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August 28th, 2011, 01:28 PM
#1
Problem name
Hi, this is odd, i'm solving a problem but i don't know its name.
The problem is the following:
There are 2 (ordered) numbers array V and W.
We have to find a number k which is the result of V[i]+W[j].
There is a literature algorithm which solves it:
i=0;
j=size(W);
while(true)
If (V(i)+W(j)==k)
return i,j;
else if V(i)+W(j)>k
j--;
else if V(i)+W(j)<k
i++
Do you remember the name of the problem or the name of the algorithm?
Thank you.
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August 29th, 2011, 06:24 PM
#2
Re: Problem name
Seems kind of simple to be a named problem... Is there a reason you think it has a formal name?
The 'literature' algorithm you cited is very simple. What is the source? Do they quote a name?
Best Regards,
BioPhysEngr
http://blog.biophysengr.net
--
All advice is offered in good faith only. You are ultimately responsible for effects of your programs and the integrity of the machines they run on.
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August 30th, 2011, 03:44 PM
#3
Re: Problem name
This problem is called, "given two sorted arrays A and B and an integer k, find the element in A and the element in B that sum to k".
Only interesting problems get names. This one is not interesting.
Also, the solution you gave is not the most efficient. If len(A) >= len(B), It is O(len(A)). If you do a binary search across A and B, you can reach the solution in O(log(len(A))) time.
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August 30th, 2011, 11:43 PM
#4
Re: Problem name
Not sure I see the time complexity you state. Assume without loss of generality that len(A) >= len(B), then the algorithm is:
Code:
foreach b in B
if k-b in A
return a, b
endfor
otherwise return false
Where if k-b in A can be done in O(log(A)) time by binary search. This gives a runtime complexity of O(B log(A)).
By contrast, the algorithm given by AEmme has complexity O(A+B) ~ O(A), which is likely less.
Agree? Or did I miss something?
Best Regards,
BioPhysEngr
http://blog.biophysengr.net
--
All advice is offered in good faith only. You are ultimately responsible for effects of your programs and the integrity of the machines they run on.
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