RGB Substring (hard version)
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".
给出长度为n的字符串，最小修改几次，达到使其中k位满足“RGB”的循环
dp转移方程：dp[i][j] 由 dp[i-1][j] 状态转移
长度为i，以 R G B结尾的最小修改次数
#include
#include
#include
#include
#include
#include
#include
#define sf(a) scanf("%d",&a)
#define rep(i,a,b) for(i=a;i<=b;i++)
#define mod 1e9+7
#define e 1e-8
using namespace std;
typedef long long ll;
const int inf=0x3f3f3f3f;
const int idata=5e5+5;
int i,j,k;
int judge,flag;
//vector step(idata);
//ll step[idata];
int dp[idata][3];
char ch[idata];
char cmp[4]="RGB";
ll n,m,t;
int maxx=0,minn=inf;
ll cnt,len,sum,ans;
int main()
{
cin>>t;
while(t--)
{
cin>>n>>k;
cin>>ch+1;
for(i=1;i<=n;i++)
{
for(j=0;j<3;j++)
{
if(ch[i]==cmp[(i+j)%3])//i为前面经过的字符个数，j为以 R G B为开头的判断循环
dp[i][j]=dp[i-1][j];
else
dp[i][j]=dp[i-1][j]+1;
}
}
minn=inf;
for(i=0;i<=n-k;i++)
for(j=0;j<3;j++)
minn=min(minn,dp[i+k][j]-dp[i][j]);
cout<