Solving the Longest Possible Set of Rows in a Table

Introduction

The problem presented involves finding the longest possible set of rows from a table based on a comparison between two columns. The table contains fields like num_index, num_val, and previous_num_val. We need to find a subset of rows where for any row with num_index = n, the value of num_val is equal to the value of previous_num_val of row num_index = n - 1.

Problem Requirements

The requirements are as follows:

The output should be the longest possible set of rows.
For every previous_num_val, there exists a row with a column having an equal num_val.
Every num_val is unique (primary key).

Approach

To solve this problem, we can use a recursive Common Table Expression (CTE) in SQL. The CTE allows us to perform recursive operations on the table.

Recursive CTE Structure

The structure of our recursive CTE will be as follows:

Select columns from the original table (t).
Concatenate num_val values separated by '-' for each iteration.
Join with the original table on num_index = num_index - 1 and num_val = previous_num_val.
Use a union to combine all possible rows.

Recursive CTE Example

The following is an example of how we can implement our recursive CTE:

with recursive cte as (
    -- Base case: Select the first row.
    select num_index, num_val, previous_num_val, cast(num_val as char(10000)) as num_vals, 1 as n
    from t
    where num_index = 5
    
    -- Recursive case: Join with the original table and concatenate values.
    union all
    select t.num_index, t.num_val, t.previous_num_val, concat_ws('-&gt;', cte.num_vals, t.num_val), n + 1
    from cte join
         t on t.num_index = cte.num_index - 1 and
            t.num_val = cte.previous_num_val
)
-- Final step: Select the rows with maximum recursion number.
select *
from (select cte.*, max(n) over () as max_n
      from cte
     ) cte
where n = max_n

How it Works

Here’s a step-by-step explanation of how our recursive CTE works:

Base Case: We start by selecting the first row where num_index = 5. This is our base case for the recursion.
Recursive Step: For each iteration, we join with the original table on num_index = num_index - 1 and num_val = previous_num_val. This allows us to compare the current value of num_val with the value in the previous row (previous_num_val).
Concatenation: We concatenate the values of num_val separated by '-' for each iteration. This ensures that we have a unique set of rows where every previous_num_val has a corresponding row with an equal num_val.
Final Step: After all iterations, we select the rows with maximum recursion number (n). This gives us the longest possible set of rows based on our requirements.

SQL Implementation

We can implement this solution in various SQL dialects such as MySQL, PostgreSQL, or SQLite. However, since the provided answer uses CTE in MySQL, I’ll provide a brief overview of how to implement it in those dialects:

MySQL

with recursive cte as (
    -- Base case: Select the first row.
    select num_index, num_val, previous_num_val, concat_ws('-', num_val, concat_ws('-', 0, previous_num_val)) as num_vals, 1 as n
    from t
    where num_index = 5
    
    -- Recursive case: Join with the original table and concatenate values.
    union all
    select t.num_index, t.num_val, t.previous_num_val, concat_ws('-&gt;', cte.num_vals, t.num_val), n + 1
    from cte join
         t on t.num_index = cte.num_index - 1 and
            t.num_val = cte.previous_num_val
)
-- Final step: Select the rows with maximum recursion number.
select *
from (select cte.*, max(n) over () as max_n
      from cte
     ) cte
where n = max_n

PostgreSQL

WITH RECURSIVE cte AS (
    -- Base case: Select the first row.
    SELECT num_index, num_val, previous_num_val, array_agg(num_val ORDER BY num_val) AS num_vals, 1 AS n
    FROM t
    WHERE num_index = 5
    
    UNION ALL
    
    -- Recursive case: Join with the original table and concatenate values.
    SELECT t.num_index, t.num_val, t.previous_num_val, cte.num_vals || '->' || t.num_val, n + 1
    FROM cte JOIN t ON t.num_index = cte.num_index - 1 AND t.num_val = cte.previous_num_val
)
-- Final step: Select the rows with maximum recursion number.
SELECT *
FROM (SELECT cte.*, max(n) OVER () AS max_n FROM cte) cte
WHERE n = max_n;

SQLite

Unfortunately, SQLite does not support recursive CTE. However, we can use a temporary table to achieve similar results:

-- Create a temporary table to store the rows.
CREATE TEMPORARY TABLE temp (
    num_index INTEGER,
    num_val TEXT,
    previous_num_val TEXT,
    num_vals TEXT,
    n INTEGER
);

-- Base case: Insert the first row.
INSERT INTO temp (num_index, num_val, previous_num_val, num_vals, n)
VALUES (5, 'value1', 'value2', 'value1', 1);

-- Recursive case: Join with the original table and concatenate values.
UPDATE t AS sub
SET num_index = cte.num_index,
    num_val = cte.previous_num_val,
    previous_num_val = cte.num_vals,
    n = (SELECT MAX(n) FROM temp WHERE num_index < cte.num_index)
FROM (
    SELECT num_index, num_val, previous_num_val, concat_ws('-', num_val, concat_ws('-', 0, previous_num_val)) AS num_vals
    FROM t
    WHERE num_index = 5
    UNION ALL
    SELECT sub.num_index, sub.previous_num_val, cte.num_vals, cte.num_vals || '->' || sub.num_val
    FROM temp sub JOIN (
        SELECT num_index, num_val, previous_num_val, array_agg(num_val ORDER BY num_val) AS num_vals
        FROM t
        WHERE num_index = 5
    ) cte ON sub.previous_num_val = cte.num_vals AND sub.num_index < cte.num_index
);

-- Final step: Select the rows with maximum recursion number.
SELECT *
FROM temp
WHERE n = (SELECT MAX(n) FROM temp);

Conclusion

In this solution, we use a recursive Common Table Expression to solve the problem. The CTE allows us to perform recursive operations on the table and find the longest possible set of rows based on our requirements.

The SQL implementation for each dialect is provided separately to make it easier for users to adapt this solution to their specific database management system.

Last modified on 2023-10-09