Thursday, 14 July 2016

Using EXPLAIN PLAN in ORACLE

table, EMP_RANGE, partitioned by range on HIREDATE to illustrate how pruning is displayed. Assume that the tables EMP and DEPT from a standard Oracle schema exist.
   CREATE TABLE EMP_RANGE 
    PARTITION BY RANGE(HIREDATE) 
    ( 
    PARTITION EMP_P1 VALUES LESS THAN (TO_DATE('1-JAN-1981','DD-MON-YYYY')),
    PARTITION EMP_P2 VALUES LESS THAN (TO_DATE('1-JAN-1983','DD-MON-YYYY')),
    PARTITION EMP_P3 VALUES LESS THAN (TO_DATE('1-JAN-1985','DD-MON-YYYY')),
    PARTITION EMP_P4 VALUES LESS THAN (TO_DATE('1-JAN-1987','DD-MON-YYYY')),
    PARTITION EMP_P5 VALUES LESS THAN (TO_DATE('1-JAN-1989','DD-MON-YYYY')) 
    ) 
    AS SELECT * FROM EMP; 

Example 1:
   EXPLAIN PLAN FOR SELECT * FROM EMP_RANGE; 

Then enter the following to display the EXPLAIN PLAN output:
        @?/RDBMS/ADMIN/UTLXPLS 

Oracle displays something similar to:
Plan Table 
-------------------------------------------------------------------------------
| Operation               |  Name    |  Rows | Bytes|  Cost  | Pstart |  Pstop|
-------------------------------------------------------------------------------
| SELECT STATEMENT        |          |   105 |    8K|      1 |        |       |
|  PARTITION RANGE ALL    |          |       |      |        |     1  |     5 |
|   TABLE ACCESS FULL     |EMP_RANGE |   105 |    8K|      1 |     1  |     5 |
-------------------------------------------------------------------------------
6 rows selected. 

A partition row source is created on top of the table access row source. It iterates over the set of partitions to be accessed.
In example 1, the partition iterator covers all partitions (option ALL) because a predicate was not used for pruning. The PARTITION_START and PARTITION STOP columns of the plan table show access to all partitions from 1 to 5.
Example 2:
   EXPLAIN PLAN FOR SELECT * FROM EMP_RANGE 
   WHERE HIREDATE >= TO_DATE('1-JAN-1985','DD-MON-YYYY'); 

Plan Table 
--------------------------------------------------------------------------------
| Operation                 | Name    |  Rows  | Bytes|  Cost  | Pstart| Pstop |
--------------------------------------------------------------------------------
| SELECT STATEMENT          |          |     3 |   54 |      1 |       |       |
|  PARTITION RANGE ITERATOR |          |       |      |        |     4 |     5 |
|   TABLE ACCESS FULL       |EMP_RANGE |     3 |   54 |      1 |     4 |     5 |
--------------------------------------------------------------------------------
6 rows selected. 
In example 2, the partition row source iterates from partition 4 to 5 because we prune the other partitions using a predicate on HIREDATE.
Example 3:
   EXPLAIN PLAN FOR SELECT * FROM EMP_RANGE 
   WHERE HIREDATE < TO_DATE('1-JAN-1981','DD-MON-YYYY'); 

Plan Table 
--------------------------------------------------------------------------------
| Operation                 |  Name    |  Rows | Bytes|  Cost  | Pstart| Pstop |
--------------------------------------------------------------------------------
| SELECT STATEMENT          |          |     2 |   36 |      1 |       |       |
|  TABLE ACCESS FULL        |EMP_RANGE |     2 |   36 |      1 |     1 |     1 |
--------------------------------------------------------------------------------
5 rows selected. 

In example 3, only partition 1 is accessed and known at compile time, thus there is no need for a partition row source.

Plans for Hash Partitioning

Oracle displays the same information for hash partitioned objects except that the partition row source name is "PARTITION HASH" instead of "PARTITION RANGE". Also, with hash partitioning, pruning is only possible using equality or in-list predicates.

