00
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010100
011100
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00
In this case, when i and j are 1, your algorithm will set s = 3. The if
statement will fail, and it will never notice that it could have formed a
square with s=2.
Don
On Sunday, March 30, 2014 9:49:21 AM UTC-4, atul007 wrote:
@don : According to
The sort is what makes this O(n*log n). The processing after the sort is
O(N).
So to describe the algorithm, after sorting A, it steps through the sorted
array, determining what the value of K would have to be at that point such
that setting the remaining values to K would cause the sum to be
@Don: what is the point of considering s=2 when we have already found
s=3.As question says find the maximum subsquare. Which is of size 3 and
this the expected outcome.
On Mon, Mar 31, 2014 at 11:28 PM, Don dondod...@gmail.com wrote:
00
00
010100
011100
01
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In this
The only square is found when s=2. Your program will look at s=3 and not
find a square, so it will never find the square.
Try it and you will see what I mean..
Don
On Monday, March 31, 2014 2:42:13 PM UTC-4, atul007 wrote:
@Don: what is the point of considering s=2 when we have already
