1],f[i]);
ans=0;
for(i=1;i<=n;i++)
ans=max(ans,f[i]+g[i+f[i]]);
printf("%d",ans);
return 0;
}
