A15074. I Might Be Wrong

You are given a binary string $S$ of length $n$ indexed from $1$ to $n$ . You can perform the following operation any number of times (possibly zero):

- Choose two integers $l$ and $r$ ( $1 \le l \le r \le n$ ). Let $cnt_0$ be the number of times 0 occurs in $S[l \ldots r]$ and $cnt_1$ be the number of times 1 occurs in $S[l \ldots r]$ . You can pay $|cnt_0 - cnt_1| + 1$ coins and sort the $S[l \ldots r]$ . (by $S[l \ldots r]$ we mean the substring of $S$ starting at position $l$ and ending at position $r$ )

For example if $S = $ 11001, we can perform the operation on $S[2 \ldots 4]$ , paying $|2 - 1| + 1 = 2$ coins, and obtain $S = $ 10011 as a new string.

Find the minimum total number of coins required to sort $S$ in increasing order.

The first line contains a single integer $t$ ( $1 \le t \le 1000$ ) — the number of test cases. The description of test cases follows.

The first line of each test case contains a single integer $n$ ( $1 \le n \le 2 \cdot 10^5$ ) — the size of $S$ .

The second line of each test case contains a binary string $S$ of $n$ characters $S_1S_2 \ldots S_n$ . ( $S_i = $ 0 or $S_i = $ 1 for each $1 \le i \le n$ )

It is guaranteed that the sum of $n$ over all test cases doesn't exceed $2 \cdot 10^5$ .

For each test case, output the minimum total number of coins required to sort $S$ in increasing order.

In the first test case, $S$ is already sorted.

In the second test case, it's enough to apply the operation with $l = 1, r = 2$ .

In the third test case, it's enough to apply the operation with $l = 1, r = 2$ .