Codeforces 1928B Equalize

problem link

Given the feature of numerically adding a permutation to a sequence, elements of different values can become the same as long as their difference is strictly less than $n$.

This conclusion can be easily proven if we construct the worst case scenario: how to turn $n$ consecutive numbers into the same value by adding a permutation. To do that, simply add the permutation in reverse order to achieve unanimous value. Moreover, it is evident that any subproblem of this form (less than $n$ elements within the value range of $n$) can be solved in the same way.

Therefore, we can greedily implement a two-pointer algorithm that maintains an interval where the values differ less than $n$.

Additionally, the above method fails completely when the value of each element is not unique (i.e. duplicates). However, notice that duplicates can never result in the same value after adding a permutation. So, we simply get rid of all duplicates before the starting the above process.

#include
#include
#include
#include
using namespace std;
const int Maxn=2e5+10;
int a[Maxn];
int n;
inline int read()
{
	int s=0,w=1;
	char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')w=-1;ch=getchar();}
	while(ch>='0' && ch<='9')s=(s<<3)+(s<<1)+(ch^48),ch=getchar();
	return s*w;
}
int main()
{
	// freopen("in.txt","r",stdin);
	int T=read();
	while(T--)
	{
		n=read();
		for(int i=1;i<=n;++i)
		a[i]=read();
		sort(a+1,a+1+n);
		int r=1,ans=1;
		int cnt=unique(a+1,a+1+n)-a-1;
		for(int i=1;i<cnt;++i)
		{
			r=max(r,i);
			while(r<cnt && a[r+1]-a[i]<n)++r;
			ans=max(ans,r-i+1);
		}
		printf("%d\n",ans);
	}
	return 0;
}