@amrit : inner most loop is running kadane algo :-
i am not writing whole algo , just giving necessary information ...
you can google kadane algo...
/**
for(k=0;kcol;k++)
{
sum_till_here+=mat[i][k] - *(mat[j][k])*; // here i am
excluding row = j and finding the
@amit : After understanding given example for 1D array ...you can consider
each column as 1D array ...then i guess you will get it.
On Fri, Jan 20, 2012 at 2:32 PM, atul anand atul.87fri...@gmail.com wrote:
@amrit : inner most loop is running kadane algo :-
i am not writing whole algo , just
since there are N leaves...
N/2 leaves(i will call nodes) will share only root. why? (they are on the
other side of the root)
N/4 leaves share 2 nodes
N/8 leaves share 3 nodes...
so on...
= there are N paths, as there are N leaves, (or N-1 to be precise..
excluding leaf v )==
so.. its N/2 * 1
what does it mean.. we cannot use an array? (a static array?)
a vector is an array..but a dynamic one... what other DS can be used? a
linked list allowed?
(each of the two algorithms can be mate to work with linked list too...
(except that it takes more time.. )
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@phoenix:
NULL character is \0 -- ascii value 0x00
Number 0 is char '0' -- ascii value is 0x30 or 48
so, 48 when encountered.. it prints.. 0, when \0 is found... it stops...
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void vsum(struct node *p ,int i)
{
if(p)
{
sum[i] = sum[i] + p-data;
vsum(p-left,i-1);
vsum(p-right,i+1);
}
}
construct an array of int sum[n] where n will be maximum no. of vertical
lines and call vsum with vsum(root,n/2)
On Fri, Jan 20, 2012 at 9:06 PM,
