Answer / vandan

in case of linear search system will search from begining.
that means Example : z table contains single field with values
1 2 3 4 5 6 7 8 9
if u r searching for a value then system will starts from
first position. if required value is founded then execution
will comes out from z table.

in case of binary search system will starts from mid point.
if value is not founded then it will search for upper half.
in that upper half it will check mid point.like that search
will takes place.

Answer / jayati saha

Linear search is ver easy and always search in sequential
way.
It starts searching from the first element upto the last
element until the search finds the number .

Let us take an example.
Let there are a list of numbers.
1,3,8,6,5,8,2,9
We have to search if 5 is there in the list or not.
So, we start from the first element from 1.
then we see 3,8,6 and then find 5.
As soon as we find the search element 5 in the list we stop
searching,otherwise we continue the search upto the last
element.
Binary search

In binary search we always start from the middle elment and
compare the middle element to the search element.
If the search element is less than the middle element then
we go to the less half otherwise we go to the larger half.
In binary search the nos should be sorted in ascending
order at first.
Let us took an example of binary search.
Assume that we are reading a book.
The book has a huge volume.
let we have to find the page no 120 in this book.
The book has 200 pages approx.
Then we do not start looking for the page from the first
page, rather we look at the middle.
If the arbitrary page is 101 then we think that 120 paghe
is more than the 101 page so we go to the half which
contains more pages. So by using this method again and
again we can finally find out the page .

Answer / asit kumsr tripathy

Linear search is ver easy and always search in sequential
way.
It starts searching from the first element upto the last
element until the search finds the number .

Let us take an example.
Let there are a list of numbers.
1,3,8,6,5,8,2,9
We have to search if 5 is there in the list or not.
So, we start from the first element from 1.
then we see 3,8,6 and then find 5.
As soon as we find the search element 5 in the list we stop
searching,otherwise we continue the search upto the last
element.
Binary search

In binary search we always start from the middle elment and
compare the middle element to the search element.
If the search element is less than the middle element then
we go to the lower half otherwise we go to the upper half.
In binary search the nos should be sorted in ascending
order at first.
Let us took an example of binary search.
Assume that we are reading a book.
The book has a huge volume.
let we have to find the page no 120 in this book.
The book has 200 pages approx.
Then we do not start looking for the page from the first
page, rather we look at the middle.
If the arbitrary page is 101 then we think that 120 paghe
is more than the 101 page so we go to the half which
contains more pages. So by using this method again and
again we can finally find out the page .

Answer / kaloo

binary search is complex whle the linear search is more easy

Answer / prabhjot singh

1.linear search is easy bt takes more time to search an
element as it compare all element sequentially...
2.elements r arranged randomly.

1. in binary search it start searching from middle, if the
searching element is not found in middle then it goes to
1/4,and vice versa..
and hence take less time than linear search.
2.elements r arranged in sorted orrder.

Answer / greeshma and arya

time complexity of linear search is O[n] whereas that of binary search is O[logn].therefore binary search is faster than linear search.

Answer / indrani chowdhury

1.linear search is easy bt takes more time to search an
element as it compare all element sequentially...
2.elements r arranged randomly.

1. in binary search it start searching from middle, if the
searching element is not found in middle then it goes to
1/4,and vice versa..
and hence take less time than linear search.
2.elements r arranged in sorted orrder.

Answer / jeyachandran.p

Binary search is an efficient search compared to linear
search.Binary search requires sorted array while linear
search operates on unsorted array.

In binary search ,we looks through half of the list i.e
the element to be search is compared with mid value, if it
equals to mid value,search operation is over.otherwise if
the search element is smaller we have to search in the first
half of the list else we have to look in second half of the
list.

But linear search or sequential search requires an complete
search over the list.As far as ,Binary search is very best
in time and efficiency.It requires O[log n} times but linear
search requires O[n]times.