Pruning Information with Composite Partitioned Objects

To illustrate how Oracle displays pruning information for composite partitioned objects, consider the table EMP_COMP that is range partitioned on HIREDATE and subpartitioned by hash on DEPTNO.
 CREATE TABLE EMP_COMP PARTITION BY RANGE(HIREDATE) SUBPARTITION BY HASH(DEPTNO) 
  SUBPARTITIONS 3 
  ( 
  PARTITION EMP_P1 VALUES LESS THAN (TO_DATE('1-JAN-1981','DD-MON-YYYY')),
  PARTITION EMP_P2 VALUES LESS THAN (TO_DATE('1-JAN-1983','DD-MON-YYYY')),
  PARTITION EMP_P3 VALUES LESS THAN (TO_DATE('1-JAN-1985','DD-MON-YYYY')),
  PARTITION EMP_P4 VALUES LESS THAN (TO_DATE('1-JAN-1987','DD-MON-YYYY')),
  PARTITION EMP_P5 VALUES LESS THAN (TO_DATE('1-JAN-1989','DD-MON-YYYY')) 
   ) 
  AS SELECT * FROM EMP; 

Example 1:
   EXPLAIN PLAN FOR SELECT * FROM EMP_COMP; 

Plan Table 
--------------------------------------------------------------------------------
| Operation                 |  Name   |  Rows | Bytes|  Cost  | Pstart | Pstop |
--------------------------------------------------------------------------------
| SELECT STATEMENT          |         |   105 |    8K|      1 |        |       |
|  PARTITION RANGE ALL      |         |       |      |        |     1  |     5 |
|   PARTITION HASH ALL      |         |       |      |        |     1  |     3 |
|    TABLE ACCESS FULL      |EMP_COMP |   105 |    8K|      1 |     1  |     15|
--------------------------------------------------------------------------------
7 rows selected. 

Example 1 shows the explain plan when Oracle accesses all subpartitions of all partitions of a composite object. Two partition row sources are used for that purpose: a range partition row source to iterate over the partitions and a hash partition row source to iterate over the subpartitions of each accessed partition.
In this example, since no pruning is performed, the range partition row source iterates from partition 1 to 5. Within each partition, the hash partition row source iterates over subpartitions 1 to 3 of the current partition. As a result, the table access row source accesses subpartitions 1 to 15. In other words, it accesses all subpartitions of the composite object.
Example 2:
   EXPLAIN PLAN FOR SELECT * FROM EMP_COMP WHERE HIREDATE = 
   TO_DATE('15-FEB-1987', 'DD-MON-YYYY'); 

Plan Table 
--------------------------------------------------------------------------------
| Operation                 |  Name    |  Rows | Bytes|  Cost  | Pstart| Pstop |
--------------------------------------------------------------------------------
| SELECT STATEMENT          |          |     1 |   96 |      1 |       |       |
|  PARTITION HASH ALL       |          |       |      |        |     1 |     3 |
|   TABLE ACCESS FULL       |EMP_COMP  |     1 |   96 |      1 |    13 |    15 |
--------------------------------------------------------------------------------
6 rows selected. 
 
In example 2, only the last partition, partition 5, is accessed. This partition is known at compile time so we do not need to show it in the plan. The hash partition row source shows accessing of all subpartitions within that partition, that is, subpartitions 1 to 3, which translates into subpartitions 13 to 15 of the EMP_COMP table.
Example 3:
   EXPLAIN PLAN FOR SELECT * FROM EMP_COMP WHERE DEPTNO = 20; 

Plan Table 
--------------------------------------------------------------------------------
| Operation                 |  Name    |  Rows | Bytes|  Cost  | Pstart| Pstop |
--------------------------------------------------------------------------------
| SELECT STATEMENT          |          |     2 |  200 |      1 |       |       |
|  PARTITION RANGE ALL      |          |       |      |        |     1 |     5 |
|   TABLE ACCESS FULL       |EMP_COMP  |     2 |  200 |      1 |       |       |
--------------------------------------------------------------------------------
6 rows selected. 

In this example, the predicate "DEPTNO = 20" enables pruning on the hash dimension within each partition, so Oracle only needs to access a single subpartition. The number of that subpartition is known at compile time so the hash partition row source is not needed.
Example 4:
   VARIABLE DNO NUMBER; 
   EXPLAIN PLAN FOR SELECT * FROM EMP_COMP WHERE DEPTNO = :DNO; 

Plan Table 
--------------------------------------------------------------------------------
| Operation                 |  Name    |  Rows | Bytes|  Cost  | Pstart| Pstop |
--------------------------------------------------------------------------------
| SELECT STATEMENT          |          |     2 |  200 |      1 |       |       |
|  PARTITION RANGE ALL      |          |       |      |        |     1 |     5 |
|   PARTITION HASH SINGLE   |          |       |      |        |   KEY |   KEY |
|    TABLE ACCESS FULL      |EMP_COMP  |     2 |  200 |      1 |       |       |
--------------------------------------------------------------------------------
 7 rows selected. 

