Hey ankit, i gave java code also...didn't u check it in the link...anyway i
am explaining here.........
*Note : Position count starts from 0. *
*ex: {1,2,3,4} ...position of '1' is zero............*
*
*
*In the below approach,we are checking element position in the modified
list(after deletion operation in the previous iteration).*
given array is : ar[]= {1,3,2,4,5,4,2}.
construct a list with array elements and sort it.Now list contains
:1,2,2,3,4,4,5
Now traverse through array elements from i=0 to n-1(start to end)
store result in result[] array.
list is *1*-2-2-3-4-4-5
for i=0, a[0]=1,search for a[0] position in the list(right
occurrence).a[0]=1 position in list is '0'
add '0' to result.===>result[0]=0;
remove the element a[0] from the list.now list contains 2-2-*3*-4-4-5
for i=1,a[1]=3,search for a[1] position in the list(right occurence).a[1]=3
position in list is '2'
add '2' to result====>result[1]=2
remove the element a[1] from the list.now list contains 2-*2*-4-4-5
for i=2,a[2]=2,search for a[2] position in the list(right occurence).a[2]=2
position in list is '1'
add '1' to result====>result[2]=1
remove the element a[2] from the list.now list contains 2-4-*4*-5
for i=3,a[3]=4,search for a[3] position in the list(right occurence).a[3]=4
position in list is '2'
add '2' to result====>result[3]=2
remove the element a[3] from the list.now list contains 2-4-5
for i=4,a[4]=5,search for a[4] position in the list(right occurence).a[4]=5
position in list is '2'
add '2' to result====>result[4]=2
remove the element a[4] from the list.now list contains 2-*4*
for i=5,a[5]=4,search for a[5] position in the list(right occurence).a[5]=4
position in list is '1'
add '1' to result====>result[5]=1
remove the element a[5] from the list.now list contains 2
for i=6,a[6]=2,search for a[6] position in the list(right occurence).a[6]=2
position in list is '0'
add '0' to result====>result[6]=0
remove the element a[6] from the list.now list is empty.
resultant array contains :
from all conditions :
add '0' to result.===>result[0]=0;
add '2' to result====>result[1]=2
add '1' to result====>result[2]=1
add '1' to result====>result[3]=2
add '2' to result====>result[4]=2
add '1' to result====>result[5]=1
add '0' to result====>result[6]=0
expected result {0,2,1,2,2,1,0}
actual result ==>{0,2,1,2,2,1,0}
I hope u r clear now.............Please let me know if you still have
doubts...........
U can execute the java code given in link :
http://rajeevprasanna.blogspot.com/2011/07/count-number-of-min-elements-on-right.html
I will be more happy if you give me failed cases...................
On Wed, Jul 27, 2011 at 11:25 AM, ankit sambyal <ankitsamb...@gmail.com>wrote:
> @rajeev :try the example given in the question. And explain ur algo with
> that example
>
--
Thank You
Rajeev Kumar
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