Answer / anupama singh

| Binary search | Linear search |
-------------------------------------------------------------------------------------------------------------------------
1).Data must be in a sorted order | 1).Data any order
2).Time complexity is O(log n) | 2).Time complexity is O(n).
3).Only 1 "When" condition used | 3).Any no. of "When" condition used
4).Only "=" relation operator is used | 4).any relation operator is used
5).Access is faster | 5).Access is slow
6).Only single dimensional array used | 6).single/multi dimensional array used

Answer / amit kumar thakur

liner search -data is a linear array with (n)and item is
given this variable algorith find the location (loc)of item
in data. (linear search is used when the address is find)
(Aigorith)
step .1 [insert item at the end of data ]
set data [n+1]:=item
step. 2 set loc:=1
step.3 [search for item]
repeat while data [loc] is not equal to item
set loc :=loc+1
[end of loop]
step.4 if loc =n+1, then
set loc:=0
[end if]
step.5 EXIT.

Binary search
suppose data is an array which is store in increasing
numerical order .then there is a extremly effficient
searching called binary search .
which can be used to find the loc of a given item of
information in data.
e.g.
suppose one want to find the location of some name in a
telephone directory(or some word in a dictionry) then we
use the binary search not the linear search.
ALGORITH OF BINARY SEARCH.
STEP.1 SET BEG= LB, END=UB AND MID= INT.
STEP.2 BEG<=END AND DATA [MID]!ITEM
STEP.3 IF ITEM< DATA[MID], THEN:
set end:=mid-1
step.4 set mid=int[(beg+end)/2]
[end if]
step.5 EXIT.

In SAP R/3 architecture, how many databases can exist?

2 Answers   IBM,

Can you transport variants of multiple programs in one step?

1 Answers  

Hi, My question: Iam an ABAP Programmer with 3+yrs exp would like to learn ABAP CRM . Plz let me know the best institues(coaching centers) in HYd to learn the same.

4 Answers  

What is an abap dictionary?

1 Answers  

do 4 times. sy-tabix,sy-index. end do. what is the output ?

3 Answers   TCS,

send coding to write a program to create customer quotation using BAPI 'BAPI_QUOTATION_CREATEFROMDATA2/BAPI_QUOTATION_CREATEFR OMDATA'

0 Answers   SGT,

what is the tcode for creating the invoice?

