Answer: parse_json()

The question asks which Snowflake function interprets an input string as a JSON document and produces a VARIANT value. The PARSE_JSON() function is specifically designed for this purpose - it takes a string containing JSON data and converts it into a VARIANT data type, which is Snowflake's semi-structured data type for handling JSON. The community discussion strongly supports this with 100% consensus on answer A, multiple comments referencing the official Snowflake documentation for PARSE_JSON(), and high upvote counts on comments confirming A as correct. Other options are less suitable: json_extract_path_text() extracts specific elements from existing JSON, object_construct() builds JSON objects from key-value pairs, and flatten() explodes arrays or objects into rows - none of these parse strings into VARIANT JSON documents.

Author: LeetQuiz Editorial Team