Shortest way to form strings

"""
Count length number of subsequences from 'source' to created 'target'
"""
class Solution:
    def solve(self, source: str, target: str) -> int:
        """
        Finds the minimum number of subsequences of 'source' required to form 'target'.
        Returns -1 if 'target' cannot be formed from 'source'.

        Time Complexity: O(len(source) * len(target))
        Space Complexity: O(1)
        """
        def match_subsequence(source_idx: int, target_idx: int) -> int:
            # Attempt to match characters in 'target' starting from 'target_idx'
            # by scanning 'source' from 'source_idx'
            while source_idx < source_len and target_idx < target_len:
                if source[source_idx] == target[target_idx]:
                    target_idx += 1  # Match found, move to next character in 'target'
                source_idx += 1  # Move to next character in 'source'
            return target_idx  # Return the updated index in 'target'

        source_len, target_len = len(source), len(target)
        subsequences_count = 0  # Counts the number of subsequences used
        target_idx = 0  # Current index in 'target'

        while target_idx < target_len:
            new_target_idx = match_subsequence(0, target_idx)
            if new_target_idx == target_idx:
                return -1  # No progress, character not in 'source'
            target_idx = new_target_idx
            subsequences_count += 1  # One subsequence of 'source' used
        return subsequences_count