Thank u frnds .
@Kamakshi- Tejas Network has nt yet visited bt it is going to visit our
campus soon
On Sun, Aug 28, 2011 at 7:44 PM, Kamakshii Aggarwal
wrote:
> @rohit:i got my mistake..insteat of arr[2]=x[2];
> it should be arr[2]=max{x[1],x[2]}
>
>
> so now my code goes lyk this
> let
@rohit:i got my mistake..insteat of arr[2]=x[2];
it should be arr[2]=max{x[1],x[2]}
so now my code goes lyk this
let arr[] stores the value of max amount upto each shop that they can rob.
arr[1]=x[1];
arr[2]=max{x[1],x[2]};
for(i=3;i<=n;i++)
{
arr[i]=max{(arr[i-2]+x[i]),arr[i-1]};
}
arr[n] gi
i tink logic which kamakshi gave is correct
code for same:::
#include
int max(int a,int b)
{
return (a>b?a:b);
}
int value(int a[],int l)
{
if(l<0)
return 0;
return max(a[l]+value(a,l-2),value(a,l-1));
}
int main()
{
int a[]={4,3,2,5,7,8};
printf("max::%d",value(
@kamakshi: your algo is wrong
Take the test case as {4,3,2,5,8,7}
op should be 16 (4,5,7) but ur op = 15
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This is a classic dynamic problem.
Search it on www.geeksforgeeks.org for "Finding Largest Non-Contiguous Sum"
in an array.
Sanju
:)
On Sun, Aug 28, 2011 at 6:21 AM, Kamakshii Aggarwal
wrote:
> let arr[] stores the value of max amount upto each shop that they can rob.
>
> arr[1]=x[1];
> arr[2]
let arr[] stores the value of max amount upto each shop that they can rob.
arr[1]=x[1];
arr[2]=x[2];
for(i=3;i<=n;i++)
{
arr[i]=max{(arr[i-2]+x[i]),arr[i-1]};
}
arr[n] gives the max amount that can be robbed...
correct me if i am wrong.
@prag:has tejas networks visited ur campus?
On Sun, Aug 28
