CF2252B Always Changing

You are given a binary string $ s $ of length $ n $ . A string is called alternating if no two adjacent characters are the same. For example, 0101, 1, and 01 are alternating, but 0110 is not. You want to transform $ s $ into an alternating string by performing the following operation any number of times (possibly zero): - Choose any character currently in the string and delete it. However, your sequence of operations must follow a rule: the characters you delete must strictly alternate. This means if the last character you deleted was 0, the next character you delete must be 1, and vice versa. Your very first deleted character can be either 0 or 1. Find the minimum number of operations required to make $ s $ an alternating string. If it is impossible to achieve this, output $ -1 $ .

Each test contains multiple test cases. The first line contains the number of test cases $ t $ ( $ 1 \le t \le 10^4 $ ). The description of the test cases follows. The first line of each test case contains a single integer $ n $ ( $ 1 \le n \le 2 \cdot 10^5 $ ) — the length of the string $ s $ . The second line of each test case contains the binary string $ s $ of length $ n $ , consisting only of the characters 0 and 1. It is guaranteed that the sum of $ n $ over all test cases does not exceed $ 2 \cdot 10^5 $ .

For each test case, output a single integer — the minimum number of operations required to make $ s $ an alternating string, or $ -1 $ if it is impossible.

In the first test case, the string 0101 is already alternating. No operations are needed, so the answer is $ 0 $ . In the second test case, the string is 111. To make it alternating, we must leave at most one 1, which requires deleting two 1s. However, our deletions must strictly alternate between 0 and 1. Since we have no 0s to delete, it is impossible to perform two deletions. The answer is $ -1 $ . In the third test case, the string is 100110. We can simply delete one 0 and one 1 to resolve the adjacent duplicates. Since we deleted exactly one 0 and one 1, the operations are perfectly balanced and valid. The remaining string is 1010, requiring exactly $ 2 $ operations. In the fourth test case, the string is 100010. To resolve the adjacent duplicates, we would naturally want to delete two 0s to form 1010. However, deleting two 0s and zero 1s violates the strict alternation rule (the difference in counts cannot exceed $ 1 $ ). To fix this imbalance, we are forced to additionally delete a 1 from the edge of the string. We delete two 0s and one 1 (e.g., in the order 0, 1, 0), leaving the alternating string 010. This requires $ 3 $ operations. In the fifth test case, the string is 011110. Resolving the adjacent duplicates requires deleting three 1s. To balance this, we must additionally delete at least two 0s. Since our compressed string 010 only has two 0s (at its outer edges), we are forced to delete both of them, leaving a final string of just 1. We deleted three 1s and two 0s, which takes $ 5 $ operations.