12 Solutions to 2018 Oracle SQL Puzzle of the Week #10

Top Salary Puzzle

Find highest salary in each department without using MAX function

  • Use a single SELECT statement only.
  • For an added complexity (optional): try not using ANY functions at all (neither group, nor analytic, not even scalar)

Expected Result:

DEPTNO MAX_SAL
10 5000
20 3000
30 2850

Solutions:

We will begin with a simpler problem that does allow us using functions.

Solution #1. Using MIN function

Credit to: Boobal Ganesan

MIN function can be seen as an opposite to the MAX, so it is trivial to employ it here:

SELECT deptno, -MIN(-sal) max_sal
FROM scott.emp
GROUP BY deptno
ORDER BY 1;

Solution #2. Using LISTAGG and REGEXP_SUBSTR functions

This is an “order” based approach that sorts the values within a concatenated string and then uses regular expression to cut the first token.

SELECT deptno,
       REGEXP_SUBSTR(LISTAGG(sal,',') 
                     WITHIN GROUP(ORDER BY sal DESC),'[^,]+',1,1) max_sal
FROM scott.emp
GROUP BY deptno
ORDER BY 1;

Solution #3. Using AVG(…) KEEP() group function

This is another “order” based strategy whete AVG function can be replaced with MIN or any other aggregate function that returns a single value out of a set of identical ones.

SELECT deptno, AVG(sal) KEEP(DENSE_RANK FIRST ORDER BY sal DESC) max_sal
FROM scott.emp
GROUP BY deptno
ORDER BY 1;

Solution #4. Using Analytic function and CTE

ROW_NUMBER is chosen in this approach, though other analytic functions, such as RANK, DENSE_RANK, LEAD, LAG, FIRST_VALUE, etc can be used here (with some changes) as well. ROW_NUMBER is convenient to use as it allows to avoid DISTINCT option.

WITH x AS (
SELECT deptno, sal, 
       ROW_NUMBER()OVER(PARTITION BY deptno ORDER BY sal DESC) rn
FROM scott.emp
)
SELECT deptno, sal max_sal
FROM x
WHERE rn=1
ORDER BY 1;

Solution #5. Using MATCH_RECOGNIZE clause

Credit to: KATAYAMA NAOTO

This approach is similar to the previous one if we used LAG analytic function: which would return NULL for the top record.

SELECT deptno, sal max_sal 
FROM scott.emp
MATCH_RECOGNIZE (
PARTITION BY deptno
ORDER BY sal DESC
ALL ROWS PER MATCH
PATTERN (ISNULL)
DEFINE ISNULL AS PREV(ISNULL.sal) IS NULL
);

Solution #6. CONNECT BY and CONNECT_BY_ISLEAF while avoiding Analytic functions

This approach is a bit artificial. We could have used DISTINCT and avoid START WITH clause completely.  CTEs x and y are used to simulate ROW_NUMBER analytic function.

WITH x AS (
SELECT deptno, sal
FROM scott.emp
ORDER BY 1,2
), y AS (
SELECT x.*, ROWNUM rn
FROM x
)
SELECT deptno, sal
FROM y
WHERE CONNECT_BY_ISLEAF=1
CONNECT BY deptno=PRIOR deptno
       AND rn=PRIOR rn+1
START WITH (deptno, rn) IN (SELECT deptno, MIN(rn)
                            FROM y
                            GROUP BY deptno);

Solution #7. Using MODEL clause with ROW_NUMBER function

This method is pretty much the same as in the Solution #4 above. The RETURN UPDATED ROWS and dummy measures are used to only return rows with rn=1.

SELECT deptno, max_sal
FROM scott.emp
MODEL
RETURN UPDATED ROWS
PARTITION BY (deptno)
DIMENSION BY (ROW_NUMBER() OVER(PARTITION BY deptno ORDER BY sal DESC) rn)
MEASURES(sal max_sal, 0 dummy)
RULES(
 dummy[1]=1
)
ORDER BY 1;

The following 5 solutions (##8-12) satisfy the “added complexity” term and do NOT use any functions at all.

Solution #8. Using ALL predicate

Generally speaking, >=ALL filter is identical to =(SELECT MAX() …). See my book for more detailed explanations.

