Hi
will this work ?
since we need Set S with n pairs of largest values , any such pair within
the set would (always) contain A[0] or B[0] because they maximize the value
of the pair.
so
// code snippet
typedef std::vector<int,int> Pairs;
Pairs.push(A[0],B[0])
int i = 1; // index for ListA
int j = 1; // index for list B
count = 1;
while( count <= n)
{
if( A[0] + B[j] > B[0] + A[i] )
{
Pairs.push(A[0],B[j])
j++;
}
else
{
Pairs.Add(A[i],B[0])
i++;
}
count++;
}
since there are n items in both the lists, j and i wont go beyond n in the
while loop
On Sun, May 2, 2010 at 3:44 PM, Shishir Mittal <1987.shis...@gmail.com>wrote:
> This problem has been discussed before @
> http://groups.google.co.in/group/algogeeks/browse_thread/thread/9c1e1aa8cf1ed437/d451cd9468d985f7
>
> I believe, the problem can't be solved in O(n) but only in O(nlog n).
>
> @Divya: Are you sure the interviewer explicitly asked for an O(n) time
> algorithm?
>
>
> On Sun, May 2, 2010 at 1:48 PM, vignesh radhakrishnan <
> rvignesh1...@gmail.com> wrote:
>
>> @divya You're rite. Post a solution if you have one.
>>
>> --
>> Regards,
>> Vignesh
>>
>>
>> On 2 May 2010 13:14, divya jain <sweetdivya....@gmail.com> wrote:
>>
>>> @Mohit
>>>
>>> according to ur algo if a[1], b[0] has sum greater than a[0],b[1]
>>> then i is incremented i is now 2 so for next iteration u ll compare
>>> a[2] b[0] and a[1] b1]. but what about a[0] b[1] this pair s lost forever.
>>> think for ths case also.
>>>
>>>
>>> On 2 May 2010 11:22, mohit ranjan <shoonya.mo...@gmail.com> wrote:
>>>
>>>> @Algoose Chase
>>>>
>>>> point taken
>>>> Thanks
>>>>
>>>>
>>>> Mohit Ranjan
>>>> Samsung India Software Operations.
>>>>
>>>>
>>>> On Sat, May 1, 2010 at 8:43 PM, Algoose Chase <harishp...@gmail.com>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 <shoonya.mo...@gmail.com
>>>>> > 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 <
>>>>>> shoonya.mo...@gmail.com> 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 <sweetdivya....@gmail.com>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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