Let $s$ and $t$ denote the two strings. Define $a_1,\dots,a_m$ to be the positions of the $1$s in $s$, and $b_1,\dots,b_m$ similarly for $t$.

In this problem, we only care about dance moves involving both types of cows. These dance moves are equivalent to moving a $1$ (i.e., a Holstein) to a position not containing a $1$ (i.e., a position containing a Guernsey).

**Claim 1:** The answer is at least the minimum value of
$\sum_{i=1}^m\left\lceil\frac{|a_i-c_i|}{K}\right\rceil$ over all permutations
$c$ of $b$. This summation represents the minimum cost to match $1$s in $s$ with
$1$s in $t$. The cost to pair up two $1$s equals the ceiling of the distance
between them divided by $K$.

*Proof:* Every $1$ in $s$ must pair up with (i.e., end at the same location
as) some $1$ in $t$. To move a $1$ a distance of $d$, at least
$\lceil d/K\rceil$ dance moves are required, as each dance move moves a $1$ by a
distance of at most $K$.

In fact, we will later show that claim 1 holds with equality.

*SUBTASK 1: $K=1$*

**Claim 1.5:** The answer is $\sum_{i=1}^m|a_i-b_i|$.

*Proof:* Using an exchange argument, it can be shown that
$\sum_{i=1}^m|a_i-c_i|$ is minimized when $b=c$. So the answer is at least
$\sum_{i=1}^m|a_i-b_i|$. Furthermore, we can construct a sequence of swaps with
exactly this length:

- While $a\neq b$, take the smallest $i$ such that
$a_i\neq b_i$:
- If $a_i>b_i$, move the $1$ at $a_i$ one position to the left.
- Otherwise, move the $1$ in $b_i$ one position to the left.

- Repeat until $a=b$.

Implementation 1 in Python:

from collections import deque N, K = map(int, input().split()) s = input() t = input() a = [i for i in range(N) if s[i] == "1"] b = [i for i in range(N) if t[i] == "1"] print(sum(abs(x - y) for x, y in zip(a, b)))

Another summation that gives the same answer is:

$$\sum_{i=1}^N|(\text{# ones in }s\text{ with position }\le i)-(\text{# ones in }t\text{ with position }\le i)|$$

We can compute this summation by sweeping $i$ from left to right.
Implementation 2 in Python:

from collections import deque N, K = map(int, input().split()) s = input() t = input() ans = 0 dif = 0 for i in range(N): dif += s[i] == "1" dif -= t[i] == "1" ans += abs(dif) print(ans)

As we will see later on, the full solution will look quite similar to implementation 2.

*SUBTASK 2: $m \leq 8$*

Assuming Claim 1 holds with equality, we can iterate over all permutations $c$ of $b$ and compute the expression from claim 1 in $O(m!\cdot m)$ time.

*FULL SOLUTION*

Suppose we draw an arrow $a_i\to c_i$ for each $i$. We say that the arrow has
length $|a_i-c_i|$, and points **rightward** if $a_i<c_i$, and
**leftward** otherwise. Two arrows point **in opposite directions** if one
points leftward and the other rightward. The $i$-th arrow spans the interval
$[\min(a_i,c_i),\max(a_i,c_i)]$, and we say that two arrows **cross** if
their intervals intersect.

**Claim 2:** There is a permutation $c$ of $b$ minimizing
$\sum_{i=1}^m\left\lceil\frac{|a_i-c_i|}{K}\right\rceil$ such that no two
opposite-direction arrows cross each other.

*Proof:* If two opposite-direction arrows cross, we can swap two entries of
$c$ to uncross them without increasing the summation:

- If one arrow completely covers the other, we need to verify that $\lceil (x+y)/K\rceil + 1\ge \lceil x/K\rceil+\lceil y/K\rceil$. Indeed, this inequality is true for all $x,y\ge 0$
- Otherwise, uncrossing them decreases the lengths of both arrows, so the summation does not increase.

We can repeatedly uncross opposite-direction arrow pairs until no such pairs exist. Because the sum of lengths of all arrows strictly decreases each time we uncross two arrows, this process must eventually terminate.

**Claim 3:** Claim 1 holds with equality.

*Proof:* Take $c$ satisfying claim 2 and partition the positions into
independent substrings where no two arrows point in
opposite directions within each substring.
Now consider one of those substrings. Without loss of generality, assume all
arrows point rightward within the substring (i.e., $a_i\le c_i$ for all $i$).

