we can convert all rows in an equivalent integer and then sort all numbers and
for(int i=1;i=n;i++)
{
if(sorted[i]==sorted[i-1])
prinf(sorted[i]);
}
On 6/28/11, sunny agrawal sunny816.i...@gmail.com wrote:
@Ankit
if N is large how will you hash the rows, numbers will be very large
can
if N = 64 then we can convert all rows in an equivalent integer and then
sort all numbers and print distinct No.s
else
Worst case Solution would be to to check for each pair of rows and match -
O(N^3)
one more solution i can think of is to divide and conquer solution which
goes like this
based
u can use hashing ...
hash fun can b base2 to base10
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In any case, I don't think the complexity can be improved from O(n^2)
because even in creating hash function every column element of every row
which itself is N^2 only..
On Tue, Jun 28, 2011 at 11:38 AM, Ankit Agrawal
ankitmnnit.agra...@gmail.com wrote:
u can use hashing ...
hash fun can b
@Ankit
if N is large how will you hash the rows, numbers will be very large
can you explain using given example ?
On Tue, Jun 28, 2011 at 11:38 AM, Ankit Agrawal
ankitmnnit.agra...@gmail.com wrote:
u can use hashing ...
hash fun can b base2 to base10
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Given a binary matrix of N X N of integers , you need to return only unique
rows of binary arrays
eg:
0 1 0 0 1
1 0 1 1 0
0 1 0 0 1
1 1 1 0 0
ans:
0 1 0 0 1
1 0 1 1 0
1 1 1 0 0
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