K-periodic Garland
Contest: Codeforces Round 642 (Div. 3)
Difficulty rating: 1900
Tags: brute force, dp, greedy

You are given a garland consisting of $$$n$$$ lamps. States of the lamps are represented by the string $$$s$$$ of length $$$n$$$. The $$$i$$$-th character of the string $$$s_i$$$ equals '0' if the $$$i$$$-th lamp is turned off or '1' if the $$$i$$$-th lamp is turned on. You are also given a positive integer $$$k$$$.

In one move, you can choose one lamp and change its state (i.e. turn it on if it is turned off and vice versa).

The garland is called $$$k$$$-periodic if the distance between each pair of adjacent turned on lamps is exactly $$$k$$$. Consider the case $$$k=3$$$. Then garlands "00010010", "1001001", "00010" and "0" are good but garlands "00101001", "1000001" and "01001100" are not. Note that the garland is not cyclic, i.e. the first turned on lamp is not going after the last turned on lamp and vice versa.

Your task is to find the minimum number of moves you need to make to obtain $$$k$$$-periodic garland from the given one.

You have to answer $$$t$$$ independent test cases.

Input format:
The first line of the input contains one integer $$$t$$$ ($$$1 \le t \le 25~ 000$$$) — the number of test cases. Then $$$t$$$ test cases follow.

The first line of the test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 10^6; 1 \le k \le n$$$) — the length of $$$s$$$ and the required period. The second line of the test case contains the string $$$s$$$ consisting of $$$n$$$ characters '0' and '1'.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^6$$$ ($$$\sum n \le 10^6$$$).

Output format:
For each test case, print the answer — the minimum number of moves you need to make to obtain $$$k$$$-periodic garland from the given one.

Constraints:
Time limit: 1.0 s
Memory limit: 256.0 MB

Example input:
6
9 2
010001010
9 3
111100000
7 4
1111111
10 3
1001110101
1 1
1
1 1
0

Example output:
1
2
5
4
0
0

Editorial:
Let $$$t_i$$$ be the string containing all characters of $$$s$$$ that have indices $$$i, i + k, i + 2k$$$ and so on (i.e. all such positions that have the remainder $$$i$$$ modulo $$$k$$$). Suppose we choose that all turned on lamps will have remainder $$$i$$$ modulo $$$k$$$. Then we need to remove all ones at the positions that do not belong to this remainder. Also considering the string $$$t_i$$$, we need to spend the minimum number of moves to make this string of kind "contiguous block of zeros, contiguous block of ones and again contiguous block of zeros", because considering the characters modulo $$$k$$$ will lead us to exactly this pattern (notice that some blocks can be empty).
How to calculate the answer for the string $$$t_i$$$ in linear time? Let $$$dp_i[p]$$$ be the number of moves we need to fix the prefix of $$$t_i$$$ till the $$$p$$$-th character in a way that the $$$p$$$-th character of $$$t_i$$$ is '1'. Let $$$cnt(S, l, r)$$$ be the number of ones in $$$S$$$ on the segment $$$[l; r]$$$. Notice that we can calculate all required values $$$cnt$$$ in linear time using prefix sums. Then we can calculate $$$dp_i[p]$$$ as $$$min(cnt(t_i, 0, p - 1) + [t_i[p] \ne~ '1'], dp_i[p-1] + [t_i[p] \ne~ '1'])$$$, where $$$[x]$$$ is the boolean value of the expression $$$x$$$ ($$$1$$$ if $$$x$$$ is true and $$$0$$$ otherwise). Let $$$len(S)$$$ be the length of $$$S$$$. Then the actual answer for the string $$$t_i$$$ can be calculated as $$$ans_i = \min(cnt(t_i, 0, len(t_i) - 1), \min\limits_{p=0}^{len(t_i) - 1} (dp_i[p] + cnt(t_i, p + 1, len(t_i) - 1)))$$$ (thus we consider the case when the obtained string doesn't contan ones at all and consider each position as the last position of some one).
So the actual answer can be calculated as $$$\min\limits_{i=0}^{k-1} (ans_i + cnt(s, 0, len(s) - 1) - cnt(t_i, 0, len(t_i) - 1))$$$.
Time complexity: $$$O(n)$$$.