>>> 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<amir.aav...@gmail.com> 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<sinv...@gmail.com> 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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