Use divide and conquer. take 2 array at a time and .so you are merging two
array at a time.
num_of_list=m;
length of list=n;
while(num_of_list > 1)
{
while( (num of list where length = length_of_list) >2)
{
merge two lists of length (length_of_list);
}
if(num_of_list %2==0)
num_of_list/=2;
else
num_of_list/=2+1;
length of list=n;
}
(it is just a general idea , you have to take care of the left over list
every time , the check for that i havent posted)
so time complexity is
2*n* (m/2) + 2* 2n* (m/4) .............. log(m) times.
so complexity is n*m*log(m)
On Sat, Jul 16, 2011 at 2:43 PM, aseem garg <[email protected]> wrote:
> Q2. Given m arrays of n size each, give an algorithm to combine these
> arrays into a single array with sorted elements. Also tell the time
> complexity of your solution.
> Aseem
>
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