Re: [algogeeks] Constant time solution needed

2012-08-14 Thread Kailash Bagaria
@vicky
your approach doesn't work for  (x1,y1) = (1,2)
  (x2,y2) = (2,3)

By your Approach:-
  Ans= sumArray[x2][y2] - sumArray[x1][y1] + ip[x1][y1]=66-18+6=
*54*
But Actual Ans is 6+7+10+11=*34*

On Sun, Aug 12, 2012 at 8:19 PM, ~*~VICKY~*~ venkat.jun...@gmail.comwrote:

 Lets build the array for the example you gave.

 i/p:

 0 1 2 3
 4 5  6 7
 8 9 10 11

 (x1,y1) = (0,0)
 (x2,y2) = (1,2)
 sumArray
 0  1 23
 4  10  18   28
 12 27  45  66
 (will take O(n^2) to build above array)
 So now when you get coordinates as input, you can calc the sum by

 Ans = sumArray[x2][y2] - sumArray[x1][y1] + ip[x1][y1]

 For our case 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 srisam261...@gmail.com
  wrote:


 @ Vicky

 Can yo explain with an illustration ?


 On Sat, Aug 11, 2012 at 10:07 PM, ~*~VICKY~*~ venkat.jun...@gmail.comwrote:

 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 get the coordinate as input u
 can calc the needed sum by subtracting sum of big rect and small rect which
 is not included in the given coordinates. This can be called constant time
 if u don't include the preprocessing time.


 On Sat, Aug 11, 2012 at 9:57 PM, adarsh kumar algog...@gmail.comwrote:

 Sum of the integers meaning? Do you mind giving an example test case?

 regards.

 On Sat, Aug 11, 2012 at 7:10 PM, Srividhya srisam261...@gmail.comwrote:

 hi all:)

 The coordinates of a rectangle will be specified. there is a matrix of
 integers. yo should find the sum of the integers that fall in the region
 specified by the  coordinates .

 The solution to be in constant time .

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B-tech 4th year
Computer Science  Engineering
Indian Institute of Technology, Roorkee
Roorkee, India (247667)

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Re: [algogeeks] Constant time solution needed

2012-08-14 Thread ~*~VICKY~*~
Hey I just gave an idea, I believe solve it yourself :) I appreciate your
effort.

On Tue, Aug 14, 2012 at 4:34 PM, Kailash Bagaria kbkailashbaga...@gmail.com
 wrote:

 @vicky
 your approach doesn't work for  (x1,y1) = (1,2)
   (x2,y2) = (2,3)

 By your Approach:-
   Ans= sumArray[x2][y2] - sumArray[x1][y1] +
 ip[x1][y1]=66-18+6=*54*
 But Actual Ans is 6+7+10+11=*34*

 On Sun, Aug 12, 2012 at 8:19 PM, ~*~VICKY~*~ venkat.jun...@gmail.comwrote:

 Lets build the array for the example you gave.

 i/p:

 0 1 2 3
 4 5  6 7
 8 9 10 11

 (x1,y1) = (0,0)
 (x2,y2) = (1,2)
 sumArray
 0  1 23
 4  10  18   28
 12 27  45  66
 (will take O(n^2) to build above array)
 So now when you get coordinates as input, you can calc the sum by

 Ans = sumArray[x2][y2] - sumArray[x1][y1] + ip[x1][y1]

 For our case 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 
 srisam261...@gmail.com wrote:


 @ Vicky

 Can yo explain with an illustration ?


 On Sat, Aug 11, 2012 at 10:07 PM, ~*~VICKY~*~ 
 venkat.jun...@gmail.comwrote:

 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 get the coordinate as input
 u can calc the needed sum by subtracting sum of big rect and small rect
 which is not included in the given coordinates. This can be called constant
 time if u don't include the preprocessing time.


 On Sat, Aug 11, 2012 at 9:57 PM, adarsh kumar algog...@gmail.comwrote:

 Sum of the integers meaning? Do you mind giving an example test case?

 regards.

 On Sat, Aug 11, 2012 at 7:10 PM, Srividhya srisam261...@gmail.comwrote:

 hi all:)

 The coordinates of a rectangle will be specified. there is a matrix
 of integers. yo should find the sum of the integers that fall in the 
 region
 specified by the  coordinates .

