carloscn
commented
12 months ago Analysis
pub fn find_the_longest_balanced_substring(s: &str) -> i32
{
let mut ret:i32 = 0;
if s.len() < 1 {
return ret;
}
let mut c_s:Vec<char> = s.chars().collect();
let mut count:usize = 0;
let mut i = 0;
// 00111
while i < c_s.len() {
if c_s[i] != '1' {
i += 1;
count = 0;
continue;
} else {
if i == 0 {
i += 1;
continue;
}
let mut j: i32 = (i - 1) as i32;
while j >= 0 && i < c_s.len() && c_s[j as usize] == '0' && c_s[i] == '1' {
count += 1;
j -= 1;
i += 1;
}
i += 1;
}
ret = ret.max(count as i32);
}
ret *= 2;
return ret;
}
Description
You are given a binary string s consisting only of zeroes and ones.
A substring of s is considered balanced if all zeroes are before ones and the number of zeroes is equal to the number of ones inside the substring. Notice that the empty substring is considered a balanced substring.
Return the length of the longest balanced substring of s.
A substring is a contiguous sequence of characters within a string.
Example 1:
Input: s = "01000111" Output: 6 Explanation: The longest balanced substring is "000111", which has length 6.
Example 2:
Input: s = "00111" Output: 4 Explanation: The longest balanced substring is "0011", which has length 4.
Example 3:
Input: s = "111" Output: 0 Explanation: There is no balanced substring except the empty substring, so the answer is 0.
Constraints:
1 <= s.length <= 50 '0' <= s[i] <= '1'