README.md

1888. Minimum Number of Flips to Make the Binary String Alternating

You are given a binary string s. You are allowed to perform two types of operations on the string in any sequence:

• Type-1: Remove the character at the start of the string s and append it to the end of the string.
• Type-2: Pick any character in s and flip its value, i.e., if its value is '0' it becomes '1' and vice-versa.

Return the minimum number of type-2 operations you need to perform such that s becomes alternating.

The string is called alternating if no two adjacent characters are equal.

• For example, the strings "010" and "1010" are alternating, while the string "0100" is not.

Example 1:

Input: s = "111000"
Output: 2
Explanation: Use the first operation two times to make s = "100011".
Then, use the second operation on the third and sixth elements to make s = "101010".

Example 2:

Input: s = "010"
Output: 0
Explanation: The string is already alternating.

Example 3:

Input: s = "1110"
Output: 1
Explanation: Use the second operation on the second element to make s = "1010".

Constraints:

• 1 <= s.length <= 105
• s[i] is either '0' or '1'.

Solutions (Rust)

1. Solution

impl Solution {
    pub fn min_flips(s: String) -> i32 {
        let mut count = 0;
        let mut ret = s.len();

        for (i, c) in s.chars().enumerate() {
            count += ((i % 2 == 0) ^ (c == '0')) as usize;
        }

        for c in s.chars() {
            if c == '0' {
                count = s.len() - count - s.len() % 2;
            } else {
                count = s.len() - count + s.len() % 2;
            }

            ret = ret.min(count).min(s.len() - count);
        }

        ret as i32
    }
}