 The solution to be in constant time .

 --
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 Cheers,

   Vicky

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   Vicky

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 --

 ‘Kailash Bagaria’
 B-tech 4th year
 Computer Science  Engineering
 Indian Institute of Technology, Roorkee
 Roorkee, India (247667)

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Cheers,

  Vicky

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Re: [algogeeks] Constant time solution needed

2012-08-13 Thread Srividhya Sampath
@vicky

Got it . thanks ..is it possible to ignore the preprocessing time when
calculating time complexity?


On Sun, Aug 12, 2012 at 10:02 PM, shady sinv...@gmail.com wrote:

 @venkat +1

 On Sun, Aug 12, 2012 at 9:09 PM, ~*~VICKY~*~ venkat.jun...@gmail.comwrote:

 @Arun: This approach is constant time once the array is build for any
 queries that follows. :) You know sum for all possible rectangles in the
 given 2d array thats makes it better than computing sum for each input.
 Hope it makes sense


 On Sun, Aug 12, 2012 at 9:07 PM, ~*~VICKY~*~ venkat.jun...@gmail.comwrote:

 Fine, the basic idea of using dp here is sum of each rectangle is a
 dependent sub problem. So when u find sum for smaller rectangle we can use
 it to compute sum of bigger rectangle with new coordinates added to
 previous small rectangle. So u can 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
 col sum value]



 On Sun, Aug 12, 2012 at 8:54 PM, Srividhya Sampath 
 srisam261...@gmail.com wrote:

 @vicky

 can yo explain the logic behind the 'Sum Array' computation (if
 possible elaborately )?


 On Sun, Aug 12, 2012 at 8:19 PM, ~*~VICKY~*~ 
 venkat.jun...@gmail.comwrote:

 Lets build the array for the example you gave.

 i/p:

 0 1 2 3
 4 5  6 7
 8 9 10 11

 (x1,y1) = (0,0)
 (x2,y2) = (1,2)
 sumArray
 0  1 23
 4  10  18   28
 12 27  45  66
 (will take O(n^2) to build above array)
 So now when you get coordinates as input, you can calc the sum by

 Ans = sumArray[x2][y2] - sumArray[x1][y1] + ip[x1][y1]

 For our case 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 
 srisam261...@gmail.com wrote:


 @ Vicky

 Can yo explain with an illustration ?


 On Sat, Aug 11, 2012 at 10:07 PM, ~*~VICKY~*~ 
 venkat.jun...@gmail.com 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 get the coordinate as 
 input
 u can calc the needed sum by subtracting sum of big rect and small rect
 which is not included in the given coordinates. This can be called 
 constant
 time if u don't include the preprocessing time.


 On Sat, Aug 11, 2012 at 9:57 PM, adarsh kumar algog...@gmail.comwrote:

 Sum of the integers meaning? Do you mind giving an example test
 case?

 regards.

 On Sat, Aug 11, 2012 at 7:10 PM, Srividhya 
 srisam261...@gmail.comwrote:

 hi all:)

 The coordinates of a rectangle will be specified. there is a
 matrix of integers. yo should find the sum of the integers that fall 
 in the
 region specified by the  coordinates .

 The solution to be in constant time .

 --
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 --
 Cheers,

   Vicky

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   Vicky

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Re: [algogeeks] Constant time solution needed

2012-08-13 Thread ~*~VICKY~*~
Its something which is good to have discussion with the interviewer rather
than making any assumption. :)

On Mon, Aug 13, 2012 at 6:07 PM, Srividhya Sampath
srisam261...@gmail.comwrote:

 @vicky

 Got it . thanks ..is it possible to ignore the preprocessing time when
 calculating time complexity?



 On Sun, Aug 12, 2012 at 10:02 PM, shady sinv...@gmail.com wrote:

 @venkat +1

 On Sun, Aug 12, 2012 at 9:09 PM, ~*~VICKY~*~ venkat.jun...@gmail.comwrote:

 @Arun: This approach is constant time once the array is build for any
 queries that follows. :) You know sum for all possible rectangles in the
 given 2d array thats makes it better than computing sum for each input.
 Hope it makes sense


 On Sun, Aug 12, 2012 at 9:07 PM, ~*~VICKY~*~ venkat.jun...@gmail.comwrote:

 Fine, the basic idea of using dp here is sum of each rectangle is a
 dependent sub problem. So when u find sum for smaller rectangle we can use
 it to compute sum of bigger rectangle with new coordinates added to
 previous small rectangle. So u can 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
 col sum value]



 On Sun, Aug 12, 2012 at 8:54 PM, Srividhya Sampath 
 srisam261...@gmail.com wrote:

 @vicky

 can yo explain the logic behind the 'Sum Array' computation (if
 possible elaborately )?