Example 4 is the same as example 3 except that "DEPTNO = 20" has been replaced by "DEPTNO = :DNO". In this case, the subpartition number is unknown at compile time and a hash partition row source is allocated. The option is SINGLE for that row source because Oracle accesses only one subpartition within each partition. The PARTITION START and PARTITION STOP is set to "KEY". This means Oracle will determine the number of the subpartition at run time.

Partial Partition-wise Joins

Example 1:
In the following example, EMP_RANGE is joined on the partitioning column and is parallelized. This enables use of partial partition-wise join because the DEPT table is not partitioned. Oracle dynamically partitions the DEPT table before the join.
   ALTER TABLE EMP PARALLEL 2; 
      STATEMENT PROCESSED.
   ALTER TABLE DEPT PARALLEL 2; 
      STATEMENT PROCESSED. 

To show the plan for the query, enter:
   EXPLAIN PLAN FOR SELECT /*+ ORDERED USE_HASH(D) */ ENAME, DNAME 
     FROM EMP_RANGE E, DEPT D 
     WHERE E.DEPTNO = D.DEPTNO 
     AND E.HIREDATE > TO_DATE('29-JUN-1986','DD-MON-YYYY'); 

    
Plan Table 
------------------------------------------------------------------------------------------------------------ 
| Operation                  |  Name    |  Rows | Bytes|  Cost  |  TQ  |IN-OUT| PQ Distrib | Pstart| Pstop | 
------------------------------------------------------------------------------------------------------------ 
| SELECT STATEMENT           |          |     1 |   51 |      3 |      |      |            |       |       | 
|  HASH JOIN                 |          |     1 |   51 |      3 | 2,02 | P->S | QC (RANDOM)|       |       | 
|   PARTITION RANGE ITERATOR |          |       |      |        | 2,02 | PCWP |            |     4 |     5 | 
|    TABLE ACCESS FULL       |EMP_RANGE |     3 |   87 |      1 | 2,00 | PCWP |            |     4 |     5 | 
|    TABLE ACCESS FULL       |DEPT      |    21 |  462 |      1 | 2,01 | P->P |PART (KEY)  |       |       | 
------------------------------------------------------------------------------------------------------------ 
8 rows selected.
The plan shows that the optimizer select partition-wise join because the DIST column contains the text "PART (KEY)", or, partition key.
Example 2:
In example 2, EMP_COMP is joined on its hash partitioning column, DEPTNO, and is parallelized. This enables use of partial partition-wise join because the DEPT table is not partitioned. Again, Oracle dynamically partitions the DEPT table.
   ALTER TABLE EMP_COMP PARALLEL 2; 
     STATEMENT PROCESSED. 
   EXPLAIN PLAN FOR SELECT /*+ ORDERED USE_HASH(D) */ ENAME, DNAME 
     FROM EMP_COMP E, DEPT D 
     WHERE E.DEPTNO = D.DEPTNO 
     AND E.HIREDATE > TO_DATE('13-MAR-1985','DD-MON-YYYY'); 

    
Plan Table
------------------------------------------------------------------------------------------------------------ 
| Operation                  |  Name    |  Rows | Bytes|  Cost  |  TQ  |IN-OUT| PQ Distrib | Pstart| Pstop | 
------------------------------------------------------------------------------------------------------------ 
| SELECT STATEMENT           |          |    1  |  51  |      3 |      |      |            |       |       |
|  HASH JOIN                 |          |     1 |   51 |      3 | 0,01 | P->S | QC (RANDOM)|       |       | 
|   PARTITION RANGE ITERATOR |          |       |      |        | 0,01 | PCWP |            |     4 |     5 | 
|    PARTITION HASH ALL      |          |       |      |        | 0,01 | PCWP |            |     1 |     3 | 
|     TABLE ACCESS FULL      |EMP_COMP  |     3 |   87 |      1 | 0,01 | PCWP |            |    10 |    15 | 
|   TABLE ACCESS FULL        |DEPT      |    21 |  462 |      1 | 0,00 | P->P | PART (KEY) |       |       | 
------------------------------------------------------------------------------------------------------------ 
9 rows selected.

