Snowflake DAA-C01 - SnowPro Advanced: Data Analyst Exam

To determine the correct output, a Data Analyst must understand the behavior of the STRTOK_SPLIT_TO_TABLE table function and how it interacts with different delimiters.

1. Functionality of STRTOK_SPLIT_TO_TABLE:

The STRTOK_SPLIT_TO_TABLE function tokenizes a string based on a set of specified delimiters. Unlike simple SPLIT functions that often look for a literal string as a delimiter, STRTOK treats the delimiter string as a set of individual characters. Any character present in the delimiter string will trigger a split. Crucially, it skips empty tokens—if multiple delimiters appear consecutively, it does not produce a row for the space between them.

2. Evaluating the Data and Delimiters ('*-'):

'JOHN*DOE': The * character is a delimiter. This results in two tokens: JOHN and DOE.

'ALICE-SMITH': The - character is a delimiter. This results in two tokens: ALICE and SMITH.

'BOB*-JOHNSON': Both * and - are delimiters. Because they appear consecutively (*-), the function splits at each but ignores the empty space between them. This results in two tokens: BOB and JOHNSON.

3. Analyzing the Resulting Rows:

The LATERAL join produces a row for every token generated. The columns SEQ (identifying the source row sequence) and INDEX (position of the token within that sequence) are used for sorting.

Student 1: 2 tokens (JOHN, DOE) $\rightarrow$ 2 rows.

Student 2: 2 tokens (ALICE, SMITH) $\rightarrow$ 2 rows.

Student 3: 2 tokens (BOB, JOHNSON) $\rightarrow$ 2 rows.

Evaluating the Options (image_8233d8.png):

Option A incorrectly only shows results for Student 3.

Option C incorrectly implies Student 1 and 2 produce three tokens each.

Option D incorrectly interprets the consecutive delimiters in Student 3's name as creating a different sequence count.

Option B is the 100% correct result set. It accurately shows that each student’s name is split into exactly two parts, properly handling the varying delimiters and the consecutive characters in the third record.

When working with "Big Data" at the scale of trillions of rows, calculating an exact count of unique values using COUNT(DISTINCT column) is extremely resource-intensive. This is because Snowflake must keep track of every unique value encountered to ensure no duplicates are counted, leading to high memory usage and long execution times (often referred to as "spilling to disk").

To solve this, Snowflake provides HyperLogLog (HLL) functions. HLL(*) (or specifically HLL_ACCUMULATE and HLL_ESTIMATE) allows an analyst to estimate the cardinality (the number of unique elements) with a very small, known margin of error (typically around 1%). This is significantly faster and uses far fewer credits than an exact count because it uses a probabilistic algorithm rather than a state-heavy tracking mechanism.

Evaluating the Options:

Option A is technically correct for small datasets but is highly inefficient for trillions of rows, directly contradicting the "efficiently" requirement of the question.

Option C is a distractor; while Snowflake has various SYSTEM$ functions, SYSTEM$ESTIMATE is not a standard function for cardinality.

Option D is a formula that doesn't target cardinality but rather a ratio (density).

Option B is the correct answer. The HLL family of functions is the industry standard within Snowflake for high-performance cardinality estimation on massive datasets.