CF 575div3 E

The only difference between easy and hard versions is the size of the input.

You are given a string s consisting of n characters, each character is ‘R’, ‘G’ or ‘B’.

You are also given an integer k. Your task is to change the minimum number of characters in the initial string s so that after the changes there will be a string of length k that is a substring of s, and is also a substring of the infinite string “RGBRGBRGB …”.

A string a is a substring of string b if there exists a positive integer i such that a1=bi, a2=bi+1, a3=bi+2, …, a|a|=bi+|a|−1. For example, strings “GBRG”, “B”, “BR” are substrings of the infinite string “RGBRGBRGB …” while “GR”, “RGR” and “GGG” are not.

You have to answer q independent queries.

Input
The first line of the input contains one integer q (1≤q≤2⋅105) — the number of queries. Then q queries follow.

The first line of the query contains two integers n and k (1≤k≤n≤2⋅105) — the length of the string s and the length of the substring.

The second line of the query contains a string s consisting of n characters ‘R’, ‘G’ and ‘B’.

It is guaranteed that the sum of n over all queries does not exceed 2⋅105 (∑n≤2⋅105).

Output
For each query print one integer — the minimum number of characters you need to change in the initial string s so that after changing there will be a substring of length k in s that is also a substring of the infinite string “RGBRGBRGB …”.

Example
Input
3
5 2
BGGGG
5 3
RBRGR
5 5
BBBRR
Output
1
0
3
Note
In the first example, you can change the first character to ‘R’ and obtain the substring “RG”, or change the second character to ‘R’ and obtain “BR”, or change the third, fourth or fifth character to ‘B’ and obtain “GB”.

In the second example, the substring is “BRG”.
给有n个字母的一个字符串，问得到一个k长度的有序字符串最少修改几次
分三种情况，第一个字母为B，为R或者为G
除了最开始的1到k-1个字母之外
每次更新i到i+k-1只用查询i-1和i+k-1是否符合就可以了

#include<stdio.h>
#include<algorithm>
using namespace std;
char ss[200005];
int a[200005];
int q,n,k;
int main()
{
	scanf("%d",&q);
	while(q--)
	{
		scanf("%d%d",&n,&k);
		scanf("%s",ss);
		for(int i=0;i<n;i++)
		{
			if(ss[i]=='R')
			a[i]=0;
			else if(ss[i]=='G')
			a[i]=1;
			else if(ss[i]=='B')
			a[i]=2;
		}
		int ans1=0,ans2=0,ans3=0;
		for(int i=0;i<k;i++)
		{
			if(a[i]!=(a[0]+i)%3)
			ans1++;
		}	
		int ans=ans1;	
		for(int i=1;i<=n-k;i++)
		{
			if(a[i-1]!=(a[0]+i-1)%3)
			ans1--;
			if(a[k+i-1]!=(a[0]+k+i-1)%3)
			ans1++;
			ans=min(ans1,ans);
		}
		
		for(int i=0;i<k;i++)
		{
			if(a[i]!=(a[0]+i+1)%3)
			ans2++;
		}ans=min(ans2,ans);		
		for(int i=1;i<=n-k;i++)
		{
			if(a[i-1]!=(a[0]+i)%3)
			ans2--;
			if(a[k+i-1]!=(a[0]+k+i)%3)
			ans2++;
			ans=min(ans2,ans);
		}
		
		for(int i=0;i<k;i++)
		{
			if(a[i]!=(a[0]+i+2)%3)
			ans3++;
		}ans=min(ans3,ans);		
		for(int i=1;i<=n-k;i++)
		{
			if(a[i-1]!=(a[0]+i+1)%3)
			ans3--;
			if(a[k+i-1]!=(a[0]+k+i+1)%3)
			ans3++;
			ans=min(ans3,ans);
		}
		printf("%d\n",ans);
		
    }
}