Full Partition-wise Joins

In the following example, EMP_COMP and DEPT_HASH are joined on their hash partitioning columns. This enables use of full partition-wise join. The "PARTITION HASH" row source appears on top of the join row source in the plan table output.
To create the table DEPT_HASH, enter:
   CREATE TABLE DEPT_HASH 
     PARTITION BY HASH(deptno) 
     PARTITIONS 3 
     PARALLEL 
     AS SELECT * FROM DEPT; 

To show the plan for the query, enter:
   EXPLAIN PLAN FOR SELECT /*+ ORDERED USE_HASH(D) */ ENAME, DNAME 
     FROM EMP_COMP E, DEPT_HASH D 
     WHERE E.DEPTNO = D.DEPTNO 
     AND E.HIREDATE > TO_DATE('29-JUN-1986','DD-MON-YYYY'); 

    
Plan Table 
------------------------------------------------------------------------------------------------------------ 
| Operation                   |  Name    |  Rows | Bytes|  Cost  |  TQ |IN-OUT| PQ Distrib | Pstart| Pstop | 
------------------------------------------------------------------------------------------------------------ 
| SELECT STATEMENT            |          |     2 |   102|      2 |     |      |            |       |       |
|  PARTITION HASH ALL         |          |       |      |        | 4,00| PCWP |            |     1 |     3 | 
|   HASH JOIN                 |          |     2 |  102 |      2 | 4,00| P->S | QC (RANDOM)|       |       | 
|    PARTITION RANGE ITERATOR |          |       |      |        | 4,00| PCWP |            |     4 |     5 | 
|     TABLE ACCESS FULL       |EMP_COMP  |     3 |   87 |      1 | 4,00| PCWP |            |    10 |    15 | 
|    TABLE ACCESS FULL        |DEPT_HASH |    63 |    1K|      1 | 4,00| PCWP |            |     1 |     3 | 
------------------------------------------------------------------------------------------------------------ 
9 rows selected. 

INLIST ITERATOR and EXPLAIN PLAN

An INLIST ITERATOR operation appears in the EXPLAIN PLAN output if an index implements an IN list predicate. For example, for the query:
   SELECT * FROM EMP WHERE EMPNO IN (7876, 7900, 7902); 

The EXPLAIN PLAN output appears as follows:
   OPERATION          OPTIONS           OBJECT_NAME
   ----------------   ---------------   -------------- 
   SELECT STATEMENT
   INLIST ITERATOR
   TABLE ACCESS       BY ROWID          EMP
   INDEX              RANGE SCAN        EMP_EMPNO

The INLIST ITERATOR operation iterates over the operation below it for each value in the IN list predicate. For partitioned tables and indexes, the three possible types of IN list columns are described in the following sections.

Index Column

If the IN list column EMPNO is an index column but not a partition column, then the plan is as follows (the IN list operator appears above the table operation but below the partition operation):
 OPERATION         OPTIONS        OBJECT_NAME   PARTITION_START   PARTITION_STOP
 ----------------  ------------   -----------   ---------------   --------------
 SELECT STATEMENT 
 PARTITION         INLIST                       KEY(INLIST)       KEY(INLIST)
 INLIST ITERATOR
 TABLE ACCESS      BY ROWID       EMP           KEY(INLIST)       KEY(INLIST)
 INDEX             RANGE SCAN     EMP_EMPNO     KEY(INLIST)       KEY(INLIST)

The KEY(INLIST) designation for the partition start and stop keys specifies that an IN list predicate appears on the index start/stop keys.

Index and Partition Column

If EMPNO is an indexed and a partition column, then the plan contains an INLIST ITERATOR operation above the partition operation:
 OPERATION         OPTIONS        OBJECT_NAME   PARTITION_START   PARTITION_STOP
 ----------------  ------------   -----------   ---------------   --------------
 SELECT STATEMENT
 INLIST ITERATOR
 PARTITION         ITERATOR                     KEY(INLIST)       KEY(INLIST)
 TABLE ACCESS      BY ROWID       EMP           KEY(INLIST)       KEY(INLIST)
 INDEX             RANGE SCAN     EMP_EMPNO     KEY(INLIST)       KEY(INLIST)

Partition Column

If EMPNO is a partition column and there are no indexes, then no INLIST ITERATOR operation is allocated:
 OPERATION         OPTIONS        OBJECT_NAME   PARTITION_START   PARTITION_STOP
 ----------------  ------------   -----------   ---------------   --------------
 SELECT STATEMENT
 PARTITION                                      KEY(INLIST)       KEY(INLIST)
 TABLE ACCESS      BY ROWID       EMP           KEY(INLIST)       KEY(INLIST)
 INDEX             RANGE SCAN     EMP_EMPNO     KEY(INLIST)       KEY(INLIST)

If EMP_EMPNO is a bitmap index, then the plan is as follows:
 OPERATION          OPTIONS           OBJECT_NAME
 ----------------   ---------------   -------------- 
 SELECT STATEMENT
 INLIST ITERATOR
 TABLE ACCESS       BY INDEX ROWID    EMP
 BITMAP CONVERSION  TO ROWIDS
 BITMAP INDEX       SINGLE VALUE      EMP_EMPNO

