compute the sum array by using this
>>> formula
>>>
>>> sum[i][j] = sum[i-1][j-1] + (sum[i-1][j] - sum[i-1][j-1]) + (sum[i][j-1]
>>> - sum[i-1][j-1])+ip[i][j]
>>> [smaller rect] [that row sum value][that
>>> c
se it will be Ans = 18-0+0 = 18
>
> Please lemme know if any bugs with the logic.
>
>
> On Sun, Aug 12, 2012 at 6:27 PM, Srividhya Sampath > wrote:
>
>>
>> @ Vicky
>>
>> Can yo explain with an illustration ?
>>
>>
>> On Sat, Aug 11, 2
@ Vicky
Can yo explain with an illustration ?
On Sat, Aug 11, 2012 at 10:07 PM, ~*~VICKY~*~ wrote:
> May be you can consider creating a 2d array to pre process and store all
> the rectangle sums as a dependent subproblem, the sum of larger rect will
> be currValuesAdded+OldRectSum. So when you g
say yo have 3*4 matrix...
0 1 2 3
4 5 6 7
8 9 10 11
if the co-ordinates are (0,0),(0,2),(1,0),(1,2)
the o/p should be 0+1+2+4+5+6
On Sat, Aug 11, 2012 at 9:57 PM, adarsh kumar wrote:
> Sum of the integers meaning? Do you mind giving an example test case?
>
> regards.
>
> On Sat, Aug 11, 2012
