call it as : bipartition(a,-1,2) . coz at the very first time k is being
incremented so it needs intiiale value -1.
On Sat, Sep 3, 2011 at 8:46 PM, Siddhartha Banerjee thefourrup...@gmail.com
wrote:
please check out the code, doesnt give right solution on running...
or perhaps i missed
Here is my soluion .
void bipartition(int a[],int p,int max)
{
if(p==max){return ;}
p++;
for(int i=p;imax;i++)
{
if(result[i]==1) continue;
result[i]=1;
print_bipartition(result,a,max);
bipartition(a,i,max);
result[i]=0;
}
}
void print_bipartition(int result[],int
this line is unnecessary
if(result[i]==1) continue;
I think this algo is good enough. But any improvements?
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please check out the code, doesnt give right solution on running...
or perhaps i missed something... how should you call your function? if array
is a={1,2,3} you call from main function as
bipartition(a,0,2), right???
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Piyush Has Correct Idea, If You Have N elements in Set/Array You Will Have
Maximum 2^n Subsets (Power Set), Now Problem Reduced to generate the all
such subsets , it will take O(2^n*n ) time , Now number of Valid
Bipartitions are exactly n/2 .
Note: Power Set includes 0 as well
Correct me
@shasank:
how come the valid bipartition s are n/2..
those should be at least n right?
ex:
{1,2,3}
{1},{2,3}
--{1,2},{3}
--{},{1,2,3}
If this is correct , then for printing sake it takes O(n^2) .
correct me if I'm wrong.
On Fri, Sep 2, 2011 at 2:48 PM, WgpShashank
no of valid bipartitions are 2^n, thats what he meant I guess... ( if you do
not want null set as one of the partitions then subtract 1 from answer...)
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