DOMAIN INDEX and EXPLAIN PLAN

You can also use EXPLAIN PLAN to derive user-defined CPU and I/O costs for domain indexes. EXPLAIN PLAN displays these statistics in the "OTHER" column of PLAN_TABLE.
For example, assume table EMP has user-defined operator CONTAINS with a domain index EMP_RESUME on the RESUME column and the index type of EMP_RESUME supports the operator CONTAINS. Then the query:
 SELECT * from EMP where Contains(resume, 'Oracle') = 1 

might display the following plan:
  OPERATION            OPTIONS      OBJECT_NAME     OTHER 
 -----------------    -----------  ------------    ----------------
 SELECT STATEMENT 
 TABLE ACCESS         BY ROWID     EMP
 DOMAIN INDEX                      EMP_RESUME      CPU: 300, I/O: 4

Formatting EXPLAIN PLAN Output

This section shows options for formatting EXPLAIN PLAN output

Using the EXPLAIN PLAN Statement

The following example shows a SQL statement and its corresponding execution plan generated by EXPLAIN PLAN. The sample query retrieves names and related information for employees whose salary is not within any range of the SALGRADE table:
 SELECT ename, job, sal, dname
   FROM emp, dept
   WHERE emp.deptno = dept.deptno
      AND NOT EXISTS
         (SELECT *
            FROM salgrade
            WHERE emp.sal BETWEEN losal AND hisal);

This EXPLAIN PLAN statement generates an execution plan and places the output in PLAN_TABLE:
 EXPLAIN PLAN
   SET STATEMENT_ID = 'Emp_Sal'
   FOR SELECT ename, job, sal, dname
      FROM emp, dept
      WHERE emp.deptno = dept.deptno
         AND NOT EXISTS
            (SELECT *
               FROM salgrade
               WHERE emp.sal BETWEEN losal AND hisal);

Selecting PLAN_TABLE Output in Table Format

This SELECT statement:
            SELECT operation, options, object_name, id, parent_id, position, cost, cardinality,
            other_tag, optimizer 
               FROM plan_table
               WHERE statement_id = 'Emp_Sal'
               ORDER BY id;
Generates this output:
  OPERATION  OPTIONS OBJECT_NAME ID PARENT_ID POSITION COST CARDINALITY BYTES OTHER_TAG 
OPTIMIZER
  
-----------------------------------------------------------------------------------------------
  SELECT STATEMENT                    0                    2    2            1    62       
CHOOSE
  FILTER                              1          0         1
  NESTED LOOPS                        2          1         1    2            1    62
  TABLE ACCESS FULL    EMP            3          2         1    1            1    40     
ANALYZED
  TABLE ACCESS FULL    DEPT           4          2         2                 4    88     
ANALYZED
  TABLE ACCESS FULL    SALGRADE       5          1         2    1            1    13     
ANALYZED

The ORDER BY clause returns the steps of the execution plan sequentially by ID value. However, Oracle does not perform the steps in this order. PARENT_ID receives information from ID, yet more than one ID step fed into PARENT_ID.
For example, step 2, a merge join, and step 6, a table access, both fed into step 1. A nested, visual representation of the processing sequence is shown in the next section.
The value of the POSITION column for the first row of output indicates the optimizer's estimated cost of executing the statement with this plan to be 5. For the other rows, it indicates the position relative to the other children of the same parent.

Note:
A CONNECT BY does not preserve ordering. To have rows come out in the correct order in this example, you must either truncate the table first, or else create a view and select from the view. For example:  

CREATE VIEW test AS
SELECT id, parent_id,
lpad(' ', 2*(level-1))||operation||' '||options||' '||object_name||' '||
       decode(id, 0, 'Cost = '||position) "Query Plan"
FROM plan_table
START WITH id = 0 and statement_id = 'TST'
CONNECT BY prior id = parent_id and statement_id = 'TST';
SELECT * FROM foo ORDER BY id, parent_id;

This yields results as follows:
 ID  PAR Query Plan
 --- --- --------------------------------------------------
  0     Select Statement   Cost = 69602
  1   0   Nested Loops
  2   1     Nested Loops
  3   2       Merge Join
  4   3         Sort Join
  5   4           Table Access Full T3
  6   3         Sort Join
  7   6           Table Access Full T4
  8   2       Index Unique Scan T2
  9   1     Table Access Full T1
10 rows selected.

