@juver++: Oops. Right you are! A corrected solution is just a little
more complicated. If the data is presented in an array as in the
previous posting, a[]={1,2,3,4,5,6,7,8,9,10} corresponding to the
triangle
1
2 3
4 5 6
7 8 9 10
then we let A(i,j) be the jth entry in the ith row, both 0-based, so
that in this example, A(2,1) = 5. Then the array can be treated as an
implicit DAG (directed acyclic graph) where the children of A(i,j) are
A(i+1,j) and A(i+1,j+1). Using the correspondence that A(i,j) = a[i*(i
+1)/2+j)], you can traverse the DAG the same way I described
traversing the binary tree.
Dave
On Nov 4, 3:19 pm, "juver++" <avpostni...@gmail.com> wrote:
> You are wrong. It's not a binary tree. Cause multiple elements have
> adjacent childs.
> Right solution is to apply DP:
> dp[i][j] - maximum sum that is ends into i-th row and j-th column.
> Then make only 2 transitions to the (i+1)-th row.
> Complexity is n^2.
>
> On Nov 4, 7:50 pm, Dave <dave_and_da...@juno.com> wrote:
>
>
>
> > @Piyush: Treat the data as an implicit binary tree, with the children
> > of a[i] being a[2*i] and a[2i+1] if a[] is a 1-based array, or a[2*i
> > +1] and a[2*i+2] if a[] is a 0-based array. Traverse the tree and
> > accumulate the sum as you descend. When you reach the leaves, keep
> > track of the maximum sum and its path. When the traversal is finished,
> > you will have your answer.
>
> > Dave
>
> > On Nov 4, 4:47 am, Piyush <piyushra...@gmail.com> wrote:
>
> > > Given an array of integers. Imagine it as a pyramid as below:
> > > a[]={1,2,3,4,5,6,7,8,9,10}
> > > 1
> > > 2 3
> > > 4 5 6
> > > 7 8 9 10
> > > find a path from level 1 to last level such that the sum is maximum.
> > > The path is such that its children should be reachable.
> > > Eg:
> > > 1,2,5,9 is a path
> > > 1,2,6,10 is not a path since 6 is not reachable from 2.
> > > The array is not sorted and numbers are random.- Hide quoted text -
>
> - Show quoted text -
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