K-Step Bit Flip with Retention

You are given a binary string S of length n and a positive integer k. You must apply exactly k transformations to S, one after another. Think of the transformations as being labeled with the values k, k-1, k-2, ..., 1 (the first transformation applied is labeled k, the last is labeled 1). For the transformation labeled with value v, applied to the current string: if v is odd, find the leftmost '1' in the current string — this position is the 'retained' position. If v is even, find the leftmost '0'. If the string has no character of the required type (odd's turn but the string is all '0's, or even's turn but all '1's), retain the last index of the string instead. Then produce the new string by flipping every character ('0' → '1', '1' → '0') except the retained position, which stays unchanged. Repeat for all k labeled transformations (k down to 1) and return the final string. Constraints: 1 ≤ len(s) ≤ 100; 1 ≤ k ≤ 100000. Examples: s = "1010", k = 1 → "1101"; k = 2 → "1111"; k = 3 → "1110"; k = 4 → "1111". Return the final string exactly; there is no alternate output ordering.

def k_step_bit_flip(s: str, k: int) -> str:
    """Apply labels k down to 1, retaining the required leftmost bit."""
    pass