@Lucifier:
i guess you made some editing mistake :-
Initialize the array with the following values,
1) A[0, j] = 0 for 1 = j = Wmax
2) A[i, 0] = 1 for 0 = i = Wmax // it shoud be 1 for 0 = i =N
about this equation :-
A[i , j] = A[i-1, j] | A[ i-1 , j - W[i] ]
say if W[i]=10 and j=3 , i =2;
then it would be accessing A[1][3-10] i.e A[1][-7] . dat would be
wrong.
On Dec 7 2011, 6:40 pm, Lucifer sourabhd2...@gmail.com wrote:
I have modified some part of the above post below to avoid confusion
regarding the generation of all subsets:
Say, we need to find all the subsets which led to A[N, K] = 1, to do
this we will take the following steps :
a) If ( i == 0 ) print the subset and return.
b) if A[N -1 , K] = 1,
b1) then W[N] doesn't belong to the subset , continue
recursively ( goto step a).
b2) goto step c.
else goto step c.
c) if A[N -1 , K - W[N] ] = 1, then W[N] belongs to the subset ,
continue recursively ( goto step a) .
else return.
On Dec 7, 6:29 pm, Lucifer sourabhd2...@gmail.com wrote:
I have an idea, i think it can be done in O(N * Wmax).
Let the weight array be W[N].
Take an array A[N][Wmax] , where N is the no. of weights provided and
Wmax is the max weight.
Initialize the array with the following values,
1) A[0, j] = 0 for 1 = j = Wmax
2) A[i, 0] = 1 for 0 = i = Wmax
Now, if an A[i , j] = 1, that means using the first i weights it is
possible pick a subset whose sum is j,
else if A[i , j] = 0, then it not possible to have subset of first
i weights whose sum would sum up to j.
Now, to solve the above problem we can use the following equation,
A[i , j] = A[i-1, j] or A[ i-1 , j - W[i] ]
Using the above equation calculate all values A[i, j] where 1 = i =
N and 1 = j = Wmax.
Now,
Scan A[N, j] from right to left ( Wmax = j = 0 ) till you get a
value of 1, let the found column index be K.
A[N, K] = 1, basically signifies the maximum sum that you can make
which is K.
Now that you have the maximum sum = Wmax which can be made i.e K,
the next problem will be 2 figure all the subsets.
To find all the subsets backtrack based on the equation given above
and record the weights for which A[i, j] = 1,
i.e.
Say, we need to find all the subsets which led to A[N, K] = 1, to do
this we will check for,
a) if A[N -1 , K] = 1, then W[N] doesn't belong to the subset ,
continue recursively.
b) if A[N -1 , K - W[N] ] = 1, then W[N] belongs to the subset ,
continue recursively.
Hence,
To find the max value it will take O( N * Wmax) + O( Wmax)
To find all the subsets, it will take O( X * Y) where, x is the no. of
subsets and y is the average no. of elements in it.
On Dec 5, 5:09 pm, Shashank Narayan shashank7andr...@gmail.com
wrote:
@piyuesh 3-Sum is not exponential ? its quadratic , check wiki for
refrence
?
On Mon, Dec 5, 2011 at 5:36 PM, Piyush Grover
piyush4u.iit...@gmail.comwrote:
As I mentioned earlier solution with exponential time-complexity is the
obvious one. Is there any way to solve this problem by greedy/Dynamic
algo?
On Mon, Dec 5, 2011 at 5:24 PM, WgpShashank
shashank7andr...@gmail.comwrote:
@piyuesh , Saurabh isn't 3-Sum Suffics Here ?
Another thought problem can also be thought by generating power set of
given set e.g. if set has n elemnts its power set has 2^n elements ,
then
finds the set that has sum up yo given weight isn't it ? hope you
know how
to find power set efficiently ?
correct if is missed anything ?
Thanks
Shashank
Computer Science
BIT Mesra
http://www.facebook.com/wgpshashank
http://shashank7s.blogspot.com/
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