Menorah
Contest: Codeforces Global Round 18
Difficulty rating: 1600
Tags: brute force, graphs, greedy, math

There are $$$n$$$ candles on a Hanukkah menorah, and some of its candles are initially lit. We can describe which candles are lit with a binary string $$$s$$$, where the $$$i$$$-th candle is lit if and only if $$$s_i=1$$$.

Initially, the candle lights are described by a string $$$a$$$. In an operation, you select a candle that is currently lit. By doing so, the candle you selected will remain lit, and every other candle will change (if it was lit, it will become unlit and if it was unlit, it will become lit).

You would like to make the candles look the same as string $$$b$$$. Your task is to determine if it is possible, and if it is, find the minimum number of operations required.

Input format:
The first line contains an integer $$$t$$$ ($$$1\le t\le 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow.

The first line of each test case contains a single integer $$$n$$$ ($$$1\le n\le 10^5$$$) — the number of candles.

The second line contains a string $$$a$$$ of length $$$n$$$ consisting of symbols 0 and 1 — the initial pattern of lights.

The third line contains a string $$$b$$$ of length $$$n$$$ consisting of symbols 0 and 1 — the desired pattern of lights.

It is guaranteed that the sum of $$$n$$$ does not exceed $$$10^5$$$.

Output format:
For each test case, output the minimum number of operations required to transform $$$a$$$ to $$$b$$$, or $$$-1$$$ if it's impossible.

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

Note:
In the first test case, the two strings are already equal, so we don't have to perform any operations.

In the second test case, we can perform a single operation selecting the second candle to transform $$$01$$$ into $$$11$$$.

In the third test case, it's impossible to perform any operations because there are no lit candles to select.

In the fourth test case, we can perform the following operations to transform $$$a$$$ into $$$b$$$:

1. Select the $$$7$$$-th candle: $$$100010{\color{red}1}11\to 011101{\color{red} 1}00$$$.
2. Select the $$$2$$$-nd candle: $$$0{\color{red} 1}1101100\to 1{\color{red} 1}0010011$$$.
3. Select the $$$1$$$-st candle: $$${\color{red}1}10010011\to {\color{red}1}01101100$$$.

In the fifth test case, we can perform the following operations to transform $$$a$$$ into $$$b$$$:

1. Select the $$$6$$$-th candle: $$$00101{\color{red}1}011\to 11010{\color{red}1}100$$$
2. Select the $$$2$$$-nd candle: $$$1{\color{red}1}0101100\to 0{\color{red}1}1010011$$$
3. Select the $$$8$$$-th candle: $$$0110100{\color{red}1}1\to 1001011{\color{red}1}0$$$
4. Select the $$$7$$$-th candle: $$$100101{\color{red}1}10\to 011010{\color{red}1}01$$$

Example input:
5
5
11010
11010
2
01
11
3
000
101
9
100010111
101101100
9
001011011
011010101

Example output:
0
1
-1
3
4

Editorial:
First, let's define the "type" of a candle $$$i$$$ to be the string $$$a_ib_i$$$. For example, if a candle is currently unlit and must be lit in the end, its type is $$$01$$$. It's useful to think about the number of candles of each type because the position of a candle is irrelevant in this problem.
Now, let's consider what happens when we do two operations consecutively. All candles except the ones we select will flip twice, so let's focus on what happens to the candles we select. If we select the same candle twice, nothing changes. And if we select two different candles, it's equivalent to swapping a $$$1$$$ and a $$$0$$$ in the string.
Since we have a really nice description of two consecutive operations, any sequence of an even number of operations is just a sequence of swaps. So it's possible in an even number of operations as long as the number of $$$1$$$'s in both strings is the same, and the minimum number of operations will be the number of positions where the strings differ.
Now, what about an odd number of operations? Well, it's the same as performing one operation and then reducing it to the case of an even number of operations. There are two types of candles we can select on the first operation: type $$$10$$$ and type $$$11$$$. Simply try both options if they're present, and find the minimum number of operations after it's reduced to the even case.
Finally, there are some strings where it's possible to do in either an even or an odd number of operations, so in those cases we have to take the minimum of both options. The total complexity is $$$O(n)$$$.
Bonus: Can you prove that in the case of an odd number of operations, it is never necessary to select a candle of type $$$10$$$?