RGB Substring (hard version)
Contest: Codeforces Round 575 (Div. 3)
Difficulty rating: 1600
Tags: data structures, dp, implementation, two pointers

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 $$$a_1 = b_i$$$, $$$a_2 = b_{i + 1}$$$, $$$a_3 = b_{i + 2}$$$, ..., $$$a_{|a|} = b_{i + |a| - 1}$$$. 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 format:
The first line of the input contains one integer $$$q$$$ ($$$1 \le q \le 2 \cdot 10^5$$$) — the number of queries. Then $$$q$$$ queries follow.

The first line of the query contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le k \le n \le 2 \cdot 10^5$$$) — 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 $$$2 \cdot 10^5$$$ ($$$\sum n \le 2 \cdot 10^5$$$).

Output format:
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 ...".

Constraints:
Time limit: 2.0 s
Memory limit: 256.0 MB

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".

Example input:
3
5 2
BGGGG
5 3
RBRGR
5 5
BBBRR

Example output:
1
0
3

Editorial:
In this problem you should make the same as in the previous one but faster. Let's consider three offsets of string "RGB": "RGB", "GBR" and "BRG". Let's copy the current offset of the string so that it will has the length $$$n$$$ (possibly, without some trailing characters) and save it in the string $$$t$$$.
Then let's compare the string $$$s$$$ with this offset of length $$$n$$$ and build an array $$$diff$$$ of length $$$n$$$ where $$$diff_i = 1$$$ if $$$s_i \ne t_i$$$. Then let's iterate over all possible continuous subsegments of this array $$$diff$$$ and maintain the variable $$$cur$$$ denoting the current answer.
Firstly, for the current position $$$i$$$ let's add $$$diff_i$$$ to $$$cur$$$. Then if the current position $$$i$$$ is greater than or equal to $$$k-1$$$ ($$$0$$$-indexed) let's decrease $$$cur$$$ by $$$diff_{i-k}$$$. So now we have the continuous subsegment of the array $$$diff$$$ of length no more than $$$k$$$. Then if the current position $$$i$$$ is greater than or equal to $$$k$$$ ($$$0$$$-indexed again) (the current subsegment has the length $$$k$$$) then let's update the answer with $$$cur$$$. Then let's do the same with two remaining offsets.