Solution Explanation for Counting Experiments

This problem requires generating a report showing the number of experiments conducted on each platform for each experiment type, even if the count is zero for a particular combination. The solution uses a combination of SQL techniques to achieve this.

Approach

The solution employs a strategy of generating all possible combinations of platforms and experiment names first, then joining this with the actual experiment data to count occurrences. If a combination doesn't exist in the Experiments table, the LEFT JOIN ensures it's still included in the result with a count of zero.

SQL Solution (MySQL)

The MySQL solution is broken down into three Common Table Expressions (CTEs):

1. P (Platforms): This CTE simply generates a list of all possible platforms ('Android', 'IOS', 'Web').

2. Exp (Experiments): This CTE generates a list of all possible experiment names ('Reading', 'Sports', 'Programming').

3. T (All Combinations): This CTE uses a Cartesian product (, in the FROM clause) to create all possible combinations of platforms and experiment names. This ensures we have a row for every possible platform-experiment pair.

The final SELECT statement then performs a LEFT JOIN between the T CTE and the Experiments table:

• LEFT JOIN Experiments USING (platform, experiment_name): This joins the generated combinations with the actual experiment data based on matching platform and experiment name. Crucially, all rows from T are included, even if there's no match in Experiments.

• COUNT(experiment_id): This counts the number of experiments for each combination. If there's no match in Experiments, experiment_id will be NULL, and COUNT will treat it as zero.

• GROUP BY 1, 2: This groups the results by platform and experiment name, providing the final counts.

Time Complexity Analysis

The time complexity is dominated by the LEFT JOIN operation. In the worst-case scenario, where there are many experiments, the complexity is O(N*M), where N is the number of rows in the T CTE (which is a constant 9 in this case: 3 platforms * 3 experiments) and M is the number of rows in the Experiments table. However, since N is a constant, the effective time complexity is considered O(M). The GROUP BY operation adds some overhead, but it's typically linear in the size of the data being grouped.

The creation of the CTEs (P, Exp, T) involves relatively small amounts of data and therefore contributes negligibly to the overall runtime. The COUNT aggregation is also generally linear in the size of the grouped data.

Therefore, the overall time complexity of this SQL query is O(M), where M is the number of rows in the Experiments table. The space complexity is also O(M) because of the intermediate data structures used during the LEFT JOIN and GROUP BY operations.