 On Sun, Aug 12, 2012 at 8:19 PM, ~*~VICKY~*~ 
 venkat.jun...@gmail.comwrote:

 Lets build the array for the example you gave.

 i/p:

 0 1 2 3
 4 5  6 7
 8 9 10 11

 (x1,y1) = (0,0)
 (x2,y2) = (1,2)
 sumArray
 0  1 23
 4  10  18   28
 12 27  45  66
 (will take O(n^2) to build above array)
 So now when you get coordinates as input, you can calc the sum by

 Ans = sumArray[x2][y2] - sumArray[x1][y1] + ip[x1][y1]

 For our case 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 
 srisam261...@gmail.com wrote:


 @ Vicky

 Can yo explain with an illustration ?


 On Sat, Aug 11, 2012 at 10:07 PM, ~*~VICKY~*~ 
 venkat.jun...@gmail.com 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 get the 
 coordinate as
 input u can calc the needed sum by subtracting sum of big rect and 
 small
 rect which is not included in the given coordinates. This can be called
 constant time if u don't include the preprocessing time.


 On Sat, Aug 11, 2012 at 9:57 PM, adarsh kumar 
 algog...@gmail.comwrote:

 Sum of the integers meaning? Do you mind giving an example test
 case?

 regards.

 On Sat, Aug 11, 2012 at 7:10 PM, Srividhya srisam261...@gmail.com
  wrote:

 hi all:)

 The coordinates of a rectangle will be specified. there is a
 matrix of integers. yo should find the sum of the integers that fall 
 in the
 region specified by the  coordinates .

 The solution to be in constant time .

 --
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 Cheers,

   Vicky

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Re: [algogeeks] Constant time solution needed

2012-08-12 Thread Srividhya Sampath
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 algog...@gmail.com wrote:

 Sum of the integers meaning? Do you mind giving an example test case?

 regards.

 On Sat, Aug 11, 2012 at 7:10 PM, Srividhya srisam261...@gmail.com wrote:

 hi all:)

 The coordinates of a rectangle will be specified. there is a matrix of
 integers. yo should find the sum of the integers that fall in the region
 specified by the  coordinates .

 The solution to be in constant time .

 --
 You received this message because you are subscribed to the Google Groups
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 https://groups.google.com/d/msg/algogeeks/-/qHSmXBshmS4J.
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Re: [algogeeks] Constant time solution needed

2012-08-12 Thread Srividhya Sampath
@ Vicky

Can yo explain with an illustration ?

On Sat, Aug 11, 2012 at 10:07 PM, ~*~VICKY~*~ venkat.jun...@gmail.comwrote:

 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 get the coordinate as input u
 can calc the needed sum by subtracting sum of big rect and small rect which
 is not included in the given coordinates. This can be called constant time
 if u don't include the preprocessing time.


 On Sat, Aug 11, 2012 at 9:57 PM, adarsh kumar algog...@gmail.com wrote:

 Sum of the integers meaning? Do you mind giving an example test case?

 regards.

 On Sat, Aug 11, 2012 at 7:10 PM, Srividhya srisam261...@gmail.comwrote:

 hi all:)

 The coordinates of a rectangle will be specified. there is a matrix of
 integers. yo should find the sum of the integers that fall in the region
 specified by the  coordinates .

 The solution to be in constant time .

 --
 You received this message because you are subscribed to the Google
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 Cheers,

   Vicky

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Re: [algogeeks] Constant time solution needed

2012-08-12 Thread ~*~VICKY~*~
Lets build the array for the example you gave.

i/p:

0 1 2 3
4 5  6 7
8 9 10 11

(x1,y1) = (0,0)
(x2,y2) = (1,2)
sumArray
0  1 23
4  10  18   28
12 27  45  66
(will take O(n^2) to build above array)
So now when you get coordinates as input, you can calc the sum by

Ans = sumArray[x2][y2] - sumArray[x1][y1] + ip[x1][y1]

For our case 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
srisam261...@gmail.comwrote:


 @ Vicky

 Can yo explain with an illustration ?