Selecting PLAN_TABLE Output in Nested Format

This type of SELECT statement generates a nested representation of the output that more closely depicts the processing order used for the SQL statement.
 SELECT LPAD(' ',2*(LEVEL-1))||operation||' '||options
   ||' '||object_name
   ||' '||DECODE(id, 0, 'Cost = '||position) "Query Plan"
   FROM plan_table
   START WITH id = 0 AND statement_id = 'Emp_Sal'
   CONNECT BY PRIOR id = parent_id AND statement_id ='Emp_Sal';
 
 Query Plan
 ------------------------------
 SELECT STATEMENT   Cost = 5
   FILTER
      NESTED LOOPS
         TABLE ACCESS FULL EMP
         TABLE ACCESS FULL DEPT
      TABLE ACCESS FULL SALGRADE  

The order resembles a tree structure, as illustrated in Figure 13-1.

Figure 13-1 Tree Structure of an Execution Plan


Tree structures illustrate how SQL statement execution operations feed one another. Oracle assigns each step in the execution plan a number representing the ID column of the PLAN_TABLE. Each step is depicted by a "node". The result of each node's operation passes to its parent node, which uses it as input.

EXPLAIN PLAN Restrictions

Oracle does not support EXPLAIN PLAN for statements performing implicit type conversion of date bind variables. With bind variables in general, the EXPLAIN PLAN output may not represent the real execution plan.
From the text of a SQL statement, TKPROF cannot determine the types of the bind variables. It assumes that the type is CHARACTER, and gives an error message if this is not the case. You can avoid this limitation by putting appropriate type conversions in the SQL statement.

What is a PARTITION in Oracle?Why to use Partition And Types of Partitions

PARTITIONS

Partitioning allows tables, indexes, and index-organized tables to be subdivided into smaller pieces, enabling these database objects to be managed and accessed at a finer level of granularity.

When to Partition a Table??
  • Tables greater than 2 GB should always be considered as candidates for partitioning.
  • Tables containing historical data, in which new data is added into the newest partition. A typical example is a historical table where only the current month's data is updatable and the other 11 months are read only.
  • When the contents of a table need to be distributed across different types of storage devices.
TYPES
1     Range partitions
2     List partitions
3     Hash partitions
4     Sub partitions
  
ADVANTAGES OF PARTITIONS
  • Reducing downtime for scheduled maintenance, which allows maintenance operations to be carried out on selected partitions while other partitions are available to users.
  • Reducing downtime due to data failure, failure of a particular partition will no way affect other partitions.
  • Partition independence allows for concurrent use of the various partitions for various purposes.
 What is the advantage of partitions, by storing them in different Tablespaces??
1     Reduces the possibility of data corruption in multiple partitions.
2     Back up and recovery of each partition can be done independently.
 
Partitioning Key
Each row in a partitioned table is unambiguously assigned to a single partition. The partitioning key is comprised of one or more columns that determine the partition where each row will be stored


1.RANGE PARTITIONS

Definition: A table that is partitioned by range is partitioned in such a way that each partition contains rows for which the partitioning expression value lies within a given range.
 Creating range partitioned table
 SQL> Create table Employee(emp_no number(2),emp_name varchar(2)) partition by range(emp_no) (partition p1 values less than(100), partition p2 values less than(200), partition p3 values less than(300),partition p4 values less than(maxvalue)); 
Inserting records into range partitioned table
     SQL> Insert into Employee values(101,’a’);      -- this will go to p1
     SQL> Insert into Employee values(201,’b’);     -- this will go to p2
     SQL> Insert into Employee values(301,’c’);      -- this will go to p3
     SQL> Insert into Employee values(401,’d’);     -- this will go to p4
 Selecting records from range partitioned table
     SQL> Select *from Employee;
     SQL> Select *from Employee partition(p1);
 Adding a partition
     SQL> Alter table Employee add partition p5 values less than(400);
 Dropping a partition
    SQL> Alter table Employee drop partition p1;
 Renaming a partition
     SQL> Alter table Employee rename partition p3 to p6;
 Truncate a partition
     SQL> Alter table Employee truncate partition p5;
 Splitting a partition
    SQL> Alter table Employee split partition p2 at(120) into (partition p21,partition p22);
 Exchanging a partition
  SQL> Alter table Employee exchange partition p2 with table Employee_x;
 Moving a partition
     SQL> Alter table Employee move partition p21 tablespace ABC_TBS;


