Give a string `s`

, count the number of non-empty (contiguous) substrings that have the same number of 0's and 1's, and all the 0's and all the 1's in these substrings are grouped consecutively.

Substrings that occur multiple times are counted the number of times they occur.

**Example 1:**

Input:"00110011"Output:6Explanation:There are 6 substrings that have equal number of consecutive 1's and 0's: "0011", "01", "1100", "10", "0011", and "01".

Notice that some of these substrings repeat and are counted the number of times they occur.

Also, "00110011" is not a valid substring becauseallthe 0's (and 1's) are not grouped together.

**Example 2:**

Input:"10101"Output:4Explanation:There are 4 substrings: "10", "01", "10", "01" that have equal number of consecutive 1's and 0's.

**Note:**

`s.length`

will be between 1 and 50,000.`s`

will only consist of "0" or "1" characters.```
struct Solution;
impl Solution {
fn count_binary_substrings(s: String) -> i32 {
let mut prev: usize = 0;
let mut curr: usize = 0;
let mut curr_c: Option<char> = None;
let mut res = 0;
for c in s.chars() {
if let Some(cc) = curr_c {
if cc == c {
curr += 1;
} else {
res += usize::min(prev, curr);
curr_c = Some(c);
prev = curr;
curr = 1;
}
} else {
curr_c = Some(c);
curr = 1;
}
}
res += usize::min(prev, curr);
res as i32
}
}
#[test]
fn test() {
let s = "00110011".to_string();
assert_eq!(Solution::count_binary_substrings(s), 6);
let s = "10101".to_string();
assert_eq!(Solution::count_binary_substrings(s), 4);
}
```