 On Sat, Aug 11, 2012 at 10:07 PM, ~*~VICKY~*~ venkat.jun...@gmail.comwrote:

 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 get the coordinate as input u
 can calc the needed sum by subtracting sum of big rect and small rect which
 is not included in the given coordinates. This can be called constant time
 if u don't include the preprocessing time.


 On Sat, Aug 11, 2012 at 9:57 PM, adarsh kumar algog...@gmail.com wrote:

 Sum of the integers meaning? Do you mind giving an example test case?

 regards.

 On Sat, Aug 11, 2012 at 7:10 PM, Srividhya srisam261...@gmail.comwrote:

 hi all:)

 The coordinates of a rectangle will be specified. there is a matrix of
 integers. yo should find the sum of the integers that fall in the region
 specified by the  coordinates .

 The solution to be in constant time .

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Re: [algogeeks] Constant time solution needed

2012-08-12 Thread Arun Kindra
You can traverse in spiral order and add each element with the specified
co-ordinate.

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Re: [algogeeks] Constant time solution needed

2012-08-12 Thread Arun Kindra
*within

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Re: [algogeeks] Constant time solution needed

2012-08-12 Thread ~*~VICKY~*~
@Arun: How do you claim your solution to be constant time? :P

On Sun, Aug 12, 2012 at 8:46 PM, Arun Kindra arunkin...@gmail.com wrote:

 *within

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Re: [algogeeks] Constant time solution needed

2012-08-12 Thread Srividhya Sampath
@vicky

can yo explain the logic behind the 'Sum Array' computation (if possible
elaborately )?

On Sun, Aug 12, 2012 at 8:19 PM, ~*~VICKY~*~ venkat.jun...@gmail.comwrote:

 Lets build the array for the example you gave.

 i/p:

 0 1 2 3
 4 5  6 7
 8 9 10 11

 (x1,y1) = (0,0)
 (x2,y2) = (1,2)
 sumArray
 0  1 23
 4  10  18   28
 12 27  45  66
 (will take O(n^2) to build above array)
 So now when you get coordinates as input, you can calc the sum by

 Ans = sumArray[x2][y2] - sumArray[x1][y1] + ip[x1][y1]

 For our case 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 srisam261...@gmail.com
  wrote:


 @ Vicky

 Can yo explain with an illustration ?


 On Sat, Aug 11, 2012 at 10:07 PM, ~*~VICKY~*~ venkat.jun...@gmail.comwrote:

 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 get the coordinate as input u
 can calc the needed sum by subtracting sum of big rect and small rect which
 is not included in the given coordinates. This can be called constant time
 if u don't include the preprocessing time.


 On Sat, Aug 11, 2012 at 9:57 PM, adarsh kumar algog...@gmail.comwrote:

 Sum of the integers meaning? Do you mind giving an example test case?

 regards.

 On Sat, Aug 11, 2012 at 7:10 PM, Srividhya srisam261...@gmail.comwrote:

 hi all:)

 The coordinates of a rectangle will be specified. there is a matrix of
 integers. yo should find the sum of the integers that fall in the region
 specified by the  coordinates .

 The solution to be in constant time .

 --
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Re: [algogeeks] Constant time solution needed

2012-08-12 Thread ~*~VICKY~*~
Fine, the basic idea of using dp here is sum of each rectangle is a
dependent sub problem. So when u find sum for smaller rectangle we can use
it to compute sum of bigger rectangle with new coordinates added to
previous small rectangle. So u can 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 col
sum value]



On Sun, Aug 12, 2012 at 8:54 PM, Srividhya Sampath
srisam261...@gmail.comwrote:

 @vicky

 can yo explain the logic behind the 'Sum Array' computation (if possible
 elaborately )?