SELECT deptno, sal max_sal
FROM scott.emp a
WHERE sal>=ALL(SELECT sal
               FROM scott.emp
               WHERE deptno=a.deptno)
GROUP BY deptno, sal
ORDER BY 1;

Solution #9. Using NOT EXISTS predicate

See Chapter 10 of my book for details.

SELECT deptno, sal max_sal
FROM scott.emp a
WHERE NOT EXISTS(SELECT 1
                 FROM scott.emp
                 WHERE deptno=a.deptno
                   AND sal>a.sal)
GROUP BY deptno, sal
ORDER BY 1;

Solution #10. Using Outer-Join with IS NULL filter

This approach is also covered very deeply in my book, Chapter 10.

SELECT a.deptno, a.sal max_sal
FROM scott.emp a LEFT JOIN scott.emp b ON a.deptno=b.deptno
                                      AND b.sal>a.sal 
WHERE b.empno IS NULL
GROUP BY a.deptno, a.sal
ORDER BY 1;

Solution #11. Using MINUS and ANY predicate

MINUS serves 2 purposes: it removes non-top rows and eliminates duplicates, so no DISTINCT option (or GROUP BY) is required.

SELECT deptno, sal max_sal 
FROM scott.emp
MINUS
SELECT deptno, sal
FROM scott.emp a
WHERE sal<ANY(SELECT sal 
              FROM scott.emp
              WHERE deptno=a.deptno);

Solution #12. Using MINUS and EXISTS predicate

Last two approaches covered in the drill from the Chapter 10 of my book.

SELECT deptno, sal max_sal 
FROM scott.emp
MINUS
SELECT deptno, sal
FROM scott.emp a
WHERE EXISTS(SELECT 1 
             FROM scott.emp
             WHERE deptno=a.deptno
               AND sal>a.sal);

You can execute the above SQL statements in Oracle Live SQL environment.
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Simulate LEAD and LAG functions using other analytic functions

Puzzle: Simulate LEAD and LAG functions using other analytic functions

Level: Intermediate

Solution:

Let’s say that we need to see every employee’s name and 2 more employees who were hired right before. The expected result may look like this:

ENAME      HIREDATE  EMPL1      EMPL2
---------- --------- ---------- ---------
SMITH      17-DEC-80
ALLEN      20-FEB-81 SMITH
WARD       22-FEB-81 ALLEN      SMITH
JONES      02-APR-81 WARD       ALLEN
BLAKE      01-MAY-81 JONES      WARD
CLARK      09-JUN-81 BLAKE      JONES
TURNER     08-SEP-81 CLARK      BLAKE
MARTIN     28-SEP-81 TURNER     CLARK
KING       17-NOV-81 MARTIN     TURNER
JAMES      03-DEC-81 KING       MARTIN
FORD       03-DEC-81 JAMES      KING
MILLER     23-JAN-82 FORD       JAMES
SCOTT      19-APR-87 MILLER     FORD
ADAMS      23-MAY-87 SCOTT      MILLER

It is a no-brainer task if we employ LAG function:

SELECT ename, hiredate, 
              LAG(ename,1) OVER(ORDER BY hiredate) empl1, 
              LAG(ename,2) OVER(ORDER BY hiredate) empl2
FROM emp;

One of the possible approaches is to use ROWS window attribute with MIN/MAX analytic functions:

SELECT ename, hiredate, 
       MAX(ename) OVER(ORDER BY hiredate ROWS BETWEEN 1 PRECEDING AND 1 PRECEDING) empl1,
       MAX(ename) OVER(ORDER BY hiredate ROWS BETWEEN 2 PRECEDING AND 2 PRECEDING) empl2
FROM emp;

Common rule is:

LAG(value_expr,offset,default) OVER ([partition_clause] order_by_clause )

is the same as

NVL(MIN(value_expr)OVER ([partition_clause] order_by_clause  
   ROWS BETWEEN offset PRECEDING AND offset PRECEDING), default)

and

LEAD(value_expr,offset,default) OVER ([partition_clause] order_by_clause )

is the same as

NVL(MIN(value_expr)OVER ([partition_clause] order_by_clause  
  ROWS BETWEEN offset FOLLOWING AND offset FOLLOWING), default)

This substitution becomes essential in other RDBMS where MIN/MAX analytic functions are supported while LEAD/LAG are not. Teradata is one of the examples.

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For more tricks and cool techniques check my book “Oracle SQL Tricks and Workarounds” for instructions.