Here is my soluion .
void bipartition(int a[],int p,int max)
{
if(p==max){return ;}
p++;
for(int i=p;i<max;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 a[],int max)
{
printf("{");
for(int k=0;k<max;k++)
if(result[k]==1)
printf("%d ",a[k]);
printf("} \t");
printf("{");
for(int k=0;k<max;k++)
if(result[k]==0)
printf("%d ",a[k]);
printf("}\n");
return ;
}
Can someone help me reduce the time complexity?
On Sat, Sep 3, 2011 at 10:07 AM, Siddhartha Banerjee <
thefourrup...@gmail.com> wrote:
> 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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