Understanding the Problem

The problem at hand involves updating a value in Presto SQL based on certain conditions. We are given a table with two columns, X and Y, and we need to find the output that satisfies a specific logic.

The Logic

The logic states that we set the Y value of the first row as the start point and look ahead at subsequent rows. If all subsequent rows have values within 2 units of the start point, we assign the first row’s X value as the output; otherwise, we update the start point with the Y value of the current row.

Example

Let’s take the provided table as an example:

Initially, the start value is set to 1 since the first and second rows have values less than 3 (start + 2), so we assign the output as 1. However, when we encounter the third row with a Y value of 3, which exceeds the criteria, the start point is updated to 3.

Approaching the Problem

Our goal is to find a scalable solution in Presto SQL that can achieve this desired output.

Initial Approach

One possible approach could involve using array aggregation and reduction functions to process the data. However, this initial attempt led to an incomplete solution, which highlights the complexity of solving such problems.

Solution Overview

After exploring various options, we found a way to solve the problem by:

1. Using array_agg with distinct to create a new column for each unique Y value.
2. Creating an additional column called start, which contains the minimum Y values in the form of an array.
3. Applying the reduction function to update the start point based on the difference between consecutive elements.

Solution Breakdown

Step 1: Creating an Array Aggregation Column

{
  <highlight language="sql">
    CREATE TABLE my_table (
      X,
      Y,
      A ARRAY<X>
    );
  </highlight>
}

First, we create a table with three columns: X, Y, and A. The column A will be used to store the array aggregation of unique Y values.

{
  <highlight language="sql">
    INSERT INTO my_table (X, Y)
    VALUES ('dummy', 1), ('dummy', 2), ('dummy', 3), ('dummy', 4), ('dummy', 5);
  </highlight>
}

Next, we insert some sample data into the table.

{
  <highlight language="sql">
    SELECT X, Y,
           array_agg(DISTINCT Y) AS A
    FROM my_table;
  </highlight>
}

Then, we select all columns from the table and calculate the A column using array_agg(DISTINCT Y).

{
  <highlight language="sql">
    SELECT X, Y,
           array_agg(DISTINCT Y) AS A
    FROM my_table;
  </highlight>
}

Step 2: Creating a Start Column

Now that we have the array aggregation column, let’s create a new column called start, which contains the minimum Y values in the form of an array.

{
  <highlight language="sql">
    SELECT X, Y,
           array_agg(DISTINCT Y) AS A,
           array_agg(Y ORDER BY Y) OVER () AS start;
  </highlight>
}

We use array_agg(DISTINCT Y) to calculate the unique Y values and then order them in ascending order using the ORDER BY clause. This will give us an array of minimum Y values.

{
  <highlight language="sql">
    SELECT X, Y,
           array_agg(DISTINCT Y) AS A,
           array_agg(Y ORDER BY Y) OVER () AS start;
  </highlight>
}

Step 3: Applying the Reduction Function

Next, we apply a reduction function to update the start point based on the difference between consecutive elements.

{
  <highlight language="sql">
    SELECT X, Y,
           array_agg(DISTINCT Y) AS A,
           reduce(A, start, case when x - s[cardinality(s)] &lt; 2 then s || s[cardinality(s)] else s || x end, s-&gt; s) as b;
  </highlight>
}

We use the reduce function to calculate the cumulative product of elements in array A. The condition inside the case statement checks if the difference between consecutive elements is less than 2; if true, it appends the last element of the array (s[cardinality(s)]) to the current array; otherwise, it appends the current element (x) to the array.

{
  <highlight language="sql">
    SELECT X, Y,
           reduce(A, start, case when x - s[cardinality(s)] &lt; 2 then s || s[cardinality(s)] else s || x end, s-&gt; s) as b;
  </highlight>
}

Step 4: Exploding and Joining

Finally, we explode the array aggregation column b after applying the reduction function and join it with the original table using a proper condition.

{
  <highlight language="sql">
    SELECT X, Y,
           explode(array_distinct(b)) AS b;
  </highlight>
}

We use explode(array_distinct(b)) to expand the array into individual elements. This will give us our final output.

Conclusion

In this solution, we break down a complex problem by using various SQL functions such as array_agg, reduce, and explode. We create an additional column called start which contains minimum values in the form of arrays. The main step is applying reduction function to update start point based on difference between consecutive elements. This approach provides us scalable way of solving similar problems.

We can further improve this solution by adding error checking, handling missing values, or optimizing performance for larger datasets.

With this solution, we demonstrate how SQL functions like array_agg, reduce, and explode can be used to process complex data sets and solve real-world problems efficiently.

Last modified on 2025-03-23