 On Sun, Aug 12, 2012 at 8:19 PM, ~*~VICKY~*~ venkat.jun...@gmail.comwrote:

 Lets build the array for the example you gave.

 i/p:

 0 1 2 3
 4 5  6 7
 8 9 10 11

 (x1,y1) = (0,0)
 (x2,y2) = (1,2)
 sumArray
 0  1 23
 4  10  18   28
 12 27  45  66
 (will take O(n^2) to build above array)
 So now when you get coordinates as input, you can calc the sum by

 Ans = sumArray[x2][y2] - sumArray[x1][y1] + ip[x1][y1]

 For our case 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 
 srisam261...@gmail.com wrote:


 @ Vicky

 Can yo explain with an illustration ?


 On Sat, Aug 11, 2012 at 10:07 PM, ~*~VICKY~*~ 
 venkat.jun...@gmail.comwrote:

 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 get the coordinate as input
 u can calc the needed sum by subtracting sum of big rect and small rect
 which is not included in the given coordinates. This can be called constant
 time if u don't include the preprocessing time.


 On Sat, Aug 11, 2012 at 9:57 PM, adarsh kumar algog...@gmail.comwrote:

 Sum of the integers meaning? Do you mind giving an example test case?

 regards.

 On Sat, Aug 11, 2012 at 7:10 PM, Srividhya srisam261...@gmail.comwrote:

 hi all:)

 The coordinates of a rectangle will be specified. there is a matrix
 of integers. yo should find the sum of the integers that fall in the 
 region
 specified by the  coordinates .

 The solution to be in constant time .

 --
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Re: [algogeeks] Constant time solution needed

2012-08-12 Thread ~*~VICKY~*~
@Arun: This approach is constant time once the array is build for any
queries that follows. :) You know sum for all possible rectangles in the
given 2d array thats makes it better than computing sum for each input.
Hope it makes sense

On Sun, Aug 12, 2012 at 9:07 PM, ~*~VICKY~*~ venkat.jun...@gmail.comwrote:

 Fine, the basic idea of using dp here is sum of each rectangle is a
 dependent sub problem. So when u find sum for smaller rectangle we can use
 it to compute sum of bigger rectangle with new coordinates added to
 previous small rectangle. So u can 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 col
 sum value]



 On Sun, Aug 12, 2012 at 8:54 PM, Srividhya Sampath srisam261...@gmail.com
  wrote:

 @vicky

 can yo explain the logic behind the 'Sum Array' computation (if possible
 elaborately )?


 On Sun, Aug 12, 2012 at 8:19 PM, ~*~VICKY~*~ venkat.jun...@gmail.comwrote:

 Lets build the array for the example you gave.

 i/p:

 0 1 2 3
 4 5  6 7
 8 9 10 11

 (x1,y1) = (0,0)
 (x2,y2) = (1,2)
 sumArray
 0  1 23
 4  10  18   28
 12 27  45  66
 (will take O(n^2) to build above array)
 So now when you get coordinates as input, you can calc the sum by

 Ans = sumArray[x2][y2] - sumArray[x1][y1] + ip[x1][y1]

 For our case 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 
 srisam261...@gmail.com wrote:


 @ Vicky

 Can yo explain with an illustration ?


 On Sat, Aug 11, 2012 at 10:07 PM, ~*~VICKY~*~ 
 venkat.jun...@gmail.comwrote:

 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 get the coordinate as 
 input
 u can calc the needed sum by subtracting sum of big rect and small rect
 which is not included in the given coordinates. This can be called 
 constant
 time if u don't include the preprocessing time.


 On Sat, Aug 11, 2012 at 9:57 PM, adarsh kumar algog...@gmail.comwrote:

 Sum of the integers meaning? Do you mind giving an example test case?

 regards.

 On Sat, Aug 11, 2012 at 7:10 PM, Srividhya srisam261...@gmail.comwrote:

 hi all:)

 The coordinates of a rectangle will be specified. there is a matrix
 of integers. yo should find the sum of the integers that fall in the 
 region
 specified by the  coordinates .

 The solution to be in constant time .

 --
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Re: [algogeeks] Constant time solution needed

2012-08-12 Thread shady
a small question, if matrix has 'r' rows and 'c' columns, how many
different rectangles can be there for this problem ?
Space Complexity = O( (r*r)*(c*c) )

On Sat, Aug 11, 2012 at 7:10 PM, Srividhya srisam261...@gmail.com wrote:

 hi all:)

 The coordinates of a rectangle will be specified. there is a matrix of
 integers. yo should find the sum of the integers that fall in the region
 specified by the  coordinates .

