its a simple hashing problem i suppose.Beware there are some traps
though.It took me 9 submissions to get it fixed.
Just remember there are 1000 numbers so the array would be a[1001]
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@Akshata
I managed to get AC with majority algorithm.
suggestions:
1) You can try to read input and find candidate in same loop
2) Use scanf and printf
Thanks,
Balaji.
On Tue, Feb 15, 2011 at 4:30 PM, jai gupta wrote:
> #include
> #include
> #include
> void work() {
> int n,max,maxpos,x,
#include
#include
#include
void work() {
int n,max,maxpos,x,i;
scanf("%d",&n);
int *arr=(int*) malloc(sizeof(int)*2005);
memset(arr,0,2005);
max=maxpos=0;
for(i=0;imax)
{
max=arr[x+1000];
maxpos=x;
}
}
if(max>n/2)
print
I simply used arrays to count the frequency of each numberMy
accepted solution is:
#include
int main()
{
int t,n,i,p,x,chk,k;
int pos[1000];
int neg[1000];
int nos[100];
scanf("%d",&t);
while(t--)
{
chk=0;
A few problems with your code :
1)Very Unclear (sorry !):- Either use a camelCase like java or use the
c++ underscores style .Paste ur code on pastebin etc.
2)Too many loops :- It is O(n) , but perhaps O(4000*n) , much worse
than O(n^2) in this case.
3)The problem is based on majority element con
Try scanf/printf instead of cin/cout.
For huge data set, the IO time matters.
On 2011-2-14 20:09, Akshata Sharma wrote:
link to problem: http://www.spoj.pl/problems/MAJOR/
On Feb 14, 5:03 pm, Akshata Sharma wrote:
I am trying to submit my solution but its giving TLE. My implemetation is
O(n).
try replacing cin, cout by printf,scanf
On Mon, Feb 14, 2011 at 5:39 PM, Akshata Sharma
wrote:
> link to problem: http://www.spoj.pl/problems/MAJOR/
>
> On Feb 14, 5:03 pm, Akshata Sharma wrote:
> > I am trying to submit my solution but its giving TLE. My implemetation is
> > O(n).. and i am not
link to problem: http://www.spoj.pl/problems/MAJOR/
On Feb 14, 5:03 pm, Akshata Sharma wrote:
> I am trying to submit my solution but its giving TLE. My implemetation is
> O(n).. and i am not able to think a faster algo than this for the problem.
> The problem is based on finding the majority ele