Re: [algogeeks] matrix question ???!!!!!!!!!!??????????
>>> it can be done in O(3n). in worst case one row will have max and anothr
>>> row
>>> will have min so the third row will be your o/p to print
Why is that? The min/max element can occur in every row:
1 4 2 10
2 1 10 4
2 3 4 5
1 2 8 10
?
Besides that, you need to scan the whole array to find the min/max value
in the matrix.
On 08/14/2011 12:20 PM, shady wrote:
no it is 3*n only read it again
On Mon, Aug 15, 2011 at 12:45 AM, Amir Aavani wrote:
On 08/14/2011 11:46 AM, aditya kumar wrote:
it can be done in O(3n). in worst case one row will have max and anothr
row
will have min so the third row will be your o/p to print
Do you mean O(n^3)?
Consider this { O(n^2) }:
1- Scan the whole matrix and find minimum and maximum entries in the
matrix. Let Delta be the difference between maximum and minimum.
2- For each row, find the minimum and maximum entries in that row. If
their difference is exactly Delta, then print that row.
Amir
On Mon, Aug 15, 2011 at 12:00 AM, Karthikeyan palani<
karthikeyan...@gmail.com> wrote:
sorry O(n^2) s the time complexity
On 14 August 2011 23:56, shady wrote:
how can it be O(n) when there are itself n*n elements..
PS : no sharing of code, else the inevitable
On Sun, Aug 14, 2011 at 11:51 PM, Karthikeyan palani<
karthikeyan...@gmail.com> wrote:
Given a n x n matrix. .number are randomly placed. .print any one row
which doesn’t have min
and max elements. Time Complexity : 0(n)
if anyone know the code.. pls share!!!
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Re: [algogeeks] matrix question ???!!!!!!!!!!??????????
On 08/14/2011 11:46 AM, aditya kumar wrote:
it can be done in O(3n). in worst case one row will have max and anothr row
will have min so the third row will be your o/p to print
Do you mean O(n^3)?
Consider this { O(n^2) }:
1- Scan the whole matrix and find minimum and maximum entries in the
matrix. Let Delta be the difference between maximum and minimum.
2- For each row, find the minimum and maximum entries in that row. If
their difference is exactly Delta, then print that row.
Amir
On Mon, Aug 15, 2011 at 12:00 AM, Karthikeyan palani<
karthikeyan...@gmail.com> wrote:
sorry O(n^2) s the time complexity
On 14 August 2011 23:56, shady wrote:
how can it be O(n) when there are itself n*n elements..
PS : no sharing of code, else the inevitable
On Sun, Aug 14, 2011 at 11:51 PM, Karthikeyan palani<
karthikeyan...@gmail.com> wrote:
Given a n x n matrix. .number are randomly placed. .print any one row
which doesn’t have min
and max elements. Time Complexity : 0(n)
if anyone know the code.. pls share!!!
--
karthikeyankkn
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Re: [algogeeks] AMAZON Q
Here is an algorithm which solves this problem in O(nlog^2 n):
Create a set of pairs B= {: i= 1..n}.
Sort B.
Now use the following Divide and Conquer algorithm (Ar_Low is a global
array):
Fill (Start, End, Low_Array)
if Start= End then
Low_Array [Start]= 0
else
Initialize Low_Array with zero.
Fill (Start, (Start+ End)/ 2, Low_Array)
Fill ((Start+ End)/ 2+ 1, End, Low_Array)
Scan B (the list of sorted pairs) and select the pairs whose
second value is in range (End/ 2+ 1, End), call the resulting list B'
For each i in (Start, (Start+ End)/ 2):
Lets j be the position of arr [i] in B'
Set Low_Array [i]= Low_Array [i]+ j
The running time of this algorithm is:
n\log n for sorting B+
T(n)= 2T(n/2)+ n\log n => T(n)= n\log^2 n
So, the total running time is: n\log^2 n .
Amir
On 07/26/2011 06:48 AM, Piyush Sinha wrote:
You have an array like ar[]= {1,3,2,4,5,4,2}. You need to create
another array ar_low[] such that ar_low[i] = number of elements lower
than or equal to ar[i] in ar[i+1:n-1].
So the output of above should be {0,2,1,2,2,1,0}
Time complexity : O(nlogn)
use of extra space allowed.
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