 The solution to be in constant time .

 --
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Re: [algogeeks] Constant time solution needed

2012-08-12 Thread shady
@venkat +1

On Sun, Aug 12, 2012 at 9:09 PM, ~*~VICKY~*~ venkat.jun...@gmail.comwrote:

 @Arun: This approach is constant time once the array is build for any
 queries that follows. :) You know sum for all possible rectangles in the
 given 2d array thats makes it better than computing sum for each input.
 Hope it makes sense


 On Sun, Aug 12, 2012 at 9:07 PM, ~*~VICKY~*~ venkat.jun...@gmail.comwrote:

 Fine, the basic idea of using dp here is sum of each rectangle is a
 dependent sub problem. So when u find sum for smaller rectangle we can use
 it to compute sum of bigger rectangle with new coordinates added to
 previous small rectangle. So u can 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
 col sum value]



 On Sun, Aug 12, 2012 at 8:54 PM, Srividhya Sampath 
 srisam261...@gmail.com wrote:

 @vicky

 can yo explain the logic behind the 'Sum Array' computation (if possible
 elaborately )?


 On Sun, Aug 12, 2012 at 8:19 PM, ~*~VICKY~*~ venkat.jun...@gmail.comwrote:

 Lets build the array for the example you gave.

 i/p:

 0 1 2 3
 4 5  6 7
 8 9 10 11

 (x1,y1) = (0,0)
 (x2,y2) = (1,2)
 sumArray
 0  1 23
 4  10  18   28
 12 27  45  66
 (will take O(n^2) to build above array)
 So now when you get coordinates as input, you can calc the sum by

 Ans = sumArray[x2][y2] - sumArray[x1][y1] + ip[x1][y1]

 For our case 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 
 srisam261...@gmail.com wrote:


 @ Vicky

 Can yo explain with an illustration ?


 On Sat, Aug 11, 2012 at 10:07 PM, ~*~VICKY~*~ venkat.jun...@gmail.com
  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 get the coordinate as 
 input
 u can calc the needed sum by subtracting sum of big rect and small rect
 which is not included in the given coordinates. This can be called 
 constant
 time if u don't include the preprocessing time.


 On Sat, Aug 11, 2012 at 9:57 PM, adarsh kumar algog...@gmail.comwrote:

 Sum of the integers meaning? Do you mind giving an example test case?

 regards.

 On Sat, Aug 11, 2012 at 7:10 PM, Srividhya 
 srisam261...@gmail.comwrote:

 hi all:)

 The coordinates of a rectangle will be specified. there is a matrix
 of integers. yo should find the sum of the integers that fall in the 
 region
 specified by the  coordinates .

 The solution to be in constant time .

 --
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[algogeeks] Constant time solution needed

2012-08-11 Thread Srividhya
hi all:)

The coordinates of a rectangle will be specified. there is a matrix of 
integers. yo should find the sum of the integers that fall in the region 
specified by the  coordinates .

The solution to be in constant time .

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Re: [algogeeks] Constant time solution needed

2012-08-11 Thread adarsh kumar
Sum of the integers meaning? Do you mind giving an example test case?

regards.

On Sat, Aug 11, 2012 at 7:10 PM, Srividhya srisam261...@gmail.com wrote:

 hi all:)

 The coordinates of a rectangle will be specified. there is a matrix of
 integers. yo should find the sum of the integers that fall in the region
 specified by the  coordinates .

 The solution to be in constant time .

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Re: [algogeeks] Constant time solution needed

2012-08-11 Thread ~*~VICKY~*~
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 get the coordinate as input u
can calc the needed sum by subtracting sum of big rect and small rect which
is not included in the given coordinates. This can be called constant time
if u don't include the preprocessing time.

On Sat, Aug 11, 2012 at 9:57 PM, adarsh kumar algog...@gmail.com wrote:

 Sum of the integers meaning? Do you mind giving an example test case?

 regards.

 On Sat, Aug 11, 2012 at 7:10 PM, Srividhya srisam261...@gmail.com wrote:

 hi all:)

 The coordinates of a rectangle will be specified. there is a matrix of
 integers. yo should find the sum of the integers that fall in the region
 specified by the  coordinates .

 The solution to be in constant time .

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Cheers,

  Vicky

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