We use induction. Take any such $c$ minimizing the number of $i$ such that $a_i=c_i$, and consider the maximum $i$ satisfying $a_i<c_i$. We can move $a_i$ to the right in such a way that $\sum_i\left\lceil\frac{|a_i-c_i|}{K}\right\rceil$ decreases by one:

- If $a_i$ and $c_i$ are within $K$ of each other, then $c_i$ must contain a zero, and we can move a one from $a_i$ to $c_i$ in one move.
- Otherwise, move $a_i$ to $c_i+K$, which must contain a zero. (If it doesn't, then there is a permutation of $c$ with fewer $i$ such that $a_i=c_i$, which is a contradiction.) Then, apply the inductive hypothesis.

It remains to describe how to compute the expression from claim 1 efficiently. One (overpowered) way to pass subtask 3 is to use the Hungarian algorithm for minimum-cost matching, which runs in $O(N^3)$ time. However, there is a much simpler way using a greedy algorithm.

Firstly, there is a minimum-cost matching pairing all $a_i$ and $b_j$ such that $a_i=b_j$. It remains to pair two types of positions:

**Type 1 positions**, where $s_i = 1$ and $t_i = 0$.**Type 2 positions**, where $s_i = 0$ and $t_i = 1$.

In the case where all arrows point rightward, every type 1 position is paired with a type 2 position to the right of it, and we can use the following greedy strategy that sweeps from left to right:

- Maintain a list of type 1 positions not yet paired.
- Iterate over all $i$ from $1$ to $N$.
- For each position in the list, if it is less than $i$, add $K$ to it and increase the answer by one, because it must be moved rightward at least once more.
- If $i$ is type 1, add it to the list of type 1 positions to be matched.
- If $i$ is type 2, we can pair it with any type 1 position from the list. To minimize the cost, we should pair it with the minimum element from the list. (This statement can be proven formally using an exchange argument.)

The time complexity is $O(N^2/K)$ if step 2.1 is implemented naively, or $O(N)$ if the list is maintained as a queue consisting of pairs of the form {value, count} in strictly increasing order of value. When all arrows point leftward, the solution is the same except types 1 and 2 are swapped.

The $O(N)$ implementation follows. Note that it is written in such a way such that it handles both leftward and rightward arrows symmetrically.

My Python code:

from collections import deque N, K = map(int, input().split()) s = input() t = input() d = deque() ans = 0 for i in range(N): if len(d) > 0 and d[0][0] < i: # step 2.1 ans += abs(d[0][1]) d.append((d[0][0] + K, d[0][1])) d.popleft() if s[i] != t[i]: dif = 1 if s[i] == "1" else -1 if len(d) == 0 or abs(d[0][1] + dif) > abs(d[0][1]): # 2.2 if not (len(d) > 0 and d[0][0] == i): d.appendleft((i, 0)) d[0] = d[0][0], d[0][1] + dif else: # 2.3 d[0] = d[0][0], d[0][1] + dif if d[0][1] == 0: d.popleft() print(ans)

$O(N\log N)$ solutions using a sorted map rather than a deque were also fast enough to pass.

Andi Qu's C++ code:

#include <iostream> #include <map> typedef long long ll; using namespace std; int main() { cin.tie(0)->sync_with_stdio(0); int n, k; cin >> n >> k; string s1, s2; cin >> s1 >> s2; ll ans = 0; map<int, int> mismatches; for (int i = 0; i < n; i++) { int d = s1[i] - s2[i]; // d = 0 if match; +1 if 0|1; -1 if 1|0 // First, move any mismatches *before* the sweep-line forward // and increment the answer accordingly while (mismatches.size() && mismatches.begin()->first < i) { pair<int, int> curr = *mismatches.begin(); mismatches[curr.first + k] += curr.second; ans += abs(curr.second); mismatches.erase(curr.first); } // Next, resolve any reachable mismatches from the current position if (mismatches.size() && mismatches.begin()->second * d < 0) { mismatches.begin()->second += d; if (mismatches.begin()->second == 0) mismatches.erase(mismatches.begin()); } else if (d) mismatches[i] += d; } cout << ans << '\n'; return 0; }