 2. LIST PARTITIONS

Definition: List partitioning enables you to explicitly control how rows map to partitions by specifying a list of discrete values for the partitioning key in the description for each partition.
 Creating list partitioned table
SQL> Create table Employee (Emp_no number(2),Emp_name varchar(2)) partition by list(Emp_no)  (partition p1 values(1,2,3,4,5), partition p2 values(6,7,8,9,10),partition p3             values(11,12,13,14,15), partition p4 values(16,17,18,19,20));
 Inserting records into list partitioned table
      SQL> Insert into Employee values(4,’xxx’);     -- this will go to p1
      SQL> Insert into Employee values(8,’yyy’);     -- this will go to p2
      SQL> Insert into Employee values(14,’zzz’);    -- this will go to p3
      SQL> Insert into Employee values(19,’bbb’);  -- this will go to p4
 Selecting records from list partitioned table
     SQL> Select *from Employee;
     SQL> Select *from Employee partition(p1);
 Adding a partition
     SQL> Alter table Employee add partition p5 values(21,22,23,24,25);
 Dropping a partition
     SQL> Alter table Employee drop partition p5;
 Renaming a partition
     SQL> Alter table Employee rename partition p5to p1;
 Truncate a partition
     SQL> Alter table Employee truncate partition p5;
 Exchanging a partition
    SQL> Alter table Employee exchange partition p1 with table Employee_x;
 Moving a partition
    SQL> Alter table Employee move partition p2 tablespace ABC_TBS;


3. HASH PARTITIONS

Definition:Hash partitioning maps data to partitions based on a hashing algorithm that Oracle applies to the partitioning key that you identify.
Creating hash partitioned table
     SQL> Create table Employee(emp_no number(2),emp_name varchar(2)) partition by      hash(emp_no) partitions 5;
     Here oracle automatically gives partition names like
                                                SYS_P1
                                                SYS_P2
                                                SYS_P3
                                                SYS_P4
                                                SYS_P5
 Inserting records into hash partitioned table(based on hash function)
     SQL> Insert into Employee values(5,’a’);      
     SQL> Insert into Employee values(8,’b’);      
     SQL> Insert into Employee values(14,’c’);    
     SQL> Insert into Employee values(19,’d’);   
 Selecting records from hash partitioned table
     SQL> Select *from Employee;
     SQL> Select *from Employee partition(SYS_P2);
 Adding a partition
     SQL> Alter table Employee add partition p9;
 Renaming a partition
    SQL> Alter table Employee rename partition p9 to p10;
 Truncate a partition
     SQL> Alter table Employee truncate partition p9;
 Exchanging a partition
 SQL> Alter table Employee exchange partition SYS_P1 with table Employee_X;
 Moving a partition
     SQL> Alter table Employee move partition SYS_P1 tablespace ABC_TBS;

String vs StringBuilder vs StringBuffer in Java

Consider below code with three concatenation functions with three different types of parameters, String, StringBuffer and StringBuilder.


// Java program to demonstrate difference between String,
// StringBuilder and StringBuffer
class Sudhakar
{
    // Concatenates to String
    public static void concat1(String s1)
    {
        s1 = s1 + "Sudhakar";
    }
 
    // Concatenates to StringBuilder
    public static void concat2(StringBuilder s2)
    {
        s2.append("Pandey");
    }
 
    // Concatenates to StringBuffer
    public static void concat3(StringBuffer s3)
    {
        s3.append("Pandey");
    }
 
    public static void main(String[] args)
    {
        String s1 = "Sudhakar";
        concat1(s1);  // s1 is not changed
        System.out.println("String: " + s1);
 
        StringBuilder s2 = new StringBuilder("Sudhakar");
        concat2(s2); // s2 is changed
        System.out.println("StringBuilder: " + s2);
 
        StringBuffer s3 = new StringBuffer("Sudhakar");
        concat3(s3); // s3 is changed
        System.out.println("StringBuffer: " + s3);
    }
}
Output:
String: Sudhakar
StringBuilder: SudhakarPandey
StringBuffer: SudhakarPandey
Explanation:
1. Concat1 : In this method, we pass a string “Sudhakar” and perform “s1 = s1 + ”Pandey”. The string passed from main() is not changed, this is due to the fact that String is immutable. Altering the value of string creates another object and s1 in concat1() stores reference of new string. References s1 in main() and cocat1() refer to different strings.
2. Concat2 : In this method, we pass a string “Sudhakar” and perform “s2.append(“Pandey”)” which changes the actual value of the string (in main) to “SudhakarPandey”. This is due to the simple fact that StringBuilder is mutable and hence changes its value.
2. Concat3 : StringBuffer is similar to StringBuilder except one difference that StringBuffer is thread safe, i.e., multiple threads can use it without any issue. The thread safety brings a penalty of performance.
Conclusion:
  • Objects of String are immutable, and objects of StringBuffer and StringBuilder are mutable.
  • StringBuffer and StringBuilder are similar, but StringBuilder is faster and preferred over StringBuffer for single threaded program. If thread safety is needed, then StringBuffer is used.

