Medical Parity

The clean way to look at this problem is to stop treating $x$ and $y$ like two freely chosen strings. They are glued together by prefix parity.

For a valid pair, $y_i$ is the parity of the number of ones in $x_1,\dots,x_i$. So when we move from position $i-1$ to position $i$, the parity changes exactly when $x_i=1$.

That gives the important relation:

$$x_i = y_i \oplus y_{i-1}$$

with a fake starting value

$$y_0 = 0$$

because before reading any bits, the count of ones is $0$, which is even.

So if we choose the true parity string $y$, the true test string $x$ is completely forced. No need to guess both. Guessing both would be doing extra cardio for no reason.

Reformulating the cost

We are given recorded strings $x'$ and $y'$. We need the closest valid pair $(x,y)$.

Using the relation above, choosing $y$ gives:

$$x_i = y_i \oplus y_{i-1}$$

At position $i$, the cost is:

• $1$ if the chosen parity bit $y_i$ differs from the recorded bit $y'_i$;
• $1$ if the forced test bit $y_i \oplus y_{i-1}$ differs from the recorded bit $x'_i$.

So the total cost is:

$$\sum_{i=1}^{n} \left([y_i \ne y'_i] + [(y_i \oplus y_{i-1}) \ne x'_i]\right)$$

Now each position only depends on the previous parity bit. That screams two-state DP.

DP definition

Let:

$$dp[b]$$

be the minimum cost after processing the current prefix, assuming the last chosen parity bit is $b$.

Initially, before processing anything, the previous parity is fixed as $0$, so:

$$dp[0] = 0, \quad dp[1] = \infty$$

For each position $i$, we choose the current parity bit $b \in \{0,1\}$ and transition from the previous parity bit $p \in \{0,1\}$.

The extra cost is:

$$[b \ne y'_i] + [(p \oplus b) \ne x'_i]$$

because:

• $b$ is the true $y_i$;
• $p \oplus b$ is the true $x_i$.

Transition:

$$ndp[b] = \min(ndp[b], dp[p] + [b \ne y'_i] + [(p \oplus b) \ne x'_i])$$

After all positions, the last parity can be either $0$ or $1$, so the answer is:

$$\min(dp[0], dp[1])$$

Why this is correct

Every valid pair $(x,y)$ corresponds to exactly one choice of $y$, because $x_i = y_i \oplus y_{i-1}$.

Every binary string $y$ also creates exactly one valid $x$ using that same formula. So we are optimizing over all valid pairs without missing any and without counting invalid nonsense.

The DP checks all possible parity strings implicitly. Since the cost at position $i$ depends only on $y_{i-1}$ and $y_i$, keeping the best cost for each possible last parity bit is enough. Classic shortest-path-on-a-line stuff. Very chill.

Complexity

There are $n$ positions and only $4$ transitions per position.

Time complexity per test case:

$$O(n)$$

Memory complexity:

$$O(1)$$

The total input length is at most $10^6$, so this comfortably fits. The computer will not even break a sweat.