@Algoose Chase

point taken
Thanks


Mohit Ranjan
Samsung India Software Operations.


On Sat, May 1, 2010 at 8:43 PM, Algoose Chase <[email protected]> wrote:

> @mohit
>
> The idea of DP is fine.
> When you find the Max i dont think you need to include A[i+1]+B[j+1]
> because it can never be greater than both A[i+1]+B[j] and A[i]+B[j+1] since
> both the lists are sorted in decreasing order.
>
>
> On Fri, Apr 30, 2010 at 8:47 PM, mohit ranjan <[email protected]>wrote:
>
>> oops
>>
>> Sorry didn't read properly
>> last algo was for array sorted in ascending order
>>
>> for this case, just reverse the process
>>
>>
>> A[n] and B[n] are two array
>>
>> loop=n, i=0, j=0;
>>
>>
>> while(loop>0)  // for n largest pairs
>> {
>>   print A[i]+B[j];          // sum of first index from both array will be
>> max
>>
>>   foo = MAX ( A[i+1]+B[j], A[i+1]+B[j+1], A[i]+B[j+1] )     // using DP,
>> moving forward
>>
>>   if foo==A[i+1]+B[j]; i++   // only increment A
>>   if foo==A[i+1]+B[j+1]; i++; j++   // increment both A and B
>>   if foo==A[i]+B[j+1]; j++  // increment only B
>>
>> }
>>
>>
>>
>> Mohit Ranjan
>> Samsung India Software Operations.
>>
>>
>> On Fri, Apr 30, 2010 at 8:40 PM, mohit ranjan <[email protected]>wrote:
>>
>>> Hi Divya,
>>>
>>>
>>> A[n] and B[n] are two array
>>>
>>> loop=n, i=n-1, j=n-1;
>>>
>>> while(loop>0)  // for n largest pairs
>>> {
>>>   print A[i]+B[j];          // sum of last index from both array will be
>>> max
>>>
>>>   foo = MAX ( A[i-1]+B[j], A[i-1]+B[j-1], A[i]+B[j-1] )     // using DP
>>> moving backward
>>>
>>>   if foo=A[i-1]+B[j]; i--   // only reduce A
>>>   if foo=A[i-1]+B[j-1]; i--; j--   // reduce both A and B
>>>   if foo=A[i]+B[j-1]; j--  // reduce only B
>>> }
>>>
>>> Time: O(n)
>>>
>>>
>>> Mohit Ranjan
>>>
>>>
>>>
>>> On Fri, Apr 30, 2010 at 5:35 PM, divya <[email protected]> wrote:
>>>
>>>> Given two sorted postive integer arrays A(n) and B(n) (W.L.O.G, let's
>>>> say they are decreasingly sorted), we define a set S = {(a,b) | a \in
>>>> A
>>>> and b \in B}. Obviously there are n^2 elements in S. The value of such
>>>> a pair is defined as Val(a,b) = a + b. Now we want to get the n pairs
>>>> from S with largest values. The tricky part is that we need an O(n)
>>>> algorithm.
>>>>
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>>>>
>>>
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