dp[i][flag] - max amount up to the i-th house (i-th house is included into
final result if flag = true).
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static int MaxLoot(int[] A, int n)
{
int[] S = new int[n];
S[0] = A[0];
S[1] = A[1];
S[2] = S[0] + A[2];
int maxloot = Math.Max(S[1], S[2]);
for (int i = 3; i n; i++)
{
S[i] =
@^ Just check ur solution for boundary case ( n = 2) .. M$ is *generally*
very strict about such mistakes :)
Programmers should realize their critical importance and responsibility in a
world gone digital. They are in many ways similar to the priests and monks
of Europe's Dark Ages; they are the
I dnt get the iterative version. Can u explain it. I can do it top down in
O(n) with state index
I am at.
On Thu, Jan 20, 2011 at 11:30 PM, Avi Dullu avi.du...@gmail.com wrote:
@^ Just check ur solution for boundary case ( n = 2) .. M$ is *generally*
very strict about such mistakes :)