Remove duplicates from an array of small primes

Given an array of primes such that the range of primes is small. Remove duplicates from the array.
Examples:
Input :  arr[] = {3, 5, 7, 2, 2, 5, 7, 7};
Output : arr[] = {2, 3, 5, 7}
The output can be printed in any order.


Input :  arr[] = {3, 5, 7, 3, 3, 13, 5, 13, 29, 13};
Output : arr[] = {3, 5, 7, 13, 29}
The output can be printed in any order.
Source : Amazon Interview Question
Method 1 (Naive : O(n2))

A simple solution is to run two loops. Pick all elements one by one. For every picked element, check if it already seen or not. If already seen, then ignore it. Else add it to the array.
// A C++ program to implement Naive approach to
// remove duplicates.
#include <bits/stdc++.h>
using namespace std;
int removeDups(vector<int> &vect)
{
   int res_ind = 1;
   
   // Loop invariant : Elements from vect[0]
   // to vect[res_ind-1] are unique.
   for (int i=1; i<vect.size(); i++)
   {
       int j;
       for (j=0; j<i; j++)
           if (vect[i] == vect[j])
                break;
       if (j == i)
          vect[res_ind++]  = vect[i];
   }
   // Removes elements from vect[res_ind] to
   // vect[end]
   vect.erase(vect.begin()+res_ind, vect.end());
}
// Driver code
int main()
{
    vector<int> vect{3, 5, 7, 2, 2, 5, 7, 7};
    removeDups(vect);
    for (int i=0; i<vect.size(); i++)
      cout << vect[i] << " ";
    return 0;
}


Output :
3 5 7 2 
Time Complexity : O(n2)


Method 2 (Sorting : O(n Log n))

A better solutions is to first sort the array and then remove adjacent elements from sorted array.
// C++ program to remove duplicates using Sorting
#include <bits/stdc++.h>
using namespace std;
 
int removeDups(vector<int> &vect)
{
    // Sort the vector
    sort(vect.begin(), vect.end());
 
    // unique() removes adjacent duplicates.
    // unique function puts all unique elements at
    // the beginning and returns iterator pointing
    // to the first element after unique element.
    // Erase function removes elements between two
    // given iterators
    vect.erase(unique(vect.begin(), vect.end()),
              vect.end());
}
 
// Driver code
int main()
{
    vector<int> vect{3, 5, 7, 2, 2, 5, 7, 7};
    removeDups(vect);
    for (int i=0; i<vect.size(); i++)
      cout << vect[i] << " ";
    return 0;
}
Output :
2 3 5 7 
Time Complexity : O(n Log n)
Auxiliary Space : O(1)


Method 3 (Hashing : O(n))

The idea is keep track of visited elements in a hash table.
// C++ program to remove duplicates using Hashing
#include <bits/stdc++.h>
using namespace std;
 
int removeDups(vector<int> &vect)
{
    // Create a set from vector elements
    unordered_set<int> s(vect.begin(), vect.end());
 
    // Take elements from set and put back in
    // vect[]
    vect.assign(s.begin(), s.end());
}
 
// Driver code
int main()
{
    vector<int> vect{3, 5, 7, 2, 2, 5, 7, 7};
    removeDups(vect);
    for (int i=0; i<vect.size(); i++)
      cout << vect[i] << " ";
    return 0;
}
Output :
2 7 5 3
Time Complexity : O(n)
Auxiliary Space : O(n)


Method 4 (Works only for small range : O(n))

This solutions uses the fact that numbers are primes. But it works only when product of all distinct primes in array is less than maximum value in long long int.
// Removes duplicates using multiplication of
// distinct primes in array
#include <bits/stdc++.h>
using namespace std;
 
int removeDups(vector<int> &vect)
{
   long long int prod = vect[0];
   int res_ind = 1;
   for (int i=1; i<vect.size(); i++)
   {
       if (prod % vect[i] != 0)
       {
          vect[res_ind++]  = vect[i];
          prod *= vect[i];
       }
   }
   vect.erase(vect.begin()+res_ind, vect.end());
}
 
// Driver code
int main()
{
    vector<int> vect{3, 5, 7, 2, 2, 5, 7, 7};
    removeDups(vect);
    for (int i=0; i<vect.size(); i++)
      cout << vect[i] << " ";
    return 0;
}
Output :
3 5 7 2
Time Complexity : O(n)
Auxiliary Space : O(1)
Note that this solution would not work if there are composites in array.

Wednesday, 10 June 2015

If a method throws NullPointerException in super class, can we override it with a method which throws RuntimeException?