Minimum number of flips to make the binary string alternating

impl Solution {
  pub fn min_flips(s: String) -> i32 {
    let n = s.len();
    // Work on doubled string
    let doubled: Vec<u8> = s.bytes().chain(s.bytes()).collect();
    let len2 = doubled.len();

    // For each position in doubled string, mismatch with pattern "010101..."
    // and with pattern "101010..."
    // Pattern 0 expected: 0 at even, 1 at odd -> expected = i%2
    // Pattern 1 expected: 1 at even, 0 at odd -> expected = (i+1)%2

    let mut mismatch0 = 0i32; // mismatches with pattern starting with 0
    let mut mismatch1 = 0i32; // mismatches with pattern starting with 1
    let mut ans = i32::MAX;

    // Initialize window [0..n)
    for i in 0..n.min(len2) {
      let b = if doubled[i] == b'1' { 1 } else { 0 };
      let exp0 = (i % 2) as i32; // 0101...
      let exp1 = ((i + 1) % 2) as i32; // 1010...
      if b != exp0 { mismatch0 += 1; }
      if b != exp1 { mismatch1 += 1; }
    }
    ans = ans.min(mismatch0.min(mismatch1));

    // Slide window
    for i in n..len2 {
      // Add new element at i
      let b_new = if doubled[i] == b'1' { 1 } else { 0 };
      let exp0_new = (i % 2) as i32;
      let exp1_new = ((i + 1) % 2) as i32;
      if b_new != exp0_new { mismatch0 += 1; }
      if b_new != exp1_new { mismatch1 += 1; }

      // Remove element at i-n
      let remove = i - n;
      let b_old = if doubled[remove] == b'1' { 1 } else { 0 };
      let exp0_old = (remove % 2) as i32;
      let exp1_old = ((remove + 1) % 2) as i32;
      if b_old != exp0_old { mismatch0 -= 1; }
      if b_old != exp1_old { mismatch1 -= 1; }

      ans = ans.min(mismatch0.min(mismatch1));
    }

    ans
  }
}