6 Answers  

correct all performance problem without changing any functionality REPORT z_assignment_01. DATA: gt_user TYPE TABLE OF usr02 WITH HEADER LINE, gt_user_role TYPE TABLE OF agr_users WITH HEADER LINE, ********** DO NOT CHANGE THIS SECTION ********** g_begintime TYPE i, g_endtime TYPE i, g_tottime TYPE i. DATA: BEGIN OF gt_output OCCURS 0, bname LIKE usr02-bname, name_text LIKE adrp-name_text, agr_name LIKE agr_users-agr_name, text LIKE agr_texts-text, role_count TYPE i, END OF gt_output. TYPE-POOLS: slis. START-OF-SELECTION. GET RUN TIME FIELD g_begintime. ********** END NO CHANGE ********** SELECT * INTO TABLE gt_user FROM usr02 WHERE class > 'E'. SELECT * INTO TABLE gt_user_role FROM agr_users FOR ALL ENTRIES IN gt_user WHERE uname = gt_user-bname. PERFORM combine_tables. PERFORM get_addl_data. ********** DO NOT CHANGE THIS SECTION ********** GET RUN TIME FIELD g_endtime. g_tottime = ( g_endtime / 1000000 ) - ( g_begintime / 1000000 ). MESSAGE s398(00) WITH g_tottime. PERFORM output. ********** END NO CHANGE ********** *&---------------------------------------------------------------------* *& Form combine_tables *&---------------------------------------------------------------------* * Combine user and role tables *----------------------------------------------------------------------* FORM combine_tables. DATA: ls_usr21 TYPE usr21, ls_adrp TYPE adrp, ls_texts TYPE agr_texts. LOOP AT gt_user. * Get user name SELECT * INTO ls_usr21 FROM usr21 WHERE bname = gt_user-bname. SELECT * INTO ls_adrp FROM adrp WHERE persnumber = ls_usr21-persnumber. gt_output-name_text = ls_adrp-name_text. ENDSELECT. ENDSELECT. LOOP AT gt_user_role WHERE uname = gt_user-bname. * Only process roles that are currently active CHECK gt_user_role-from_dat <= sy-datum AND gt_user_role-to_dat >= sy-datum. * Get role text SELECT * INTO ls_texts FROM agr_texts WHERE agr_name = gt_output-agr_name AND spras = sy-langu AND line = 0. gt_output-text = ls_texts-text. ENDSELECT. gt_output-bname = gt_user-bname. gt_output-agr_name = gt_user_role-agr_name. APPEND gt_output. ENDLOOP. ENDLOOP. ENDFORM. " combine_tables *&---------------------------------------------------------------------* *& Form get_addl_data *&---------------------------------------------------------------------* * Get extra fields *----------------------------------------------------------------------* FORM get_addl_data. DATA: l_count TYPE i. * Count number of roles per user SORT gt_user_role BY uname. LOOP AT gt_user_role. ADD 1 TO l_count. AT END OF uname. LOOP AT gt_output WHERE bname = gt_user_role-uname. gt_output-role_count = l_count. MODIFY gt_output. CLEAR l_count. ENDLOOP. ENDAT. ENDLOOP. ENDFORM. " get_addl_data *&---------------------------------------------------------------------* *& Form output *&---------------------------------------------------------------------* * Output report *----------------------------------------------------------------------* FORM output. DATA: lt_fieldcat TYPE slis_t_fieldcat_alv, ls_fieldcat TYPE slis_fieldcat_alv, l_repid LIKE sy-repid, ls_layout TYPE slis_layout_alv, ls_variant TYPE disvariant. l_repid = sy-repid. CALL FUNCTION 'REUSE_ALV_FIELDCATALOG_MERGE' EXPORTING i_program_name = l_repid i_internal_tabname = 'GT_OUTPUT' i_inclname = l_repid CHANGING ct_fieldcat = lt_fieldcat EXCEPTIONS inconsistent_interface = 1 program_error = 2 OTHERS = 3. IF sy-subrc <> 0. MESSAGE ID sy-msgid TYPE sy-msgty NUMBER sy-msgno WITH sy-msgv1 sy-msgv2 sy-msgv3 sy-msgv4. ENDIF. ls_fieldcat-seltext_l = 'Role Count'. ls_fieldcat-seltext_m = 'Role Count'. ls_fieldcat-seltext_s = 'Role Count'. ls_fieldcat-reptext_ddic = 'Role Count'. MODIFY lt_fieldcat FROM ls_fieldcat TRANSPORTING seltext_l seltext_m seltext_s reptext_ddic WHERE fieldname = 'ROLE_COUNT'. ls_layout-zebra = 'X'. ls_layout-colwidth_optimize = 'X'. CALL FUNCTION 'REUSE_ALV_GRID_DISPLAY' EXPORTING is_layout = ls_layout it_fieldcat = lt_fieldcat i_save = 'A' is_variant = ls_variant TABLES t_outtab = gt_output EXCEPTIONS program_error = 1 OTHERS = 2. IF sy-subrc <> 0. MESSAGE ID sy-msgid TYPE sy-msgty NUMBER sy-msgno WITH sy-msgv1 sy-msgv2 sy-msgv3 sy-msgv4. ENDIF. ENDFORM. " output

0 Answers  

by which function module we are going to put data into sapscript ?

3 Answers   IBM,

what are the main technical differences between oracle and sap? it would be more helpful to me if get this answer...

0 Answers  

How to add background picture/graphics in alv report?

1 Answers   Accenture,

What are the functional areas? User groups?

0 Answers