what about taking sum of the boundaries of the rectangle and omiting the
boundary with least sum and including the rest of rectangle in the sub-set?


On Mon, Jan 16, 2012 at 8:51 PM, atul anand <atul.87fri...@gmail.com> wrote:

>
> find cumulative sum of each column.
> now for each arr[x][y] = sum of arr[i=0 to x] [j] ;
>
> a1 b1 c1 d1
> a2 b2 c2 d2
> a3 b3 c3 d3
> a4 d4 c4 d4
>
> now we have reduced this problem to find max-subarray . which can
> be efficiently calculated using kadane's algo for each row.
>
> NOTE: now suppose if row = 0  does not participate in calculating max sum
> matrix so u need to subtract a1 from a2,a3,a4 .... similarly for other
> element in row 1,2,3.
>
> now updated matrix considered i.e from (row =1 to row =3 ). for this
> updated matrix,
>  each[x][y] is the sum of arr[i=1 to x] [j].
>
> similarly do for other elements.
>
> On Mon, Jan 16, 2012 at 6:55 AM, Ashish Goel <ashg...@gmail.com> wrote:
>
>> given a m*n matrix, find the subset rectangle with max sum (any other
>> rectangle taken would have lesser sum)
>> Best Regards
>> Ashish Goel
>> "Think positive and find fuel in failure"
>> +919985813081
>> +919966006652
>>
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-- 
Umer

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