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Snowflake SnowPro Advanced: Data Analyst Certification Exam Sample Questions (Q49-Q54):

NEW QUESTION # 49
A Directed Acyclic Graph (DAG) of 12 tasks (ETLDAILY) failed because one of the tasks failed (STEP10), therefore the dependent tasks (STEP11 and STEP12) did not run. Which command is needed to re-run task STEP10 and the two dependent tasks?

Answer: C

Explanation:
Managing complex pipelines involving multiple dependent tasks requires an understanding of Task Graphs (DAGs). In Snowflake, a DAG is a series of tasks with parent-child relationships. If a parent task (like STEP10) fails, Snowflake automatically cancels the execution of all child tasks (like STEP11 and STEP12) to prevent data corruption or inconsistent states.
To recover from such a failure, Snowflake provides the RETRY LAST clause within the EXECUTE TASK command. The critical nuance here is that you do not target the individual failed task; instead, you target the root task of the graph (in this case, ETLDAILY). When you run EXECUTE TASK <root_task> RETRY LAST;, Snowflake identifies the most recent failed run of that graph and intelligently resumes execution starting from the failed task (STEP10) and continues through all subsequent dependent tasks that were previously skipped.
Evaluating the Options:
* Option A is incorrect because attempting to execute just STEP10 would treat it as a standalone run and would not automatically trigger the dependent logic of the ETLDAILY graph structure.
* Options C and D are incorrect because RESUME is a DDL command used to change a task's state from SUSPENDED to STARTED (enabled). It does not trigger an immediate execution or a retry of a failed run; it merely allows the task to run on its next scheduled interval.
* Option B is the 100% correct answer. It utilizes the dedicated Snowflake mechanism for graph recovery, ensuring that the entire pipeline is completed starting from the point of failure without re- running successfully completed earlier steps.


NEW QUESTION # 50
When employing window functions versus table functions in Snowflake, how do they differ in their application and output?

Answer: B

Explanation:
Table functions in Snowflake process data within specified partitions or frames, whereas window functions generate aggregate results and operate on entire datasets, differing in their scope and operation.


NEW QUESTION # 51
You're working with a Snowflake database containing sales transaction data'. The 'SALES' table includes columns 'transaction id' , product id', 'customer id', 'transaction_date', and 'sales amount'. The Bl team needs to analyze sales trends, identify top-selling products, and segment customers based on their purchase behavior. They also need to support ad-hoc queries with various filtering and aggregation criteria'. The data volume is significant, and query performance is a key concern. Consider the following Snowflake table definition (simplified):

Which of the following SQL queries, combined with appropriate data modeling techniques, will provide the BEST performance for analyzing monthly sales trends by product category, assuming a separate 'PRODUCTS table exists with 'product_id' and 'category' columns?

Answer: A

Explanation:
Option B provides the best performance because it uses a materialized view to pre- compute the monthly sales trends by product category. Subsequent queries to the materialized view will be significantly faster than re-calculating the aggregation each time. Option A performs the join and aggregation every time the query is executed. Option C uses a correlated subquery, which is generally less performant than a join. Option D uses , which might provide the same result but doesn't address the performance issue of recalculating the aggregation every time. Option E uses a temporary table which is unncessary because a materialized view can be created directly from the SALES and PRODUCTS table.


NEW QUESTION # 52
You need to load data from a set of CSV files located in an AWS S3 bucket into a Snowflake table called 'SALES DATA. The CSV files have a header row, and you want to automatically detect the data types of the columns. However, some files might have inconsistent data types (e.g., a column that is usually an integer might occasionally contain a string). Furthermore the file are compressed using GZIP. Which of the following COPY INTO command options can you use to achieve this goal while handling potential data type mismatches?

Answer: D

Explanation:
The 'ON_ERROR = 'SKIP_RECORD" option allows the COPY INTO command to skip individual records that have data type mismatches or other errors during the load process. This ensures that the load process continues even if some records fail validation. 'SKIP_FILE' would skip entire file, 'CONTINUE' will only continue the process with the given data, and 'SKIP_FILE' skips the file. The VALIDATE UTF8 isn't directly relevant to data type issues


NEW QUESTION # 53
A Data Analyst has a very large table with columns that contain country and city names. Which query will provide a very quick estimate of the total number of different values of these two columns?

Answer: B

Explanation:
When working with very large tables, calculating the exact number of unique combinations of columns (cardinality) using COUNT(DISTINCT ...) is a resource-intensive operation. It requires the Snowflake query engine to keep an exhaustive list of every unique pair encountered in memory, which can lead to high credit consumption and performance bottlenecks.
To provide a "very quick estimate," Snowflake utilizes the HyperLogLog (HLL) algorithm. The function HLL(column1, column2, ...) returns an HLL state (a binary representation) that can be used to estimate the number of distinct values with a high degree of accuracy and minimal computational overhead. This is part of Snowflake's suite of approximate aggregation functions, which are essential for Data Analysis on massive datasets where a 1% margin of error is acceptable in exchange for significantly faster results.
Evaluating the Options:
* Option A is syntactically incorrect; DISTINCT cannot be used in that position with COUNT.
* Option C is a valid query but will be significantly slower and more expensive than an HLL estimate on a "very large table".
* Option D simply counts the total number of non-null rows, which does not represent the "number of different values" (cardinality).
* Option B is the 100% correct answer. It specifically addresses the requirement for a "quick estimate" using the industry-standard probabilistic counting method built into Snowflake.


NEW QUESTION # 54
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