Codeforces Round #575 (Div. 3) D2. RGB Substring (hard version) 水題

D2. RGB Substring (hard version)

inputstandard input
outputstandard output
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".

題目大意

q次詢問，每次詢問給你一個n個長度的字符串和長度k，要求你在n長度的字符串裏面找到一個長度爲k的子串，使得修改最少爲RGBRGB....的某個字串

題解

水題，枚舉匹配位置，而後暴力匹配便可，而後再求長度爲k的前綴和最小值

代碼

#include<bits/stdc++.h>
using namespace std;

const int maxn = 2e5+7;
int n,k;
string ori = "RGB";
string s;
int a[maxn],sum[maxn];
int solve(int st) {
  for(int i=0;i<s.size();i++){
    a[i]=(s[i]==ori[(st+i)%3]?0:1);
    sum[i]=(i>0?a[i]+sum[i-1]:a[i]);
  }
  int res = s.size();
  for(int i=k-1;i<s.size();i++){
    res = min(res, sum[i]-(i-k>=0?sum[i-k]:0));
  }
  return res;
}
void solve(){
  scanf("%d%d",&n,&k);
  cin>>s;
  int ans = s.size();
  for(int st=0;st<3;st++){
    ans = min(ans, solve(st));
  }
  printf("%d\n", ans);
}
int main(){
  int t;
  scanf("%d",&t);
  while(t--){
    solve();
  }
  return 0;
}